D.1 at SL covers Newton's universal law of gravitation, F = Gm₁m₂/r², for bodies treated as point masses, and the conditions under which an extended body can be treated as one. Gravitational field strength is the force per unit mass, g = GM/r², and fields are drawn with field lines.
Combining gravitation with circular motion lets you analyse orbits, and Kepler's three laws describe planetary motion. Gravitational potential, potential energy −GMm/r and escape speed are Higher Level.
32 questions
197 marks
Paper 1A: 16
Paper 1B: 5
Paper 2: 11
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32 practice questions on D.1 Gravitational fields
1D-1A-01
Newton's law of gravitation·D.1 Gravitational fields
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksCalculate
Two small spheres, each of mass 5.0 kg, have their centres 0.20 m apart. G = 6.67 × 10−11 N m² kg−2.
AThe separation has not been squared. The law is an inverse-square law.
BThis is the force at 0.40 m, twice the stated separation.
CThis adds the two masses instead of multiplying them, using 10 kg in place of 25 kg².
DCorrect: F = 6.67 × 10⁻¹¹ × 25/0.040 = 4.2 × 10⁻⁸ N.
Syllabus understandingD.1 — Newton's universal law of gravitation as given by F = Gm1m2/r² for bodies treated as point masses Command term: Calculate
2D-1A-02
Gravitational field strength·D.1 Gravitational fields
Paper 1AMedium1 mark
Multiple choice · 1 mark1 step to full marksIdentify
Which statement about the gravitational field strength g at a point is correct?
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Notes
Step 1g = F/m is defined as force per unit mass on a small point mass. Doubling the test mass doubles F, so F/m is unchanged: the field belongs to the point, not to the test mass.
✓ 1
Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThis confuses the field strength with the force. The force doubles with the mass, but the force per unit mass does not.
BThe acceleration equals g only in free fall. If other forces act, for example a normal force, the acceleration is different.
CCorrect: g = F/m is the force per unit mass on a small test mass, so the test mass cancels.
DThe unit is right, but g is a vector. It has the direction of the force on a test mass, towards the attracting body.
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r² Command term: Identify
3D-1A-03
Variation of g with distance·D.1 Gravitational fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
The gravitational field strength at the surface of a planet of radius R is g.
Point X is a distance 3R from the centre of the planet. Point Y is a height 3R above the surface of the planet.
Which row gives the gravitational field strength at X and at Y?
Field strength at XField strength at Y
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Notes
Step 1g ∝ 1/r², with r measured from the centre: X is at r = 3R, and Y is at r = R + 3R = 4R.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Field at X = g/3² = g/9; field at Y = g/4² = g/16.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis uses 1/r instead of 1/r² for both points (3R and 4R).
BThis measures both distances from the surface (2R for X and 3R for Y) instead of from the centre.
CThis treats the height 3R as a distance 3R from the centre; Y is 4R from the centre.
DCorrect: X is 3R and Y is 4R from the centre, so the fields are g/9 and g/16.
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r² Command term: Determine
4D-1A-04
Kepler's laws·D.1 Gravitational fields
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksIdentify
Which statement about the planets of the Solar System agrees with Kepler's laws of orbital motion?
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Notes
Step 1Kepler's first law: each orbit is an ellipse with the Sun at one focus. (The second law implies the speed varies; the third law, T² ∝ r³, gives longer periods farther out.)
✓ 1
Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: this is Kepler's first law.
BThe Sun is at a focus, not at the centre. Only for a circle do the focus and the centre coincide.
CEqual areas in equal times means the planet moves faster when it is nearer the Sun, so its speed is not constant on an ellipse.
DThis inverts Kepler's third law: T² ∝ r³, so the period increases with distance from the Sun.
Syllabus understandingD.1 — Kepler's three laws of orbital motion Command term: Identify
5D-1A-05
Gravitational field lines·D.1 Gravitational fields
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksIdentify
Which diagram best represents the gravitational field lines around an isolated, uniform spherical planet?
Diagrams NOT accurately drawnShow mark scheme
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Step 1The force on a small test mass points straight towards the centre of a uniform sphere, so the lines are radial and point inwards; being radial, they crowd together near the surface where the field is strongest.
✓ 1
Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThe arrows point outwards. Gravity is always attractive, so the field points towards the planet.
BClosed loops are a feature of magnetic fields. Gravitational field lines end on the mass.
CCorrect: radial lines pointing inwards, closer together near the surface where the field is strongest.
DThe lines curve round the planet. For a uniform sphere the force on a test mass points straight at the centre, so the lines must be radial.
Syllabus understandingD.1 — gravitational field lines Command term: Identify
6D-1A-06
Kepler's third law·D.1 Gravitational fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
Two moons, X and Y, move in circular orbits around the same planet. The orbital radius of Y is four times the orbital radius of X.
What is (orbital period of Y)/(orbital period of X)?
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Notes
Step 1Kepler's third law: T² ∝ r³, so T ∝ r3/2.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Period ratio = 43/2 = 8.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis is 41/2: it takes T ∝ r1/2.
BThis takes the period to be proportional to the radius.
CCorrect: T ∝ r3/2 and 43/2 = 8.
DThis is 4²: it takes T ∝ r², squaring instead of raising to the power 3/2.
Syllabus understandingD.1 — Kepler's three laws of orbital motion Command term: Determine
7D-1A-07
Variation of g with distance·D.1 Gravitational fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksIdentify
The gravitational field strength g is measured at several distances r from the centre of a planet. All the points are outside the planet.
Which graph shows the variation of g with 1/r²?
Sketch graphs, not to scaleShow mark scheme
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Notes
Step 1Outside the planet g = GM/r², so g is directly proportional to 1/r².
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2A graph of g against 1/r² is therefore a straight line of gradient GM through the origin (g → 0 as r → ∞).
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AA positive intercept would mean a non-zero field infinitely far away (1/r² = 0).
BThis curve (g ∝ √(1/r²)) is what g ∝ 1/r would give. The field follows an inverse-square law.
CA negative gradient treats "inverse" as "decreasing". Plotted against 1/r², the field increases.
DCorrect: g = GM × (1/r²), a straight line through the origin with gradient GM.
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r² Command term: Identify
8D-1A-08
Combining two gravitational fields·D.1 Gravitational fields
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Two spheres of mass M and 9M have their centres a distance d apart. At point P on the line joining their centres, between the spheres, the resultant gravitational field strength is zero.
What is the distance of P from the centre of the sphere of mass M?
Diagram NOT accurately drawnShow mark scheme
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Notes
Step 1At P the two fields are equal and opposite: GM/x² = G(9M)/(d − x)².
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Taking square roots: d − x = 3x.
—
Step 3x = d/4 (closer to the smaller mass).
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis sets d − x = 9x, using the mass ratio instead of its square root.
BCorrect: (d − x)/x = √9 = 3, so x = d/4.
CThe midpoint is the null point only for equal masses.
DThis is the distance of P from the larger sphere, not from the sphere of mass M.
Syllabus understandingD.1 — that gravitational field strength g at a point is given by g = F/m = GM/r²; the resultant field is restricted to points along the line joining two bodies (guidance) Command term: Determine
9D-1A-09
Extended bodies as point masses·D.1 Gravitational fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksExplain
Newton's law of gravitation is stated for point masses, yet it is used to calculate the force between the Earth and the Moon.
Which statement justifies this?
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Notes
Step 1A spherically symmetric body produces the same external field as a point mass of the same total mass at its centre.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The Earth and the Moon are very nearly spherically symmetric and far apart compared with their sizes, so the point-mass law applies.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
ALarge masses give large forces, but that says nothing about whether the point-mass model is valid.
BWeakness is not the issue — the model is used for very precise orbit calculations.
CThe shape of the orbit has nothing to do with whether a body can be treated as a point mass.
DCorrect: a spherically symmetric body attracts external bodies as though all its mass were at its centre.
Syllabus understandingD.1 — conditions under which extended bodies can be treated as point masses Command term: Explain
10D-1B-01
Gravitational field strength·D.1 Gravitational fields
Paper 1BMedium10 marks
Data-based question10 steps to full marksDetermine
A student uses a published satellite catalogue to test how the gravitational field strength g of the Earth varies with the distance r from the Earth's centre. For five satellites in near-circular orbits, the catalogue lists the mean altitude h above the Earth's surface and the orbital period T.
For a satellite in a circular orbit, g at the orbit is equal to the centripetal acceleration a = 4π²r/T². The radius of the Earth is 6.37 × 106 m.
The graph shows a against 1/r² for four of the satellites.
Satellite
h / km
T / min
r / 106 m
a / m s−2
1/r² / 10−15 m−2
A
410
92.6
6.78
8.67
21.8
B
1336
112.2
7.71
6.71
16.8
C
8062
287.6
D
20180
717.6
26.6
0.565
1.42
E
35786
1435.7
42.2
0.224
0.563
Graph drawn to scale
(a)
Show that a for satellite B is about 6.7 m s−2.
(1)
(b)
Complete the table for satellite C.
(2)
(c)
Plot the data point for satellite C on the graph and draw the line of best fit for the data.
(2)
(d)
Determine the gradient of the line. State its unit.
(2)
(e)
Determine the mass of the Earth. Compare your answer with the accepted value of 5.97 × 1024 kg.
(2)
(f)
Suggest one advantage of plotting lg a against lg r for these data.
(1)
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Notes
Part (a)
r = 6.37 × 106 + 1.336 × 106 = 7.706 × 106 m AND a = 4π² × 7.706 × 106/(112.2 × 60)² = 6.71 m s−2
✓ 1
Must see full substitution with T in seconds OR answer to at least 3 s.f.
Part (b)
r = 14.4 × 106 m AND 1/r² = 4.80 × 10−15 m−2
✓ 1
Accept 1/r² in the range 4.79–4.81.
a = 4π² × 14.43 × 106/(287.6 × 60)² = 1.91 m s−2
✓ 1
Must be given to 3 s.f., consistent with the rest of the table.
Part (c)
Point plotted at (4.80, 1.91) to within half a small square
✓ 1
Single straight line passing through (or very close to) all five points and the origin
✓ 1
Do not accept a line joining the dots or a curve.
Part (d)
Gradient calculated from a triangle using at least half the length of the line
✓ 1
gradient = 3.99 × 1014 m³ s−2
✓ 1
Accept 3.90 × 1014 to 4.10 × 1014. Accept N m² kg−1. Unit required for this mark.
Part (e)
M = gradient/G = 3.99 × 1014/6.67 × 10−11 = 5.98 × 1024 kg
✓ 1
Allow ECF from (d). Accept 5.85 × 1024 to 6.15 × 1024 kg.
Differs from the accepted value by about 0.1 %, so the two agree / the data support g = GM/r²
✓ 1
The comparison must be consistent with the candidate's value; a percentage difference or a clear statement of closeness is required. OWTTE
Part (f)
The points would be spread more evenly across the graph (here four of the five points lie near the ends of the range) OR the gradient gives the power of r directly, so the inverse-square relationship (gradient −2) is tested without being assumed
✓ 1
OWTTE
Answers: (a) a = 6.71 m s−2 · (b) r = 14.4 × 106 m; a = 1.91 m s−2; 1/r² = 4.80 × 10−15 m−2 · (d) 3.99 × 1014 m³ s−2 · (e) M = 5.98 × 1024 kg (the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r² Command term: Determine
11D-1B-02
Variation of g with height·D.1 Gravitational fields
Paper 1BHard11 marks
Data-based question11 steps to full marksDetermine
A student investigates how the gravitational field strength g varies with height h above the ground. A portable gravimeter is carried by lift to different floors of a tall office building. On each floor it measures the change Δg in g relative to a reference point at street level. Each value of Δg is the mean of several readings and has an uncertainty of ±10 μm s−2; the uncertainty in h is negligible.
For h much smaller than the radius R of the Earth, g = GM/(R + h)² leads to Δg = −(2g0/R)h, where g0 = 9.81 m s−2 is the field strength at street level.
The graph shows the data with error bars.
h / m
Δg / μm s−2
40
−111
80
−236
120
−345
160
−467
200
−576
240
−701
Graph drawn to scale
(a)
The gravimeter displays readings to the nearest 0.1 μm s−2. Suggest why the uncertainty in Δg is much larger than this.
(1)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine the gradient of the line and its absolute uncertainty. Draw lines of maximum and minimum gradient on the graph.
(3)
(d)
Determine, from your answer to (c), the radius R of the Earth and its absolute uncertainty.
(2)
(e)
The accepted value of R is 6.37 × 106 m. Comment on the student's result.
(1)
(f)
The building has a very large mass. Explain how the gravitational attraction of the building affects the measured values of Δg and hence the value of R obtained.
(2)
(g)
Suggest one improvement to the investigation.
(1)
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Notes
Part (a)
Repeated readings vary/fluctuate (e.g. because of vibration or sway of the building, lifts, traffic, wind), so the spread of the readings, not the resolution, sets the uncertainty
✓ 1
OWTTE
Part (b)
Single straight line passing through all the error bars with the points scattered on both sides
✓ 1
Line does not need to pass through the origin.
Part (c)
Best gradient = −2.92 × 10−6 s−2
✓ 1
Accept −2.87 to −2.97 × 10−6 s−2 (μm s−2 per m). Sign not required.
Max and min gradient lines drawn through all the error bars, with gradients of about −3.00 × 10−6 and −2.85 × 10−6 s−2
✓ 1
Uncertainty = ½(max − min) ≈ ±0.07 × 10−6 s−2
✓ 1
Accept ±0.05 to ±0.12 × 10−6 s−2. Allow the larger of (max − best) and (best − min).
Part (d)
R = 2g0/|gradient| = 2 × 9.81/2.92 × 10−6 = 6.7 × 106 m
✓ 1
Allow ECF from (c).
ΔR = R × (Δgradient/gradient) ≈ 0.2 × 106 m, so R = (6.7 ± 0.2) × 106 m
✓ 1
MP2 is for the uncertainty with matching precision of value and uncertainty. Accept an uncertainty range 6.54–6.88 × 106 m from the max/min gradients.
Part (e)
The accepted value lies (just) outside the student's range, so the result is not consistent with it: there is a systematic error / the result is too large
✓ 1
Allow ECF from (d).
Part (f)
On the upper floors most of the building's mass is below the gravimeter and pulls downwards (at street level it is above and pulls upwards), so Δg is less negative than for the Earth alone
✓ 1
OWTTE
The magnitude of the gradient is too small, so R = 2g0/|gradient| is too large — consistent with (e)
✓ 1
Award MP2 only if the direction of the effect on the gradient is correct.
Part (g)
Repeat the measurements on a structure of much smaller mass (e.g. an open lattice mast) OR correct the readings for the building's own attraction OR extend the range of heights to reduce the percentage uncertainty in the gradient
✓ 1
Accept any specific, relevant improvement. Do not accept “repeat the readings” without further detail.
Answers: (c) −2.92 × 10−6 s−2 ± 0.07 × 10−6 s−2 · (d) R = (6.7 ± 0.2) × 106 m (the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r² Command term: Determine
12D-2-01
Gravitational field strength·D.1 Gravitational fields
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDetermine
A rover on the surface of Mars carries a simple pendulum of length 0.800 m. A camera on the rover records that the pendulum completes 20 small oscillations in 58.4 s.
The radius of Mars is 3.39 × 106 m.
Diagram NOT to scale
(a)
(i)
State what is meant by the gravitational field strength at a point.
(1)
(ii)
Show that the gravitational field strength at the surface of Mars is about 3.7 N kg−1.
(2)
(b)
Calculate the mass of Mars.
(2)
(c)
Diagram 1 shows Mars. Diagram 2 shows a small region of space just above the surface near the rover, greatly magnified.
(i)
Draw, on Diagram 1, the gravitational field lines of Mars.
(2)
(ii)
Draw, on Diagram 2, the gravitational field lines in this region. Outline why they differ from the lines you drew on Diagram 1.
(2)
(d)
A pendulum clock keeps correct time on the Earth, where the gravitational field strength is 9.81 N kg−1. The clock is taken to Mars. Determine whether the clock runs fast or slow on Mars, and by how many minutes it is wrong after one hour of real time.
(3)
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Notes
Part (a)(i)
«gravitational» force per unit mass acting on a small «point» mass placed at that point
✓ 1
Accept g = F/m with the symbols defined. Do not accept «acceleration due to gravity» alone.
Part (a)(ii)
T = 58.4/20 = 2.92 s
✓ 1
g = 4π²l/T² = 4π² × 0.800/2.92² = 3.70 N kg−1
✓ 1
Must see full substitution OR answer to at least 3 s.f. Unrounded answer 3.704 N kg−1.
Part (b)
M = gR²/G = 3.70 × (3.39 × 106)²/6.67 × 10−11
✓ 1
Allow ECF from (a)(ii).
M = 6.38 × 1023 kg
✓ 1
Award [2] for CNA. Accept 6.3–6.4 × 1023 kg.
Part (c)(i)
straight radial lines, evenly spaced around the planet, meeting the surface at right angles
✓ 1
At least four lines. Lines end at the surface; they must not cross.
arrows on the lines pointing towards Mars «towards its centre»
✓ 1
Do not award if any arrow points away from the planet.
Part (c)(ii)
parallel, equally spaced, vertical lines with arrows pointing down towards the surface
✓ 1
the region is very small compared with the radius of Mars, so the radial lines are almost parallel there and the field strength is almost the same everywhere in it «uniform field»
✓ 1
OWTTE
Part (d)
T ∝ 1/√g, so TMars/TEarth = √(9.81/3.70) = 1.63
✓ 1
Allow ECF from (a)(ii).
the period is longer on Mars, so the clock ticks less often and runs slow «loses time»
✓ 1
MP2 only scores if the period on Mars is stated or shown to be longer.
in one hour the clock records 3600/1.63 = 2210 s, so it is about 23 minutes slow
✓ 1
Accept 22–24 minutes.
Answers: (a)(ii) g = 3.70 N kg−1 · (b) M = 6.38 × 1023 kg · (d) runs slow by about 23 min (the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r²; gravitational field lines (linked to C.1 — T = 2π√(l/g)) Command term: Determine
13D-2-02
Variation of g with distance·D.1 Gravitational fields
Paper 2Medium11 marks
Short answer & extended response11 steps to full marksSketch
Mercury may be treated as a uniform sphere of radius R = 2.44 × 106 m. The graph shows how the gravitational field strength g due to Mercury varies with the distance r from the centre of Mercury, for r ≥ R.
G = 6.67 × 10−11 N m² kg−2.
Graph drawn to scale
(a)
Determine, using the graph, the mass of Mercury.
(2)
(b)
Planet X, a uniform sphere, has the same mass as Mercury but twice its radius.
(i)
Sketch, on the graph, the variation of the gravitational field strength due to planet X with distance from its centre, from the surface of X outwards.
(2)
(ii)
Explain why the two graphs are the same at distances greater than 4.88 × 106 m.
(2)
(iii)
Deduce (mean density of planet X)/(mean density of Mercury).
(1)
(c)
A lander of mass 620 kg approaches Mercury.
(i)
Determine the magnitude of the gravitational force on the lander when it is at a height of 2R above the surface of Mercury.
(3)
(ii)
State the magnitude and direction of the gravitational force that the lander exerts on Mercury at this instant.
(1)
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Notes
Part (a)
a point read correctly from the curve, e.g. g = 1.38 N kg−1 at r = 4.0 × 106 m OR g = 3.70 N kg−1 at r = 2.44 × 106 m
✓ 1
Any correct reading; r must be in metres.
M = gr²/G = 1.38 × (4.0 × 106)²/6.67 × 10−11 = 3.3 × 1023 kg
✓ 1
Accept 3.2–3.4 × 1023 kg. Award [2] for CNA.
Part (b)(i)
line starts at r = 4.88 × 106 m «2R» with g ≈ 0.92 N kg−1 «one quarter of Mercury's surface value»
✓ 1
Accept a start between 0.85 and 1.0 N kg−1.
for r > 4.88 × 106 m the line lies on Mercury's curve; no line drawn for r < 4.88 × 106 m
✓ 1
Do not award MP2 for a separate curve that lies above or below Mercury's.
Part (b)(ii)
outside a uniform «spherically symmetric» body, the body acts as a point mass at its centre
✓ 1
OWTTE
g = GM/r² depends only on M and r, and the two masses are equal «so the radius of the body does not matter outside it»
✓ 1
Part (b)(iii)
1/8 «the volume is 2³ = 8 times larger for the same mass»
✓ 1
Accept 0.125.
Part (c)(i)
distance from the centre r = 3R = 7.32 × 106 m
✓ 1
Do not accept 2R as the distance.
g = 3.70/3² = 0.411 N kg−1
✓ 1
Accept a value read from the graph, 0.40–0.42 N kg−1, or GM/r² with ECF from (a).
F = 620 × 0.411 = 255 N
✓ 1
Accept 248–260 N. Award [1 max] for 574 N «r = 2R used».
Part (c)(ii)
255 N directed towards the lander «away from the centre of Mercury»
✓ 1
Allow ECF from (c)(i). Both magnitude and direction needed. Newton's third law.
Answers: (a) 3.3 × 1023 kg · (b)(iii) 1/8 · (c)(i) 255 N · (c)(ii) 255 N (the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point, as given by g = F/m = GM/r²; conditions under which extended bodies can be treated as point masses; Newton's universal law of gravitation (linked to A.2 — Newton's third law) Command term: Sketch
14D-2-04
Combining two gravitational fields·D.1 Gravitational fields
Paper 2Hard12 marks
Short answer & extended response12 steps to full marksSketch
Pluto and its largest moon, Charon, may be treated as uniform spheres. In this question treat both bodies as stationary and ignore the gravitational fields of all other bodies.
The distance between the centres of Pluto and Charon is 1.96 × 107 m.
mass / kg
radius / m
Pluto
1.30 × 1022
1.19 × 106
Charon
1.59 × 1021
6.06 × 105
Axes drawn to scale
(a)
Outline why there is a point on the line between the centres of Pluto and Charon at which the resultant gravitational field strength is zero.
(2)
(b)
(i)
Show that this point is about 1.45 × 107 m from the centre of Pluto.
(2)
(ii)
A probe rests on the surface of Charon at the point nearest to Pluto. Determine the magnitude and direction of the resultant gravitational field strength at the probe.
(3)
(c)
On the axes, sketch a graph to show how the resultant gravitational field strength varies along the line joining the centres, from the surface of Pluto to the surface of Charon. Take field strength directed towards Charon as positive.
(3)
(d)
A small probe is placed at rest at the point in (b)(i) and is then displaced slightly towards Charon. Explain why the probe does not return to that point.
(2)
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Part (a)
between the bodies the two fields act in opposite directions «each towards its own body»
✓ 1
each field decreases with distance from its own body «inverse square», so at some point the two magnitudes are equal
Must see full substitution OR answer to at least 3 s.f. Unrounded answer 1.452 × 107 m.
Part (b)(ii)
field of Charon at its surface = 6.67 × 10−11 × 1.59 × 1021/(6.06 × 105)² = 0.289 N kg−1
✓ 1
field of Pluto at the probe = 6.67 × 10−11 × 1.30 × 1022/(1.90 × 107)² = 2.40 × 10−3 N kg−1
✓ 1
The distance from the centre of Pluto must be 1.96 × 107 − 6.06 × 105 m.
resultant = 0.286 N kg−1 towards the centre of Charon
✓ 1
Direction required. Award [2 max] if the fields are added (0.291 N kg−1).
Part (c)
negative near the surface of Pluto «about −0.6 N kg−1» and positive near the surface of Charon «about +0.3 N kg−1», with the magnitude at Pluto clearly the larger
✓ 1
Values need not be exact.
curve crosses zero once, at about 1.45 × 107 m «closer to Charon»
✓ 1
Allow ECF from (b)(i).
curve steepest close to each surface and flatter in between «1/r² shape»
✓ 1
Do not award for straight-line segments.
Part (d)
closer to Charon, Charon's field is larger and Pluto's is smaller, so the resultant field points towards Charon
✓ 1
the resultant force accelerates the probe further away from the point «towards Charon»; the equilibrium is unstable
✓ 1
OWTTE
Answers: (b)(i) 1.45 × 107 m · (b)(ii) 0.286 N kg−1 towards the centre of Charon (the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r²; determination of the resultant gravitational field strength restricted to points along a line joining two bodies (guidance) Command term: Sketch
15D-2-15
Orbits and satellites·D.1 Gravitational fields
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksExplain
A lunar orbiter moves in a circular orbit at a height of 100 km above the surface of the Moon.
Mass of the Moon = 7.35 × 1022 kg. Radius of the Moon = 1.74 × 106 m.
(a)
(i)
Explain why the speed of the orbiter is constant even though a resultant force acts on it.
(2)
(ii)
Show that the speed of the orbiter is about 1.6 km s−1.
(2)
(iii)
Calculate the time for one orbit, in minutes.
(1)
(b)
Tracking shows that orbits this low are disturbed above some regions of the Moon, where there are large masses of unusually dense rock below the surface.
(i)
State the condition under which the Moon can be treated as a point mass at its centre.
(1)
(ii)
Suggest why these dense regions disturb a low orbit much more than an orbit far from the Moon.
(3)
(c)
The orbiter must get rid of 350 W of waste thermal energy. It uses a flat radiator panel of area 0.80 m² and emissivity 0.90, which emits from one face only.
(i)
Determine the temperature of the panel when it emits 350 W. Ignore any radiation absorbed by the panel.
(2)
(ii)
Suggest why the panel must operate at a higher temperature when its emitting face points towards the sunlit surface of the Moon.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
the gravitational force is always perpendicular to the velocity «directed towards the centre of the Moon»
✓ 1
so no work is done on the orbiter; its kinetic energy «speed» is unchanged and only the direction of its velocity changes
✓ 1
OWTTE
Part (a)(ii)
GMm/r² = mv²/r with r = 1.74 × 106 + 1.00 × 105 = 1.84 × 106 m
✓ 1
MP1 requires the orbit radius, not the height.
v = √(6.67 × 10−11 × 7.35 × 1022/1.84 × 106) = 1.63 × 103 m s−1
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (a)(iii)
T = 2π × 1.84 × 106/1630 = 7.1 × 103 s = 118 min
✓ 1
Allow ECF from (a)(ii). Accept 117–119 min.
Part (b)(i)
its mass must be distributed with spherical symmetry «density depends only on the distance from the centre»
✓ 1
Accept «uniform sphere». Accept «the distance from the Moon is very large compared with its radius». OWTTE
Part (b)(ii)
a low orbit passes close to the dense regions
✓ 1
the gravitational force decreases with the square of the distance, so at short range the extra pull of a dense region is relatively large «and is not directed towards the centre of the Moon»
✓ 1
far from the Moon the distance is large compared with the size of the Moon, so it behaves as a point mass and local differences in density have little effect
✓ 1
OWTTE
Part (c)(i)
350 = 0.90 × 5.67 × 10−8 × 0.80 × T⁴
✓ 1
T = 304 K
✓ 1
Award [2] for CNA. Accept 300–305 K.
Part (c)(ii)
the panel then absorbs infrared radiation emitted by the hot lunar surface «and reflected sunlight», so it must emit more power to remove the same 350 W
✓ 1
OWTTE
Answers: (a)(ii) v = 1.63 × 103 m s−1 · (a)(iii) 118 min · (c)(i) 304 K (the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — Newton's universal law of gravitation as given by F = Gm1m2/r² for bodies treated as point masses; conditions under which extended bodies can be treated as point masses (linked to A.2 — centripetal force, and B.2 — emissivity) Command term: Explain
16D-1A-26
Density, mass and g·D.1 Gravitational fields
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
Planet X has twice the radius of the Earth and the same mean density as the Earth.
What is the gravitational field strength at the surface of planet X, in terms of the Earth's surface value g?
Show mark scheme
Marking point
Mark
Notes
Step 1At the same density M = ρ × (4/3)πR³, so doubling R multiplies the mass by 2³ = 8.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2g = GM/R² → factor 8/2² = 2, so the field is 2g.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis keeps the mass unchanged and allows only for the larger radius. A bigger planet of the same density is more massive.
BCorrect: M ∝ R³ at fixed density, so g ∝ R³/R² = R and the field doubles.
CThis divides the mass factor 8 by R (2) instead of by R² (4).
DThis is the mass factor alone, with no allowance for the greater distance from the centre.
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r² Command term: Determine
17D-1A-27
Kepler's second law·D.1 Gravitational fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
The diagram shows the elliptical orbit of a comet around the Sun S. P and Q are the points of the orbit closest to and farthest from the Sun.
Which row identifies where the speed of the comet and the magnitude of its acceleration are greatest?
Diagram NOT accurately drawn
Greatest speedGreatest acceleration
Show mark scheme
Marking point
Mark
Notes
Step 1Kepler's second law: the line from the Sun sweeps out equal areas in equal times, so the comet moves fastest where that line is shortest, at P.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The acceleration is the field strength GM/r², which is also greatest where r is smallest, at P.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: both the speed (equal areas in equal times) and the acceleration (GM/r²) are greatest at the closest point.
BThe speed is right, but the acceleration is GM/r², largest where r is smallest, not where the comet is slowest.
CThis reverses the second law. To sweep equal areas the comet must move fastest where it is nearest the Sun.
DEqual areas in equal times does not mean equal speeds. On an ellipse the speed changes round the orbit.
Syllabus understandingD.1 — Kepler's three laws of orbital motion; Newton's universal law of gravitation as given by F = Gm1m2/r² Command term: Deduce
18D-1A-28
Combining two gravitational fields·D.1 Gravitational fields
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
Two spheres of mass M and 4M have their centres a distance d apart. Point P lies on the line through both centres, a distance d from the centre of the sphere of mass M, on the side away from the sphere of mass 4M.
What is the magnitude of the resultant gravitational field strength at P?
Diagram NOT accurately drawnShow mark scheme
Marking point
Mark
Notes
Step 1Field of M at P: GM/d². Field of 4M at P (distance 2d): 4GM/(2d)² = GM/d².
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2P is outside the pair, so both fields point from P towards the spheres and add: 2GM/d².
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe two contributions are equal, but at P they point the same way; subtracting them to get zero treats them as opposite.
BThis subtracts the fields and uses GM/(2d)² for the larger sphere, leaving out its factor 4.
CCorrect: GM/d² + 4GM/(2d)² = 2GM/d², both directed from P towards the spheres.
DThis places the sphere of mass 4M a distance d from P instead of 2d.
Syllabus understandingD.1 — that gravitational field strength g at a point is given by g = F/m = GM/r²; the resultant field is restricted to points along the line joining two bodies (guidance) Command term: Determine
19D-1A-29
Orbits and satellites·D.1 Gravitational fields
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A satellite moves in a circular orbit of radius r around a planet. The gravitational field strength of the planet at the orbit is g.
What is the orbital speed of the satellite?
Show mark scheme
Marking point
Mark
Notes
Step 1The gravitational force is the centripetal force: mg = mv²/r.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2v² = gr, so v = √(gr).
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: mg = mv²/r gives v = √(gr).
BThis rearranges g = v²/r wrongly as v² = g/r.
CThis uses ½mv² = mgr, the speed after falling a height r in a uniform field, which is not an orbit.
DThis is v², not v: the square root has been left out.
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r² Command term: Determine
Multiple choice · 1 mark2 steps to full marksExplain
An astronaut in a space station 400 km above the Earth's surface floats freely inside the station.
Which statement is correct?
Show mark scheme
Marking point
Mark
Notes
Step 1At 400 km the field is still about 8.7 N kg−1, so the astronaut has weight.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The astronaut and the station fall towards the Earth with the same acceleration, so no contact force acts between them: this is apparent weightlessness.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe field there is about 8.7 N kg−1, only about 11 % less than at the surface.
BThere is no outward force on the astronaut; the unbalanced gravitational force provides the centripetal acceleration.
CCorrect: gravity gives the astronaut and the floor the same acceleration, so no contact force acts between them.
DWeight is the gravitational force, which does not depend on whether there is air.
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r² Command term: Explain
21D-1A-31
Gravitational field strength·D.1 Gravitational fields
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksCalculate
Planet Y has five times the mass of the Earth and twice the radius of the Earth.
What is (gravitational field strength at the surface of Y)/(gravitational field strength at the surface of the Earth)?
Show mark scheme
Marking point
Mark
Notes
Step 1g = GM/R², so g ∝ M/R²: ratio = 5/2² = 1.25.
✓ 1
Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: 5/2² = 1.25.
BThis divides by the radius ratio, 2, instead of its square.
CThis allows for the mass but not for the larger radius.
DThis multiplies by the square of the radius ratio instead of dividing by it.
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r² Command term: Calculate
22D-1A-32
Gravitational field lines·D.1 Gravitational fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDescribe
Two identical spherical masses are held a fixed distance apart. A field-line diagram is drawn for the region around them.
Which statement about the diagram is correct?
Show mark scheme
Marking point
Mark
Notes
Step 1At the midpoint the two equal masses produce fields that are equal in magnitude and opposite in direction, so the resultant field is zero.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2A zero field has no direction, so no field line passes through that point.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the resultant field at the midpoint is zero, so no line can be drawn through it.
BField lines never cross: the resultant field has only one direction at each point.
CGravity is only attractive, so every line points towards a mass.
DEqual spacing would mean a uniform field; here the field varies strongly with position.
Syllabus understandingD.1 — gravitational field lines Command term: Describe
23D-1B-11
Newton's law of gravitation·D.1 Gravitational fields
Paper 1BHard11 marks
Data-based question11 steps to full marksDetermine
A student uses a school torsion balance to investigate the gravitational force between two spheres. A small lead sphere of mass 15.0 g is fixed to each end of a light beam hanging from a fine wire. A large lead sphere of mass 1.50 kg is placed near each small sphere, and the gravitational attraction twists the wire until the beam comes to rest. The balance has been calibrated so that the force F on each small sphere is found from the deflection of a laser spot.
The centre-to-centre separation r of each pair of spheres is varied. The uncertainty in r is ±0.5 mm and the uncertainty in F is ±0.20 × 10−10 N. The graph shows F against 1/r² with error bars.
r / m
1/r² / m−2
F / 10−10 N
0.046
473
6.93
0.052
5.26
0.060
278
4.07
0.070
204
2.90
0.085
138
2.07
0.100
100
1.40
Graph drawn to scale
(a)
Calculate 1/r² for r = 0.052 m and its absolute uncertainty.
(2)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine, using lines of maximum and minimum gradient, a value for the gravitational constant G and its absolute uncertainty.
(4)
(d)
The accepted value of G is 6.67 × 10−11 N m² kg−2. Comment on the student's result.
(1)
(e)
Each small sphere is also attracted by the large sphere on the far side of the beam. Explain the effect of this on the value of G obtained.
(2)
(f)
Suggest one change to the apparatus that would reduce the percentage uncertainty in F.
Accept ±7 m−2 (±7.1). Award [0] for MP2 if the fractional uncertainty is not doubled.
Part (b)
Single straight line passing through all the error bars, close to the origin
✓ 1
Part (c)
Best gradient ≈ 1.46 × 10−12 N m²
✓ 1
Accept 1.40–1.52 × 10−12 N m².
G = gradient/(Mm) = 1.46 × 10−12/(1.50 × 0.0150) = 6.5 × 10−11 N m² kg−2
✓ 1
Allow ECF from the candidate's gradient.
Max and min gradient lines drawn through all error bars, gradients ≈ 1.61 × 10−12 and 1.34 × 10−12 N m²
✓ 1
Accept any pair of lines that pass through all the error bars.
ΔG = ½(7.17 − 5.94) × 10−11 ≈ 0.6 × 10−11, so G = (6.5 ± 0.6) × 10−11 N m² kg−2
✓ 1
Accept ±0.4 to ±0.8 × 10−11. MP4 is for matching the precision of value and uncertainty.
Part (d)
The accepted value lies within the student's range of values, so the result is consistent with it (although the student's value is lower)
✓ 1
Allow ECF from (c).
Part (e)
The far large sphere pulls each small sphere in the opposite direction to the near sphere, so the net force measured is smaller than GMm/r²
✓ 1
OWTTE
The measured F values, the gradient and hence G are all too small / G is underestimated
✓ 1
Award MP2 only if the direction of the effect is correct.
Part (f)
Use large spheres of greater mass (or denser material) so the forces are larger while the absolute uncertainty in reading the deflection stays the same
✓ 1
OWTTE. Accept a longer light path to the scale so that the laser-spot deflection for a given force is larger.
Answers: (a) 1/r² = (370 ± 7) m−2 · (c) G = (6.5 ± 0.6) × 10−11 N m² kg−2(the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — Newton's universal law of gravitation as given by F = Gm1m2/r² for bodies treated as point masses Command term: Determine
24D-1B-12
Kepler's third law·D.1 Gravitational fields
Paper 1BMedium12 marks
Data-based question12 steps to full marksDetermine
A student uses a published exoplanet database to test Kepler's third law for the seven planets that orbit the star TRAPPIST-1. The table gives the orbital radius a of each planet in astronomical units (AU) and its orbital period T in days. The orbits may be treated as circular.
The student tests the relationship T = kan, where k and n are constants, by plotting lg (T / day) against lg (a / AU). Five of the points have been plotted.
Planet
a / AU
T / day
lg (a / AU)
lg (T / day)
b
0.01154
1.511
−1.938
0.179
c
0.01580
2.422
d
0.02227
4.049
−1.652
0.607
e
0.02925
6.101
−1.534
0.785
f
0.03849
9.208
−1.415
0.964
g
0.04683
12.35
h
0.06189
18.77
−1.208
1.273
Graph drawn to scale
(a)
Complete the table for planets c and g.
(2)
(b)
Plot the points for planets c and g on the graph and draw the line of best fit for the data.
(2)
(c)
Determine the gradient of the line.
(2)
(d)
Kepler's third law predicts that T² is proportional to a³. Outline whether the data support this prediction.
(1)
(e)
The relationship T² = 4π²a³/(GM) applies, where M is the mass of the star. Determine, using the data for planet e, the mass of TRAPPIST-1 in kg. (1 AU = 1.50 × 1011 m)
(2)
(f)
The database gives a for planet e as (0.02925 ± 0.00025) AU; the uncertainty in T is negligible. Determine the absolute uncertainty in your answer to (e) and discuss whether your value agrees with the published mass of 0.0898 solar masses. (Mass of the Sun = 1.99 × 1030 kg)
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
lg (a / AU): c = −1.801; g = −1.329
✓ 1
Accept 3 or 4 decimal places; do not accept 2 d.p.
lg (T / day): c = 0.384; g = 1.092
✓ 1
Accept 3 or 4 decimal places; do not accept 2 d.p.
Part (b)
Both points plotted correctly to within half a small square
✓ 1
Allow ECF from (a).
Single straight line of best fit through (or very close to) all seven points
✓ 1
Part (c)
Gradient calculated from a large triangle (at least half the length of the line)
✓ 1
gradient = 1.50
✓ 1
Accept 1.46 to 1.54. No unit.
Part (d)
Gradient ≈ 1.5 = 3/2, so T ∝ a3/2; squaring gives T² ∝ a³, which supports the prediction
✓ 1
Allow ECF from (c) if the candidate's gradient is within ±3 % of 1.5.
Part (e)
a = 0.02925 × 1.50 × 1011 = 4.388 × 109 m AND T = 6.101 × 86 400 = 5.271 × 105 s
✓ 1
Both conversions needed.
M = 4π²a³/(GT²) = 1.80 × 1029 kg
✓ 1
Accept 1.79–1.81 × 1029 kg.
Part (f)
percentage uncertainty in M = 3 × (0.00025/0.02925) × 100 = 2.6 %
✓ 1
ΔM = 0.05 × 1029 kg, so M = (1.80 ± 0.05) × 1029 kg
✓ 1
MP2 is for matching precision of value and uncertainty. Allow ECF from (e).
Published mass = 0.0898 × 1.99 × 1030 = 1.79 × 1029 kg, which lies within the range 1.75–1.85 × 1029 kg, so the values agree
✓ 1
Accept comparison in solar masses: 0.0904 ± 0.0023 M☉.
Answers: (a) c: −1.801, 0.384; g: −1.329, 1.092 · (c) 1.50 · (e) M = 1.80 × 1029 kg · (f) M = (1.80 ± 0.05) × 1029 kg (the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — Kepler's three laws of orbital motion Command term: Determine
25D-1B-19
Kepler's second law·D.1 Gravitational fields
Paper 1BMedium9 marks
Data-based question9 steps to full marksExplain
A student tests Kepler's second law using a published ephemeris (a table of predicted planetary positions) for Mars. For five 10-day intervals, each starting a given number of days after Mars passes perihelion (its closest point to the Sun), the student records the angle Δθ through which Mars moves around the Sun during the interval and the distance r of Mars from the Sun at the middle of the interval. The ephemeris gives angles to the nearest 0.01°.
For a short interval, the area swept out by the line joining Mars to the Sun is approximately ½r²Δθ, with Δθ in radians.
Start of interval / days after perihelion
r / AU
Δθ / °
1/r² / AU−2
0
1.3815
6.35
0.524
95
1.4459
5.79
181
1.5549
5.01
0.414
267
1.6411
4.50
343
1.6659
4.36
0.360
Graph drawn to scale
(a)
Complete the table.
(1)
(b)
Draw the line of best fit for the data.
(1)
(c)
Explain how the graph supports Kepler's second law.
(2)
(d)
Calculate, in m², the area swept out in 10 days starting at perihelion. (1 AU = 1.50 × 1011 m)
(2)
(e)
Explain why the student used the angle moved in 10 days rather than the angle moved in 1 day or in 100 days.
(2)
(f)
Determine the percentage difference between the values of r²Δθ at perihelion (0 days) and at 343 days, and comment on your answer.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.478 AND 0.371 AU−2
✓ 1
Both needed, to 3 s.f. (3 d.p.).
Part (b)
Single straight line through (or very close to) all the points, passing through (or very close to) the origin
✓ 1
Part (c)
The graph is a straight line through the origin, so Δθ ∝ 1/r² / r²Δθ is constant
✓ 1
The area swept out in each 10-day interval, ½r²Δθ, is therefore the same: equal areas are swept out in equal times
✓ 1
OWTTE
Part (d)
r = 1.3815 × 1.50 × 1011 = 2.072 × 1011 m AND Δθ = 6.35 × π/180 = 0.1108 rad
✓ 1
area = ½ × (2.072 × 1011)² × 0.1108 = 2.38 × 1021 m²
✓ 1
Accept 2.37–2.39 × 1021 m².
Part (e)
In 1 day Δθ is only about 0.4–0.6°, so the ±0.005° rounding gives a percentage uncertainty about ten times larger
✓ 1
OWTTE
In 100 days r changes considerably, so ½r²Δθ with a single value of r is no longer a good approximation to the area
✓ 1
OWTTE
Part (f)
r²Δθ = 12.12 and 12.10 AU² degrees, a difference of about 0.16 %, which is within the rounding uncertainty of the angles (0.005/6.35 + 0.005/4.36 ≈ 0.19 %), so r²Δθ is constant: consistent with Kepler's second law
✓ 1
Accept any comparison showing the difference is very small / within uncertainty.
Answers: (a) 0.478, 0.371 AU−2 · (d) 2.38 × 1021 m² · (f) about 0.16 % (the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — Kepler's three laws of orbital motion Command term: Explain
26D-2-16
Density, mass and g·D.1 Gravitational fields
Paper 2Hard13 marks
Short answer & extended response13 steps to full marksShow that
Many small asteroids are "rubble piles": collections of loose rocks held together only by their own gravity.
Asteroid R is modelled as a uniform sphere of radius 250 m and mean density 2.0 × 103 kg m−3. It rotates about its axis once every 3.6 hours.
Diagram NOT to scale
(a)
(i)
Show that the mass of R is about 1.3 × 1011 kg.
(1)
(ii)
Calculate the gravitational field strength at the surface of R.
(2)
(b)
A loose rock of mass 5.0 kg rests on the surface of R at its equator, as shown in the diagram.
(i)
Draw and label, on the diagram, the forces acting on the rock. The lengths of your arrows should show the relative sizes of the forces.
(2)
(ii)
Determine the normal force exerted on the rock by the surface.
(4)
(c)
(i)
Show that the shortest period of rotation that a rubble pile of mean density ρ can have without losing rocks from its equator is √(3π/(Gρ)).
(2)
(ii)
Calculate this shortest period for R, in hours.
(1)
(iii)
Surveys show that almost no asteroids wider than a few hundred metres rotate faster than once every 2.2 hours, but many smaller asteroids rotate once every few minutes. Suggest what this shows about these small, fast-rotating asteroids.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
M = ρ × (4/3)πR³ = 2.0 × 103 × (4/3)π × 250³ = 1.31 × 1011 kg
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
g = GM/R² = 6.67 × 10−11 × 1.31 × 1011/250²
✓ 1
g = 1.40 × 10−4 N kg−1
✓ 1
Award [2] for CNA.
Part (b)(i)
weight «gravitational force» towards the centre of R AND normal «reaction» force away from the centre, both labelled
✓ 1
Do not award if a «centripetal force» is added as a third force.
weight arrow longer than the normal force arrow, both starting on the rock
✓ 1
Part (b)(ii)
ω = 2π/(3.6 × 3600) = 4.85 × 10−4 rad s−1
✓ 1
ALTERNATIVE: v = 2πR/T = 0.121 m s−1, then a = v²/R.
centripetal acceleration a = ω²R = 5.88 × 10−5 m s−2
✓ 1
the resultant force is towards the centre: mg − N = ma, so N = m(g − a)
✓ 1
Allow ECF from (a)(ii).
N = 5.0 × (1.40 × 10−4 − 5.88 × 10−5) = 4.1 × 10−4 N «4.0 × 10−4 N with unrounded values»
✓ 1
Accept 4.0–4.1 × 10−4 N. Award [1 max] for N = mg = 7.0 × 10−4 N.
Part (c)(i)
at the shortest period N = 0, so gravity alone provides the centripetal force: GM/R² = ω²R
✓ 1
with M = (4/3)πρR³: ω² = 4πGρ/3, so T = 2π/ω = √(3π/(Gρ))
✓ 1
R must be seen to cancel.
Part (c)(ii)
T = √(3π/(6.67 × 10−11 × 2.0 × 103)) = 8.4 × 103 s = 2.3 h
✓ 1
Accept 2.3–2.4 h.
Part (c)(iii)
they cannot be held together by gravity alone, so they must be single solid bodies «held together by the strength of the rock»
✓ 1
OWTTE
Answers: (a)(ii) g = 1.40 × 10−4 N kg−1 · (b)(ii) N = 4.0–4.1 × 10−4 N · (c)(ii) 2.3 h (the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r² (linked to A.2 — free-body diagrams and centripetal acceleration a = ω²r) Command term: Show that
27D-2-17
Kepler's laws·D.1 Gravitational fields
Paper 2Medium14 marks
Short answer & extended response14 steps to full marksState
A spacecraft travels from the Earth to Mars on a transfer orbit around the Sun. The transfer orbit is half of an ellipse. Its closest point to the Sun, P, lies on the orbit of the Earth, 1.00 AU from the Sun. Its farthest point from the Sun, A, lies on the orbit of Mars, 1.52 AU from the Sun. The engines are switched off between P and A.
Treat the orbits of the Earth and Mars as circles. The period of the Earth's orbit is 1.00 year.
Diagram NOT to scale
(a)
(i)
State Kepler's first law as it applies to the transfer orbit.
(1)
(ii)
Draw and label arrows on the diagram to show the direction of the gravitational force on the spacecraft and the direction of its velocity when it is at X.
(2)
(iii)
Explain, with reference to your answer to (a)(ii), why the spacecraft slows down as it moves from P to A.
(2)
(b)
(i)
Calculate (gravitational field strength of the Sun at P)/(gravitational field strength of the Sun at A).
(1)
(ii)
State Kepler's second law.
(1)
(iii)
At P and at A the velocity of the spacecraft is perpendicular to the line joining it to the Sun. The speed of the spacecraft at P is 32.7 km s−1. Using Kepler's second law, determine its speed at A.
(3)
(c)
Kepler's third law also applies to an elliptical orbit if the orbital radius is replaced by half of the sum of the greatest and the smallest distances from the Sun.
(i)
Determine the time taken by the spacecraft to travel from P to A, in days.
(3)
(ii)
Suggest why the spacecraft must be launched from the Earth at a particular time.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
the transfer orbit is an ellipse with the Sun at one focus
✓ 1
Do not accept «the Sun at the centre».
Part (a)(ii)
force arrow from X pointing towards the Sun, labelled
✓ 1
velocity arrow from X along the tangent to the path, in the direction of motion «towards A», labelled
✓ 1
The velocity arrow must not be drawn perpendicular to the force arrow.
Part (a)(iii)
the angle between the force and the velocity is greater than 90°, so the force has a component opposite to the velocity
✓ 1
Allow ECF from (a)(ii).
so the gravitational force does negative work on the spacecraft and its kinetic energy decreases
✓ 1
Accept «the component of the force along the path decelerates it». OWTTE
Part (b)(i)
g ∝ 1/r², so the ratio is (1.52/1.00)² = 2.31
✓ 1
Part (b)(ii)
the line joining the Sun and an orbiting body sweeps out equal areas in equal times
✓ 1
Part (b)(iii)
in a short time Δt the area swept out is ½rvΔt, at P and at A
✓ 1
equal areas in equal times, so rPvP = rAvA
✓ 1
vA = 32.7 × 1.00/1.52 = 21.5 km s−1
✓ 1
Award [1 max] for a bare answer with no use of Kepler's second law. Do not accept 49.7 km s−1.
Part (c)(i)
half of the sum of the distances = (1.00 + 1.52)/2 = 1.26 AU
✓ 1
T² ∝ r³ compared with the Earth: T = 1.00 × 1.263/2 = 1.41 years
✓ 1
P to A is half an orbit: ½ × 1.414 × 365 = 258 days
✓ 1
Award [2 max] for a whole orbit «516 days». Accept 257–259 days.
Part (c)(ii)
Mars must reach A at the same time as the spacecraft, so the Earth and Mars must be in a particular relative position at launch
✓ 1
OWTTE
Answers: (b)(i) 2.31 · (b)(iii) 21.5 km s−1 · (c)(i) 258 days (the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — Kepler's three laws of orbital motion; that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r² (linked to A.3 — work done by a force) Command term: State
28D-2-18
Geostationary orbits·D.1 Gravitational fields
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksEstimate
A communications satellite S is in a geostationary orbit around the Earth. The period of the orbit is 8.62 × 104 s.
Mass of the Earth = 5.97 × 1024 kg. Radius of the Earth = 6.37 × 106 m.
The equipment on S needs 4.0 kW of electrical power. In sunlight the power comes from solar panels; in the Earth's shadow it comes from batteries.
Diagram NOT to scale
(a)
(i)
Draw an arrow on the diagram to show the direction of the velocity of S.
(1)
(ii)
Outline why a geostationary orbit must lie in the plane of the equator.
(2)
(iii)
Show that the radius of the orbit is about 4.2 × 107 m.
(2)
(b)
(i)
Calculate the speed of S.
(1)
(ii)
For a few weeks each year S passes through the Earth's shadow once a day. Treat the shadow as a band whose width is equal to the diameter of the Earth. Estimate the longest time that S spends in the shadow.
(2)
(iii)
The fully charged batteries can deliver 1.8 × 107 J. Determine whether they can keep the equipment working for the whole of the time in (b)(ii).
(2)
(c)
(i)
Calculate the shortest time for a radio signal to travel from a ground station directly below S up to S and back to the ground.
(2)
(ii)
Suggest one reason why some internet services use satellites in low orbits rather than geostationary satellites.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
arrow at S along the tangent to the orbit, pointing anticlockwise «the same sense as the Earth's rotation»
✓ 1
Part (a)(ii)
the gravitational force on S is directed towards the centre of the Earth, so the centre of the orbit must be the centre of the Earth
✓ 1
S must stay above one point on the rotating Earth, which moves in a circle about the axis; only the equator's circle has the centre of the Earth as its centre
✓ 1
OWTTE
Part (a)(iii)
GMm/r² = 4π²mr/T², so r³ = GMT²/4π²
✓ 1
r = (6.67 × 10−11 × 5.97 × 1024 × (8.62 × 104)²/4π²)1/3 = 4.22 × 107 m
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(i)
v = 2πr/T = 2π × 4.22 × 107/8.62 × 104 = 3.07 × 103 m s−1
✓ 1
Allow ECF from (a)(iii).
Part (b)(ii)
distance travelled in the shadow ≈ 2 × 6.37 × 106 = 1.27 × 107 m
✓ 1
t = 1.27 × 107/3.07 × 103 = 4.1 × 103 s «≈ 69 min»
✓ 1
Allow ECF from (b)(i). Accept 4.0–4.2 × 103 s.
Part (b)(iii)
energy needed = 4.0 × 103 × 4.1 × 103 = 1.6 × 107 J «1.7 × 107 J with the unrounded time»
✓ 1
Allow ECF from (b)(ii).
this is less than 1.8 × 107 J, so the batteries are «just» sufficient
✓ 1
The conclusion must be consistent with the candidate's energy.
Part (c)(i)
height = 4.22 × 107 − 6.37 × 106 = 3.58 × 107 m
✓ 1
t = 2 × 3.58 × 107/3.00 × 108 = 0.24 s
✓ 1
Award [1 max] for 0.28 s «orbit radius used instead of height».
Part (c)(ii)
the signal travels a much shorter distance, so the delay is much smaller
✓ 1
Accept another valid reason, e.g. less powerful transmitters are needed. OWTTE
Answers: (a)(iii) r = 4.22 × 107 m · (b)(i) v = 3.07 × 103 m s−1 · (b)(ii) 4.1 × 103 s · (b)(iii) 1.6–1.7 × 107 J «sufficient» · (c)(i) 0.24 s (the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — Kepler's three laws of orbital motion; Newton's universal law of gravitation as given by F = Gm1m2/r² for bodies treated as point masses (linked to A.3 — power, and C.2 — the speed of electromagnetic waves) Command term: Estimate
29D-2-19
Gravitational field strength·D.1 Gravitational fields
Paper 2Easy10 marks
Short answer & extended response10 steps to full marksCalculate
Rhea is one of the moons of Saturn. It may be treated as a uniform sphere of mass 2.31 × 1021 kg and radius 7.64 × 105 m, and it has no atmosphere. A robotic lander of mass 450 kg is designed to explore its surface.
G = 6.67 × 10−11 N m² kg−2; gravitational field strength at the surface of the Earth = 9.81 N kg−1.
(a)
Show that the gravitational field strength at the surface of Rhea is about 0.26 N kg−1.
(1)
(b)
The engine of the lander can provide a maximum upward thrust of 300 N.
(i)
Determine whether this engine could hold the lander at rest just above the surface of Rhea, and whether it could do so just above the surface of the Earth.
(2)
(ii)
Calculate the initial acceleration of the lander when it lifts off from Rhea with the engine at maximum thrust.
(2)
(c)
The lander releases a small sensor from rest at a height of 12.0 m above the surface.
(i)
Calculate the time taken for the sensor to reach the surface.
(2)
(ii)
Outline why the gravitational field strength may be treated as constant during this fall.
(1)
(d)
A rock sample collected by the lander has a weight of 0.50 N on Rhea.
weight on Rhea = 450 × 0.264 = 119 N, which is less than 300 N, so the lander can hover above Rhea
✓ 1
Allow ECF from (a).
weight on the Earth = 450 × 9.81 = 4.41 × 103 N, which is greater than 300 N, so it cannot hover above the Earth
✓ 1
A conclusion without a calculated weight scores [0] for that mark.
Part (b)(ii)
resultant «upward» force = 300 − 119 = 181 N
✓ 1
Allow ECF from (b)(i).
a = 181 / 450 = 0.40 m s−2
✓ 1
Accept 0.40–0.41 m s−2. Award [2] for CNA.
Part (c)(i)
s = ½gt², so t = √(2 × 12.0 / 0.264)
✓ 1
Allow ECF from (a).
t = 9.5 s
✓ 1
Accept 9.5–9.6 s. Award [2] for CNA. An answer of 1.6 s «uses g = 9.81 N kg−1» scores [0].
Part (c)(ii)
12.0 m is negligible compared with the radius «7.64 × 105 m», so the distance from the centre and hence g change by a negligible fraction «≈ 0.003 %»
✓ 1
OWTTE. Accept “the field near the surface is «approximately» uniform over such a small height”.
Part (d)
mass of sample = 0.50 / 0.264 = 1.89 kg «the mass is the same on both bodies»
✓ 1
Allow ECF from (a).
weight on the Earth = 1.89 × 9.81 = 18.6 N
✓ 1
Accept 18.6–18.9 N. Award [2] for CNA.
Answers: (a) 0.264 N kg−1 · (b)(i) 119 N on Rhea (yes); 4.41 × 103 N on the Earth (no) · (b)(ii) 0.40 m s−2 · (c)(i) 9.5 s · (d) 18.6 N (the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r². Links: A.1 — equations of motion for uniform acceleration; A.2 — Newton's second law of motion Command term: Calculate
30D-2-20
Kepler's third law·D.1 Gravitational fields
Paper 2Hard11 marks
Short answer & extended response11 steps to full marksDetermine
A star has a mass M = 0.80 M☉ and a luminosity of 0.35 L☉. A planet moves around the star in a circular orbit of radius r with a period T of 150 days. For a circular orbit, T² = (4π²/GM) r³.
M☉ = 1.99 × 1030 kg; L☉ = 3.83 × 1026 W; G = 6.67 × 10−11 N m² kg−2; σ = 5.67 × 10−8 W m−2 K−4.
(a)
State Kepler's third law.
(1)
(b)
Calculate r.
(2)
(c)
Show that the intensity of the star's radiation at the distance of the planet is about 1.8 × 103 W m−2.
(1)
(d)
The planet has an albedo of 0.30 and its surface emits as a black body.
(i)
Determine the equilibrium temperature of the planet's surface, ignoring any effect of an atmosphere.
(3)
(ii)
Suggest why the actual mean surface temperature of the planet could be higher than your answer to (d)(i).
(2)
(e)
The luminosity of the star is known only to within ±10 %.
Estimate the resulting absolute uncertainty in your answer to (d)(i).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the square of the orbital period is proportional to the cube of the «mean» orbital radius «for bodies orbiting the same central body»
✓ 1
Accept T² ∝ r³ with the symbols identified. Accept “semi-major axis” for radius.
Part (b)
T = 150 × 86 400 = 1.30 × 107 s AND M = 0.80 × 1.99 × 1030 = 1.59 × 1030 kg substituted into r³ = GMT²/4π²
✓ 1
r = 7.67 × 1010 m
✓ 1
Accept 7.6–7.7 × 1010 m «≈ 0.51 AU». Award [2] for CNA. Do not award MP2 if T is left in days.
mean intensity over the whole surface = I/4 «= 453 W m−2»
✓ 1
Allow ECF from (c). Use of 1.8 × 103 W m−2 is acceptable.
σT⁴ = (1 − 0.30) × I/4 «= 317 W m−2»
✓ 1
T = 273 K
✓ 1
Accept 270–276 K. Award [3] for CNA. Omitting the factor ¼ gives 387 K: award [2].
Part (d)(ii)
an atmosphere containing greenhouse gases «such as CO₂, H₂O, CH₄» absorbs infrared radiation emitted by the surface
✓ 1
OWTTE
the gases re-emit infrared in all directions, so some returns to the surface «and the surface must be hotter to lose energy at the rate it gains it»
✓ 1
Do not accept “traps heat” alone.
Part (e)
T ∝ L1/4, so the fractional uncertainty in T is ¼ × 10 % = 2.5 %
✓ 1
absolute uncertainty ≈ ±7 K
✓ 1
Accept ±6 K to ±7 K. Allow ECF from (d)(i).
Answers: (b) 7.67 × 1010 m · (c) 1.81 × 103 W m−2 · (d)(i) 273 K · (e) ±7 K (the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — Kepler's three laws of orbital motion; Newton's universal law of gravitation as given by F = Gm1m2/r² for bodies treated as point masses. Links: B.1 — apparent brightness b = L/4πd²; B.2 — albedo, mean incoming intensity and the greenhouse effect; Tools — propagation of uncertainties Command term: Determine
31D-2-21
Newton's law of gravitation·D.1 Gravitational fields
Paper 2Hard11 marks
Short answer & extended response11 steps to full marksExplain
At new moon the Moon M lies on the straight line between the Sun S and the Earth E, as shown in the diagram.
Mass of the Sun = 1.99 × 1030 kg; mass of the Earth = 5.97 × 1024 kg; mass of the Moon = 7.35 × 1022 kg; distance between the centres of the Earth and the Moon = 3.84 × 108 m; distance between the centres of the Sun and the Moon = 1.50 × 1011 m; G = 6.67 × 10−11 N m² kg−2.
Diagram NOT to scale
(a)
(i)
Show that the gravitational force exerted by the Earth on the Moon is about 2.0 × 1020 N.
(1)
(ii)
Calculate the gravitational force exerted by the Sun on the Moon.
(1)
(b)
Draw and label, on the diagram, arrows to represent the two gravitational forces acting on the Moon. The lengths of your arrows should show the relative sizes of the forces.
(2)
(c)
Determine the magnitude and direction of the resultant gravitational field strength at the centre of the Moon.
(3)
(d)
Part (a) shows that the Sun pulls on the Moon more strongly than the Earth does.
(i)
Calculate the gravitational field strength due to the Sun at the centre of the Earth at the same instant.
(1)
(ii)
Explain, using your answers to (c) and (d)(i), why the Moon nevertheless continues to orbit the Earth.
one arrow from M towards S and one from M towards E, each labelled «e.g. force of Sun, force of Earth»
✓ 1
Arrows should start at, or touch, M.
arrow towards S about twice as long as arrow towards E «ratio 2.2»
✓ 1
Accept a length ratio between 1.8 and 2.6. Allow ECF from (a).
Part (c)
field of the Sun = 5.90 × 10−3 N kg−1 towards S
✓ 1
ALTERNATIVE: resultant force from (a) divided by the mass of the Moon, for MP1 and MP2.
field of the Earth = 2.70 × 10−3 N kg−1 towards E
✓ 1
resultant = 3.20 × 10−3 N kg−1 towards the Sun
✓ 1
Accept 3.1–3.3 × 10−3 N kg−1. Direction needed for MP3. Allow ECF from (a).
Part (d)(i)
g = 6.67 × 10−11 × 1.99 × 1030 / (1.504 × 1011)² = 5.87 × 10−3 N kg−1
✓ 1
Must see the distance 1.50 × 1011 + 3.84 × 108 = 1.504 × 1011 m used. Accept 5.86–5.88 × 10−3 N kg−1.
Part (d)(ii)
the Sun gives the Earth and the Moon almost the same acceleration «≈ 5.9 × 10−3 m s−2», so the Earth and the Moon move around the Sun together
✓ 1
OWTTE
the difference between the Sun's field at the Moon and at the Earth is only about 3 × 10−5 N kg−1
✓ 1
Allow ECF from (c) and (d)(i).
this is about 1 % of the Earth's field at the Moon «2.70 × 10−3 N kg−1», so relative to the Earth the Moon's motion is controlled by the Earth's pull
✓ 1
Award MP3 only if a comparison with the Earth's field at the Moon is made.
Answers: (a)(i) 1.98 × 1020 N · (a)(ii) 4.34 × 1020 N · (c) 3.20 × 10−3 N kg−1 towards the Sun · (d)(i) 5.87 × 10−3 N kg−1(the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — Newton's universal law of gravitation as given by F = Gm1m2/r² for bodies treated as point masses; that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r² Command term: Explain
32D-2-22
Variation of g with distance·D.1 Gravitational fields
Paper 2Medium9 marks
Short answer & extended response9 steps to full marksDetermine
A geophysicist searches for buried ore with a gravimeter, an instrument that can detect a change in the gravitational field strength of 5.0 × 10−8 N kg−1. An ore body may be modelled as a uniform sphere of radius 40 m whose centre is 150 m below the surface. Its density is 2500 kg m−3 greater than that of the surrounding rock.
G = 6.67 × 10−11 N m² kg−2; g at the surface = 9.81 N kg−1; radius of the Earth R = 6.37 × 106 m.
(a)
Show that the extra mass of the ore body, compared with the same volume of surrounding rock, is about 6.7 × 108 kg.
(1)
(b)
Calculate the increase in the gravitational field strength at the surface directly above the centre of the ore body.
(2)
(c)
Determine whether the gravimeter could detect the same ore body if its centre were 1.2 km below the surface.
(2)
(d)
The gravimeter contains a mass hanging from a spring, which is extended by 0.040 m.
Calculate the change in the extension of the spring caused by the ore body in (b).
(2)
(e)
For a small height h above the surface, the gravitational field strength of the Earth decreases by approximately 2gh/R.
(i)
Determine the change in height of the gravimeter that would produce the same change in gravitational field strength as the ore body in (b).
(1)
(ii)
Suggest, using your answer to (e)(i), one precaution the geophysicist must take during the survey.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Δm = 4/3 × π × 40³ × 2500 = 6.70 × 108 «kg»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)
Δg = GΔm/d² = 6.67 × 10−11 × 6.7 × 108 / 150² «the sphere acts as a point mass at its centre»
ALTERNATIVE: Δg falls by a factor (1200/150)² = 64, giving 3.1 × 10−8 N kg−1.
this is less than 5.0 × 10−8 N kg−1, so it could not be detected
✓ 1
Conclusion must follow from a calculation. Allow ECF from (b).
Part (d)
kx = mg, so x ∝ g and Δx = x Δg/g
✓ 1
Δx = 0.040 × 1.99 × 10−6 / 9.81 = 8.1 × 10−9 m
✓ 1
Accept 8.1–8.2 × 10−9 m. Allow ECF from (b). Award [2] for CNA.
Part (e)(i)
h = Δg R / 2g = 2.0 × 10−6 × 6.37 × 106 / (2 × 9.81) = 0.65 m
✓ 1
Accept 0.63–0.66 m. Allow ECF from (b).
Part (e)(ii)
measure the height of the gravimeter at every station to within a few centimetres and correct the readings for it
✓ 1
OWTTE. Accept any precaution clearly linked to height errors.
Answers: (a) 6.70 × 108 kg · (b) 2.0 × 10−6 N kg−1 · (c) 3.1 × 10−8 N kg−1; no · (d) 8.1 × 10−9 m · (e)(i) 0.65 m (the remaining parts are explanations — see the table above)
Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r²; conditions under which extended bodies can be treated as point masses. Links: B.1 — density ρ = m/V; A.2 — Hooke's law Command term: Determine
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