D.1 Gravitational fields flashcards: IB Physics SL and HL

All 71 D.1 Gravitational fields cards

1. DefinitionSL & HL

   State Newton's law of universal gravitation.

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   Every point mass attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the separation of their centres, directed along the line joining them. Exam tip: three marking points are needed — "product of the masses", "inversely proportional to the SQUARE of the distance", and "between point masses / centres of mass". Writing "inversely proportional to the distance" or "distance between the surfaces" loses the mark. The force is a vector in newtons (N), always attractive, and the pair obeys Newton's third law: the two forces are equal in magnitude and opposite in direction whatever the masses.

2. DefinitionSL & HL

   Define a gravitational field.

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   A gravitational field is a region of space in which a mass experiences a force due to the gravitational attraction of another mass. Exam tip: the mark scheme wants "region (of space) where a mass experiences a force" — saying "where gravity acts" is circular and scores zero. The field is the model that removes action-at-a-distance: the source mass modifies the space around it, and a second mass placed there responds locally. Fields extend to infinity but weaken as 1/r². The field is a vector field, represented by field lines or by the vector quantity g at each point.

3. DefinitionSL & HL

   Define gravitational field strength.

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   Gravitational field strength is the gravitational force per unit mass experienced by a small point test mass placed at that point: g = F/m. Exam tip: the words "per unit mass" and "(small/point) test mass" both earn credit — "the force of gravity at a point" is not a definition. Unit: N kg⁻¹, which is dimensionally identical to m s⁻² (so g is also the free-fall acceleration). It is a VECTOR, directed towards the centre of the source mass. The test mass must be small so its own field does not disturb the field being measured.

4. DefinitionSL & HL

   State the rules for drawing gravitational field lines.

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   Field lines show the direction of the gravitational force on a small test mass: they point towards the source mass, are radial and evenly spaced in angle for a point/spherical mass, never cross, and their density (lines per unit area) represents the magnitude of g. Exam tip: arrows are compulsory and must point INWARDS — a sketch without arrowheads scores zero. Near the surface of a planet over a small region the lines are parallel, equally spaced and vertical (uniform field). Lines start at infinity and end on the mass; there are no gravitational "sources" of outward lines because gravity is only attractive.

5. DefinitionSL & HL

   Explain what is meant by a uniform gravitational field.

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   A uniform gravitational field is one in which the gravitational field strength has the same magnitude and the same direction at every point, so the field lines are straight, parallel and equally spaced. Exam tip: state BOTH magnitude and direction — "g is constant" alone is usually only one mark. The field close to the Earth's surface is approximately uniform because the region considered is tiny compared with the Earth's radius (6.37 × 10⁶ m), so r changes by a negligible fraction. Over a few kilometres g falls by well under 0.1 %, which justifies using the constant value 9.8 N kg⁻¹ in projectile work.

6. DefinitionSL & HL

   State what the universal gravitational constant G represents, with its value and units.

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   G is the constant of proportionality in Newton's law of gravitation; it fixes the strength of the gravitational interaction and is the same everywhere in the universe. G = 6.67 × 10⁻¹¹ N m² kg⁻² (equivalently m³ kg⁻¹ s⁻²), given in the data booklet. Exam tip: never confuse G with g. G is a universal scalar constant with units N m² kg⁻²; g is a local vector field strength in N kg⁻¹ that depends on the planet and on r. Because G is so small, gravitational forces between laboratory masses are minute — Cavendish's torsion-balance measurement is the standard Nature of Science example.

7. DefinitionSL & HL

   State the point-mass (shell) approximation used for spherical bodies in gravitation.

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   A body with a spherically symmetric mass distribution behaves, for all external points, exactly as if its entire mass were concentrated at its centre. Exam tip: this is why r in F = GMm/r² and g = GM/r² is measured from the CENTRE of the planet, not from its surface — for a satellite at altitude h use r = R + h. A second result you must quote qualitatively: at points INSIDE a uniform sphere only the mass within radius r contributes, so g falls linearly to zero at the centre and the shell outside a point exerts no net force on it.

8. DefinitionSL & HL

   State Kepler's first law.

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   The orbit of each planet is an ellipse with the Sun at one focus. Exam tip: the mark is for "ellipse" AND "Sun at ONE focus" — "the Sun at the centre" is wrong. Most planetary orbits have small eccentricity, so a circle (the special case where the two foci coincide) is an acceptable model for IB calculations; in that case the semi-major axis becomes the orbital radius r. The point of closest approach is the perihelion and the furthest is the aphelion. The law is empirical — derived from Tycho Brahe's data — and was later explained by Newton's inverse-square law.

9. DefinitionSL & HL

   State Kepler's second law.

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   The line joining a planet to the Sun sweeps out equal areas in equal intervals of time. Exam tip: the mark scheme wants "equal AREAS in equal TIMES" — do not write "equal distances". The physical consequence, which is what questions actually test, is that the planet moves fastest at perihelion (closest) and slowest at aphelion (furthest). The underlying reason is conservation of angular momentum: the gravitational force is central, so it exerts no torque about the Sun and mvr (or mr²ω) is constant. For a circular orbit the law simply means constant orbital speed.

10. DefinitionSL & HL

    State Kepler's third law.

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    The square of the orbital period of a planet is directly proportional to the cube of its mean orbital radius (semi-major axis): T² ∝ r³. Exam tip: the proportionality is given in the data booklet; the constant 4π²/GM depends only on the mass of the CENTRAL body, so the law lets you compare two satellites of the same planet with (T₁/T₂)² = (r₁/r₂)³ without knowing G or M. A frequent error is applying one planet's constant to a different central mass — Earth satellites and Sun-orbiting planets have completely different constants of proportionality.

11. DefinitionSL & HL

    Define a geostationary orbit and state the conditions a satellite must satisfy to be in one.

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    A geostationary satellite remains permanently above the same point on the Earth's equator, so it appears stationary to an observer on the ground. Conditions: period exactly equal to one sidereal day (23 h 56 min = 8.62 × 10⁴ s); orbit in the equatorial plane; direction of rotation the same as the Earth's (west to east); orbit circular. These fix the radius uniquely at r = 4.22 × 10⁷ m (altitude ≈ 3.6 × 10⁷ m). Exam tip: students often give only the period. Full credit requires the equatorial plane and same-sense rotation as well. Contrast with a polar orbit, which is low, fast and scans the whole surface.

12. DefinitionSL & HL

    Distinguish between the mass and the weight of a body.

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    Mass is the amount of matter in a body / the measure of its inertia; it is a scalar measured in kilograms and is the same everywhere. Weight is the gravitational force acting on that mass, W = mg; it is a vector measured in newtons and depends on the local field strength g. Exam tip: a 60 kg astronaut has mass 60 kg on the Moon but weight only 60 × 1.6 = 96 N there instead of 590 N on Earth. Losing marks by writing "weight in kg" is common — always quote weight in N and state the direction (towards the planet's centre).

13. DefinitionSL & HL

    Explain what is meant by the apparent weightlessness of an astronaut in an orbiting spacecraft.

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    The astronaut and the spacecraft are both in free fall: gravity is the only force acting and it supplies exactly the centripetal acceleration, so both accelerate identically at g = GM/r² and there is no contact (normal) force between them. The astronaut therefore feels weightless even though the true gravitational force on them is large. Exam tip: the fatal error is writing "there is no gravity in space". At the ISS (r ≈ 6.77 × 10⁶ m) g is about 8.7 N kg⁻¹, roughly 89 % of its surface value. The correct phrase is "apparent weightlessness — zero normal reaction because both are in free fall".

14. EquationSL & HLData booklet: Yes

    State Newton's law of gravitation in symbols, defining every quantity.

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    F = Gm₁m₂/r² (data booklet). F = magnitude of the mutual attractive force (N); G = 6.67 × 10⁻¹¹ N m² kg⁻² (universal constant); m₁, m₂ = the two interacting masses (kg); r = separation of their centres (m). Valid for point masses, or for spherically symmetric bodies at external points. Common misuse: using the altitude above the surface instead of r measured from the centre. Sanity check: two 1 kg masses 1 m apart attract with F = 6.67 × 10⁻¹¹ N — utterly negligible, which is why gravity only matters for astronomical masses. The force on each mass is equal and opposite.

15. EquationSL & HLData booklet: Yes

    State the defining equation for gravitational field strength and give its rearrangements.

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    g = F/m (data booklet). g = gravitational field strength (N kg⁻¹ = m s⁻²), a vector towards the source; F = gravitational force on the test mass (N); m = mass of the small test mass (kg). Rearranged, F = mg gives weight near a planet's surface. Valid at any point in any gravitational field; m must be small enough not to perturb the field. Common misuse: substituting the mass of the PLANET for m — m is always the test mass. Sanity check: a 2.0 kg mass weighing 19.6 N gives g = 19.6/2.0 = 9.8 N kg⁻¹.

16. EquationSL & HLData booklet: Yes

    State the equation for the gravitational field strength due to a point or spherical mass.

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    g = GM/r² (data booklet). g = field strength (N kg⁻¹); G = 6.67 × 10⁻¹¹ N m² kg⁻²; M = mass of the source body only (kg); r = distance from its CENTRE (m). Note the test mass has cancelled, so g is independent of the mass placed in the field — that is why all bodies fall with the same acceleration. Common misuse: forgetting that r is measured from the centre. Sanity check for Earth: g = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴)/(6.37 × 10⁶)² = 3.98 × 10¹⁴/4.06 × 10¹³ = 9.81 N kg⁻¹.

17. EquationSL & HLData booklet: No – derive

    State how gravitational field strength varies with altitude above a planet's surface.

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    g = GM/(R + h)², where R = planetary radius (m) and h = altitude above the surface (m); g in N kg⁻¹, M in kg. Equivalently g = g_surface × R²/(R + h)². Valid outside a spherically symmetric body. Common misuse: putting h alone in the denominator, which gives absurdly large values for low orbits. Sanity check: at the ISS altitude h = 4.0 × 10⁵ m, g = 9.81 × (6.37 × 10⁶)²/(6.77 × 10⁶)² = 9.81 × 0.885 = 8.7 N kg⁻¹ — only about 11 % less than at the surface, which is why "zero gravity in orbit" is wrong.

18. EquationSL & HLData booklet: Yes

    State the equation for the orbital speed of a satellite in a circular orbit and show where it comes from.

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    v = √(GM/r) (data booklet). v = orbital speed (m s⁻¹); M = mass of the central body (kg); r = orbital radius from the centre (m). Derivation: gravity supplies the centripetal force, GMm/r² = mv²/r, so v² = GM/r. Valid only for a circular orbit about a much more massive central body. Common misuse: using the satellite's mass for M — the satellite's mass cancels, so all satellites at the same radius have the same speed. Sanity check: low Earth orbit r = 6.77 × 10⁶ m gives v = √(3.98 × 10¹⁴/6.77 × 10⁶) = 7.7 × 10³ m s⁻¹.

19. EquationSL & HLData booklet: Yes

    State Kepler's third law as a proportionality and explain how to use it to compare two orbits.

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    T² ∝ r³ (data booklet). T = orbital period (s); r = mean orbital radius / semi-major axis (m). For two bodies orbiting the SAME central mass, (T₁/T₂)² = (r₁/r₂)³, so no constants are needed. Valid for any closed orbit about a common central body. Common misuse: mixing orbits about different central masses, or using altitude instead of radius. Sanity check: the Moon has r = 3.84 × 10⁸ m and T = 27.3 days; a geostationary satellite has r = 4.22 × 10⁷ m, and (4.22 × 10⁷/3.84 × 10⁸)³ = 1.33 × 10⁻³, whose square root × 27.3 d = 1.00 d.

20. EquationSL & HLData booklet: No – derive

    Derive the full form of Kepler's third law for a circular orbit and identify the constant.

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    T² = 4π²r³/GM (derived, not printed). Derivation: GMm/r² = m(4π²r/T²) since a = 4π²r/T², giving T² = 4π²r³/GM. T = period (s); r = orbital radius (m); M = mass of the CENTRAL body (kg); G = 6.67 × 10⁻¹¹ N m² kg⁻². The constant 4π²/GM depends only on M, which is how astronomers weigh stars and planets: M = 4π²r³/GT². Common misuse: substituting the orbiting mass for M. Sanity check: r = 4.22 × 10⁷ m, M = 5.97 × 10²⁴ kg gives T² = 39.5 × 7.51 × 10²²/3.98 × 10¹⁴, so T = 8.64 × 10⁴ s.

21. EquationSL & HLData booklet: No – derive

    State the centripetal condition for circular orbital motion in all three equivalent forms.

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    GMm/r² = mv²/r = mω²r = 4π²mr/T². Symbols: M = central mass (kg); m = orbiting mass (kg); r = orbital radius (m); v = orbital speed (m s⁻¹); ω = angular speed (rad s⁻¹); T = period (s). Valid only for uniform circular motion where gravity is the ONLY force. Cancelling m gives GM/r² = v²/r = ω²r = 4π²r/T², the starting point for every orbital derivation. Common misuse: adding a separate "centripetal force" to the free-body diagram — gravity IS the centripetal force here, so a satellite's free-body diagram shows exactly one arrow, pointing to the centre.

22. Graph/diagramSL & HL

    Describe the field-line diagram for an isolated spherical mass and what it shows.

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    Draw straight lines directed radially INWARDS towards the centre, evenly spaced around the sphere, with arrowheads on each line; lines are drawn from outside the surface only. Interpretation: the direction of a line gives the direction of g (and of the force on a test mass), while the spacing gives the magnitude — lines converge and get closer together near the surface where g is largest, and diverge far away where g is small, reproducing the 1/r² fall-off automatically. Exam tip: marks are lost for missing arrows, for lines that do not point at the centre, and for uneven angular spacing. There are no outward lines because gravity is only attractive.

23. Graph/diagramSL & HL

    Describe the field-line diagram for the region close to a planet's surface and state when the approximation fails.

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    Over a small region near the surface the lines are drawn straight, vertical, parallel and equally spaced, with arrows pointing down into the ground — a uniform field of constant magnitude g = GM/R². This is the small-scale limit of the radial pattern: over a few kilometres the radial lines have not measurably converged, since Δr/R is of order 10⁻³. The approximation fails once the vertical extent becomes comparable with R (6.37 × 10⁶ m for Earth), for example for a satellite, where the lines must be redrawn as converging radial lines and g = GM/(R + h)² used instead of a constant 9.8 N kg⁻¹.

24. Graph/diagramSL & HL

    Sketch and describe the graph of gravitational field strength g against distance r from the centre of a uniform spherical planet.

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    Axes: g/N kg⁻¹ (vertical) against r/m (horizontal). For r < R (inside) g rises LINEARLY from 0 at the centre to a maximum g_s = GM/R² at the surface, because only the mass within radius r acts. For r > R the curve falls as 1/r², reaching g_s/4 at r = 2R and tending to zero as r → ∞ without ever touching the axis. The peak occurs exactly at r = R and the graph is continuous but has a sharp change of gradient there. Exam tip: students often draw the inside portion as a curve or start the outside branch at r = 0 — the two branches must join at (R, g_s).

25. Graph/diagramSL & HL

    Describe the linearisation graph used to test Kepler's third law with satellite data.

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    Plot T²/s² on the vertical axis against r³/m³ on the horizontal axis. A straight line through the origin confirms T² ∝ r³. Since T² = 4π²r³/GM, the gradient equals 4π²/GM, so the central mass is found from M = 4π²/(G × gradient). Exam tip: state "straight line THROUGH THE ORIGIN" — proportionality needs both. A non-zero intercept suggests a systematic error in r (for example using altitude rather than radius). If instead the central mass increased, the gradient would decrease; the alternative log–log plot of lg T against lg r gives a straight line of gradient 3/2 and intercept ½ lg(4π²/GM).

26. Graph/diagramSL & HL

    Describe the linearisation graph used to find the mass of a planet from measurements of g at different distances.

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    Plot g/N kg⁻¹ on the vertical axis against (1/r²)/m⁻² on the horizontal axis. Because g = GM/r², the result is a straight line through the origin of gradient GM, so M = gradient/G. Exam tip: r must be measured from the planet's CENTRE — plotting against 1/h² where h is altitude gives a curve, and this is the standard trap. A y-intercept that is not zero indicates a systematic error, most often a zero error in the force sensor or an unsubtracted planetary radius. Uncertainty in M is found by drawing maximum and minimum gradient lines through the error bars and halving their difference.

27. Concept/explainSL & HLData booklet: No – derive

    Explain how Kepler's third law (T² ∝ r³) follows from Newton's law of gravitation for a circular orbit.

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    • Gravitation supplies the centripetal force: GMm/r² = mv²/r
    • the orbiting mass m cancels, so the result is independent of satellite mass
    • for a circular orbit v = 2πr/T, so GM/r² = 4π²r/T²
    • rearranging gives T² = 4π²r³/(GM), i.e. T² ∝ r³ with constant 4π²/(GM) depending only on the central mass
    • the booklet prints T² ∝ r³; the full form must be derived
    • a straight line through the origin on a T² against r³ graph confirms the law, gradient = 4π²/(GM). Exam tip: incomplete answers quote T² ∝ r³ without ever writing GMm/r² = mv²/r, and forget to cancel m.
28. Concept/explainSL & HLData booklet: Yes

    Outline what is meant by a gravitational field and by gravitational field strength.

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    • A gravitational field is a region of space in which a mass experiences a gravitational force
    • field strength g = F/m is the force per unit mass on a small test mass placed at that point
    • it is a vector, directed towards the mass creating the field, unit N kg⁻¹ (equivalently m s⁻²)
    • for a point (or spherical) mass g = GM/r², with r measured from the centre
    • the test mass must be small enough not to disturb the source field
    • near the Earth's surface g ≈ 9.8 N kg⁻¹ and is treated as uniform over small vertical distances. Exam tip: students lose the mark by saying force per unit mass on a mass rather than per unit mass, or omit the direction/vector nature.
29. Concept/explainSL & HLData booklet: No – derive

    Explain how the gravitational field strength varies from the centre of a uniform spherical planet out to a large distance.

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    • Outside the planet the sphere behaves as if all its mass were concentrated at its centre, so g = GM/r² and g ∝ 1/r²
    • g is a maximum at the surface, r = R
    • inside a uniform sphere only the mass within radius r contributes; the shell outside exerts zero net field
    • that enclosed mass ∝ r³, so g = GMr/R³ and g ∝ r inside
    • therefore g rises linearly from zero at the centre to GM/R² at the surface, then falls as an inverse square
    • the graph is continuous with a sharp peak at r = R. Exam tip: students often draw g rising to infinity at the centre, confusing the point-mass result with a real extended body.
30. Concept/explainSL & HLData booklet: Yes

    Explain why a satellite in a lower circular orbit moves faster but has a shorter period than one in a higher orbit.

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    • For a circular orbit GMm/r² = mv²/r gives v = √(GM/r), so v ∝ 1/√r and a smaller r means a larger orbital speed
    • the gravitational field is stronger closer in, so a larger centripetal force and hence a greater speed is needed to maintain the orbit
    • the circumference 2πr is also smaller, so the satellite covers a shorter path at a higher speed
    • combining, T = 2πr/v = 2π√(r³/GM), so T ∝ r^(3/2) and the period is much shorter
    • example: the ISS at 400 km takes about 92 minutes, a geostationary satellite 24 hours. Exam tip: many answers assert faster therefore longer period, mixing up speed with period.
31. Concept/explainSL & HL

    Explain Kepler's first and second laws and what the second law implies about a planet's speed.

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    • First law: each planet moves in an ellipse with the Sun at one focus; circular orbits are a special case of small eccentricity
    • second law: the line joining planet to Sun sweeps out equal areas in equal times
    • so when the planet is near perihelion the radius is short and it must move faster; near aphelion it moves more slowly
    • this is a consequence of the conservation of angular momentum, since the gravitational force acts along the radius and exerts no torque about the Sun
    • total energy is also constant, so kinetic energy is greatest where gravitational potential energy is most negative. Exam tip: saying the planet speeds up because gravity is stronger without linking to equal areas or angular momentum gains little credit.
32. Concept/explainSL & HLData booklet: Yes

    Discuss the key features of Newton's law of universal gravitation and the evidence supporting it.

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    • Every point mass attracts every other with F = GMm/r², directed along the line joining them
    • the force is always attractive, because mass has only one sign, unlike charge in Coulomb's law
    • the forces on the two bodies form a Newton's third law pair: equal magnitude, opposite direction, however unequal the masses
    • r is measured centre to centre for spherical bodies
    • G is a universal constant, the same everywhere, and is the least precisely known fundamental constant
    • the law unified terrestrial and celestial motion (falling apple and orbiting Moon) and was confirmed in the laboratory by Cavendish's torsion balance, which measured G and hence the mass of the Earth. Exam tip: state that r is centre-to-centre.
33. Worked problemSL & HLData booklet: Yes

    Two uniform lead spheres, each of mass 5.0 kg, have their centres 0.20 m apart. Calculate the gravitational force between them.

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    Principle: F = GMm/r² with r measured centre to centre. Substitution: F = (6.67 × 10⁻¹¹ N m² kg⁻²)(5.0 kg)(5.0 kg)/(0.20 m)². Numerator = 6.67 × 10⁻¹¹ × 25 = 1.67 × 10⁻⁹. Denominator = 0.040 m². F = 4.2 × 10⁻⁸ N, attractive, along the line joining the centres. Check/Trap: the answer is about 10⁻⁸ N, some 10⁹ times smaller than each sphere's weight of 49 N — gravitational attraction between laboratory masses is tiny, which is why Cavendish needed a torsion balance. Do not use the separation of the surfaces, and give the answer to 2 s.f. to match the data.

34. Worked problemSL & HLData booklet: Yes

    The Earth has mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m. Determine the gravitational field strength at its surface and at an altitude of 1.28 × 10⁷ m.

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    Principle: g = GM/r², r from the centre. At the surface: GM = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ = 3.98 × 10¹⁴ N m² kg⁻¹. g = 3.98 × 10¹⁴/(6.37 × 10⁶)² = 3.98 × 10¹⁴/4.06 × 10¹³ = 9.81 N kg⁻¹. At altitude 1.28 × 10⁷ m: r = 6.37 × 10⁶ + 1.28 × 10⁷ = 1.91 × 10⁷ m ≈ 3R. g = 3.98 × 10¹⁴/(1.91 × 10⁷)² = 3.98 × 10¹⁴/3.65 × 10¹⁴ = 1.09 N kg⁻¹ ≈ 1.1 N kg⁻¹. Check/Trap: r = 3R gives g/9 = 1.09 N kg⁻¹, confirming the inverse square. The commonest error is using the altitude alone as r.

35. Worked problemSL & HLData booklet: Yes

    A planet has twice the mass and twice the radius of the Earth. Deduce the gravitational field strength at its surface. (Paper 1 style)

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    Principle: g = GM/r², so g scales as M/r². Ratio: g_planet/g_Earth = (2M)/(2R)² ÷ (M/R²) = 2/4 = 0.50. Therefore g_planet = 0.50 × 9.8 = 4.9 N kg⁻¹. Answer: 4.9 N kg⁻¹, quoted to 2 s.f. Check/Trap: doubling the radius quarters g, doubling the mass only doubles it, so the net effect is a halving — students who forget to square the radius get 1 × g and choose the wrong option. In Paper 1 always work with ratios rather than substituting numerical values; it is faster and avoids arithmetic slips under the 1.5 min per question limit.

36. Worked problemSL & HLData booklet: Yes

    The International Space Station orbits at an altitude of 4.00 × 10⁵ m above the Earth's surface. Calculate its orbital speed and period. (GM_E = 3.98 × 10¹⁴ N m² kg⁻¹, R_E = 6.37 × 10⁶ m)

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    Principle: gravity provides the centripetal force, GMm/r² = mv²/r, so v = √(GM/r). Orbital radius r = 6.37 × 10⁶ + 4.00 × 10⁵ = 6.77 × 10⁶ m. v = √(3.98 × 10¹⁴/6.77 × 10⁶) = √(5.88 × 10⁷) = 7.67 × 10³ m s⁻¹. Period: T = 2πr/v = 2π(6.77 × 10⁶)/7.67 × 10³ = 4.25 × 10⁷/7.67 × 10³ = 5.55 × 10³ s ≈ 92 minutes. Check/Trap: about 7.7 km s⁻¹ and roughly 90 minutes are the standard ISS values — a useful sanity check. Adding the altitude to the radius is essential; using 4.00 × 10⁵ m alone gives an absurd 31 km s⁻¹.

37. Worked problemSL & HLData booklet: No – derive

    Determine the orbital radius and altitude of a geostationary satellite. (T = 8.62 × 10⁴ s, GM_E = 3.98 × 10¹⁴ N m² kg⁻¹, R_E = 6.37 × 10⁶ m)

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    Principle: GMm/r² = 4π²mr/T², so r = [GMT²/(4π²)]^(1/3). Substitution: T² = (8.62 × 10⁴)² = 7.43 × 10⁹ s². GMT² = 3.98 × 10¹⁴ × 7.43 × 10⁹ = 2.96 × 10²⁴. Divide by 4π² = 39.5: 7.49 × 10²² m³. Cube root: r = 4.22 × 10⁷ m. Altitude h = r − R_E = 4.22 × 10⁷ − 6.37 × 10⁶ = 3.59 × 10⁷ m. Check/Trap: r is about 6.6 Earth radii, so g there is 9.8/6.6² ≈ 0.22 N kg⁻¹ — consistent with the slow 24 h orbit. Use the sidereal day 8.62 × 10⁴ s, not 8.64 × 10⁴ s, and never forget to subtract R_E when altitude is asked for.

38. Worked problemSL & HLData booklet: Yes

    Planet X orbits a star at 4.0 times the orbital radius of planet Y. The period of Y is 90 days. Determine the period of X. (Paper 1 style)

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    Principle: Kepler's third law, T² ∝ r³, valid because both planets orbit the same star. Ratio: (T_X/T_Y)² = (r_X/r_Y)³ = 4.0³ = 64. So T_X/T_Y = √64 = 8.0. T_X = 8.0 × 90 = 7.2 × 10² days ≈ 720 days. Check/Trap: the scaling is r^(3/2), so a factor of 4 in radius gives a factor of 8, not 4 or 64, in period. Students who forget the square root quote 5760 days. A quick reality check: Mars is 1.5 AU from the Sun and takes 1.5^1.5 ≈ 1.8 years, which matches the real 687 days.

39. Worked problemSL & HLData booklet: No – derive

    A moon orbits Jupiter in a circular path of radius 4.22 × 10⁸ m with a period of 1.53 × 10⁵ s. Determine the mass of Jupiter.

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    Principle: T² = 4π²r³/(GM), so M = 4π²r³/(GT²). Substitution: r³ = (4.22 × 10⁸)³ = 7.52 × 10²⁵ m³. Numerator = 4π² × 7.52 × 10²⁵ = 39.5 × 7.52 × 10²⁵ = 2.97 × 10²⁷. Denominator = 6.67 × 10⁻¹¹ × (1.53 × 10⁵)² = 6.67 × 10⁻¹¹ × 2.34 × 10¹⁰ = 1.56. M = 2.97 × 10²⁷/1.56 = 1.90 × 10²⁷ kg. Check/Trap: this is about 320 Earth masses, correct for Jupiter. The mass of the orbiting moon cancels and is never needed — students who are told the moon's mass often try to use it. Watch the units: T must be in seconds, not days.

40. Worked problemSL & HLData booklet: Yes

    A 1200 kg satellite is in a circular orbit of radius 8.00 × 10⁶ m around the Earth. Calculate the gravitational force on it and its centripetal acceleration.

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    Principle: F = GMm/r² and a = F/m = g at that radius. F = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(1200)/(8.00 × 10⁶)² = (3.98 × 10¹⁴ × 1200)/(6.40 × 10¹³) = 4.78 × 10¹⁷/6.40 × 10¹³ = 7.47 × 10³ N. Acceleration a = F/m = 7.47 × 10³/1200 = 6.22 m s⁻², directed towards the Earth's centre. Check/Trap: a equals the local field strength g = GM/r² = 6.22 N kg⁻¹, confirming a = g in free fall. Do not use 9.8 m s⁻² for the acceleration of an orbiting body, and remember the acceleration is centripetal even though the speed is constant.

41. Exam technique/trapSL & HL

    In gravitational-field questions, what must r always be measured from, and how do IB questions exploit this?

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    The trap: r in F = GMm/r² and g = GM/r² is the distance from the CENTRE of the body, not from its surface. Questions give an altitude (for example 400 km) and students substitute it directly, or give a radius when the mark scheme wants radius plus altitude. Correct approach: write r = R + h explicitly as a first line of working, with both values in metres, then substitute. If a body's radius is given in km, convert before squaring. For two bodies, r is centre-to-centre separation, so subtracting or adding the radii of both may be needed for a surface-to-surface distance. Exam tip: state the value of r used; a correct r earns method marks even if the arithmetic slips.

42. Exam technique/trapSL & HL

    How should you structure an answer to: explain why a satellite in a stable circular orbit does not fall to Earth?

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    Explain means give reasons with physics, so a bare statement is not enough. The trap: writing that gravity is balanced by centrifugal force, which contradicts Newton's laws and scores zero. Correct approach, three linked points: (1) the satellite IS falling — the gravitational force is unbalanced and provides the centripetal force; (2) this force is perpendicular to the velocity, so it changes direction but not speed; (3) the tangential speed is such that the satellite's curved path matches the curvature of the Earth, so it never gets closer. Command-term guidance: state needs no reason, outline needs a brief one, explain needs a causal chain, deduce requires reaching a conclusion from given data.

43. Exam technique/trapSL & HL

    Describe a practical to determine g using a simple pendulum, including the graph, controlled variables and improvements.

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    Apparatus: bob on inextensible thread from a fixed clamp, metre rule, timer or photogate. Independent variable length l (centre of bob), dependent variable period T; control amplitude (< 10° so SHM holds), bob mass, same location. Method: time 20 oscillations from the centre of the swing and divide, repeat 3 times, take a mean; repeat for 6–8 lengths. Analysis: T = 2π√(l/g) so T² = (4π²/g)l — plot T² against l, gradient = 4π²/g, hence g = 4π²/gradient. Limitations: reaction-time error, air resistance, thread stretching, large-amplitude systematic error. Improvements: photogate at the equilibrium position, longer pendulum, more oscillations. Exam tip: the line should pass through the origin; a non-zero intercept indicates a systematic error in measuring l.

44. Exam technique/trapSL & HL

    Outline a free-fall method for measuring g, and distinguish the random and systematic errors involved.

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    Method: drop a steel ball through two light gates a measured distance s apart, or use an electromagnet release and timer, measuring time t of fall from rest; s = ½gt² so plot s against t² and take gradient = ½g, or plot 2s against t². Random errors: variation in timing and in measuring s — reduce by repeating and averaging, and by taking a best-fit line through many points. Systematic errors: delay in the electromagnet releasing the ball (makes t too large, g too small), gate misalignment, air resistance, zero error in the rule — these shift the line and often produce a non-zero intercept. Exam tip: precision refers to spread of repeats; accuracy refers to closeness to 9.81 m s⁻². Quote g to 3 s.f. with an uncertainty from max/min gradients.

45. Exam technique/trapSL & HL

    How is Kepler's third law tested from data, and how are uncertainties handled in that analysis?

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    Method 1: plot T² against r³ — a straight line through the origin verifies T² ∝ r³, gradient = 4π²/(GM), giving the central mass. Method 2 (log linearisation): from T = kr^n, log T = n log r + log k, so plot log T against log r; a gradient of 1.50 confirms the law and is independent of the units used. Uncertainties: convert fractional uncertainties — Δ(r³)/r³ = 3Δr/r and Δ(T²)/T² = 2ΔT/T — then draw error bars, and find the uncertainty in the gradient from the maximum and minimum gradient lines through the bars, ΔM/M following from the gradient uncertainty. Exam tip: students lose marks by forgetting the factors of 2 and 3 when propagating powers, and by omitting units on axes.

46. DefinitionHL only

    Define gravitational potential energy at a point.

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    The gravitational potential energy of a mass m at a point is the work done by an external agent in bringing that mass from infinity to the point: E_p = −GMm/r. Exam tip: the mark scheme requires "work done in moving the mass FROM INFINITY to the point", and the zero of energy must be stated as being at infinity. E_p is a SCALAR measured in joules and is always NEGATIVE for gravity, because the attractive force does positive work as the mass approaches, so external work is negative. It is a property of the PAIR of masses, not of one of them alone; it increases (becomes less negative) as r increases.

47. DefinitionHL only

    Define gravitational potential at a point.

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    Gravitational potential V_g at a point is the work done per unit mass in bringing a small test mass from infinity to that point: V_g = E_p/m = −GM/r. Exam tip: "per unit mass" and "from infinity" are both required — omitting either loses a mark. Unit: J kg⁻¹ (equivalently m² s⁻²). It is a SCALAR, so potentials from several masses simply add algebraically with no vector components. V_g is always negative, is zero only at infinity, and becomes more negative closer to the mass; its most negative value along a path corresponds to the deepest point of the potential well.

48. DefinitionHL only

    Define an equipotential surface and state how equipotentials relate to field lines.

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    An equipotential surface is a surface on which every point has the same gravitational potential, so no work is done in moving a mass along it. Equipotentials are always PERPENDICULAR to the field lines, and never cross one another. For an isolated spherical mass they are concentric spheres; near a planet's surface they are horizontal parallel planes. Exam tip: draw them for equal INCREMENTS of V_g — the spacing then shows the field strength, since g = −ΔV_g/Δr means closely spaced equipotentials indicate a strong field and widely spaced ones a weak field. Around a sphere the spacing therefore increases with distance.

49. DefinitionHL only

    Define escape speed.

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    Escape speed is the minimum speed at which a body must be projected from the surface of a planet in order to escape completely from its gravitational field, that is to just reach infinity with zero kinetic energy, with no further propulsion. Exam tip: it comes from ½mv² + (−GMm/r) = 0, and is independent of the mass and of the direction of projection (ignoring air resistance and rotation). For Earth v_esc = 11.2 km s⁻¹. Students commonly say "the speed to leave the atmosphere" — that is wrong; escape means reaching infinity. Rockets never actually reach escape speed at launch because they are continuously propelled.

50. DefinitionHL only

    Define gravitational potential difference and state the work done when a mass moves between two points.

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    Gravitational potential difference ΔV_g between two points is the work done per unit mass in moving a small test mass from one point to the other: ΔV_g = V_final − V_initial, in J kg⁻¹. The work done on a mass m is then W = mΔV_g, in joules. Exam tip: this work is independent of the path taken, because gravity is a conservative force — the standard mark-scheme phrase is "depends only on the initial and final positions". Moving away from a planet gives positive ΔV_g and therefore positive external work; moving along an equipotential gives ΔV_g = 0 and hence zero work.

51. EquationHL onlyData booklet: Yes

    State the equation for gravitational potential energy of a two-mass system.

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    E_p = −GMm/r (data booklet). E_p = potential energy of the pair (J, scalar, always negative); G = 6.67 × 10⁻¹¹ N m² kg⁻²; M and m = the two masses (kg); r = separation of centres (m); zero taken at r = ∞. Valid for point or spherical masses. Common misuse: writing E_p = mgh in orbital problems — mgh is only the near-surface approximation valid when Δr ≪ R and g is constant. Sanity check: a 1200 kg satellite at r = 7.0 × 10⁶ m has E_p = −(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 1200)/(7.0 × 10⁶) = −6.8 × 10¹⁰ J.

52. EquationHL onlyData booklet: Yes

    State the equation for gravitational potential due to a point or spherical mass.

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    V_g = −GM/r (data booklet). V_g = gravitational potential (J kg⁻¹, scalar, negative); M = source mass (kg); r = distance from its centre (m). For several masses the resultant potential is the ALGEBRAIC sum ΣV_g — no vectors involved. Related: E_p = mV_g. Valid outside a spherically symmetric body; inside a uniform sphere V_g is a parabola with value −3GM/2R at the centre. Common misuse: dropping the minus sign, which then makes escape-speed and orbital-energy calculations come out with the wrong sign. Sanity check: at Earth's surface V_g = −3.98 × 10¹⁴/6.37 × 10⁶ = −6.25 × 10⁷ J kg⁻¹.

53. EquationHL onlyData booklet: Yes

    State the equation for the work done when a mass is moved in a gravitational field.

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    W = mΔV_g (data booklet). W = work done by an external agent (J); m = mass moved (kg); ΔV_g = V_final − V_initial = gravitational potential difference (J kg⁻¹). Equivalently W = ΔE_p = (−GMm/r₂) − (−GMm/r₁), i.e. W = GMm(1/r₁ − 1/r₂). Valid for any path, because gravity is conservative. Common misuse: using ΔV_g = GM/Δr — the potentials must be evaluated separately at each radius and then subtracted. Sanity check: raising 500 kg from V_g = −6.25 × 10⁷ to −5.90 × 10⁷ J kg⁻¹ needs W = 500 × 3.5 × 10⁶ = 1.75 × 10⁹ J.

54. EquationHL onlyData booklet: Yes

    State the equation relating gravitational field strength to gravitational potential.

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    g = −ΔV_g/Δr (data booklet). g = field strength (N kg⁻¹); ΔV_g = change in potential (J kg⁻¹); Δr = change in distance along the field direction (m). The field is minus the potential GRADIENT: the minus sign shows that g points from high to low potential, i.e. towards the mass, since V_g increases outwards. Valid at every point in any gravitational field. Common misuse: reading g as V_g/r instead of as a gradient — on a V_g–r graph, g is the magnitude of the TANGENT gradient at that r, not the chord. Check: near Earth's surface a 1 m rise raises V_g by 9.8 J kg⁻¹.

55. EquationHL onlyData booklet: Yes

    State the equation for escape speed and show how it is obtained.

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    v_esc = √(2GM/r) (data booklet). v_esc = escape speed (m s⁻¹); M = planet mass (kg); r = launch distance from the centre (m), usually the planetary radius R. Derivation: total energy zero at infinity gives ½mv² − GMm/r = 0. Note v_esc = √2 × v_orbital at the same radius. Valid ignoring air resistance, planetary rotation and other bodies; independent of the escaping mass and of the launch direction. Common misuse: adding the rocket's mass. Sanity check: Earth gives v_esc = √(2 × 3.98 × 10¹⁴/6.37 × 10⁶) = √(1.25 × 10⁸) = 1.12 × 10⁴ m s⁻¹ = 11.2 km s⁻¹.

56. EquationHL onlyData booklet: No – derive

    Derive the total energy of a satellite in a circular orbit and state the relations between its kinetic, potential and total energies.

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    From GMm/r² = mv²/r, E_k = ½mv² = +GMm/2r. With E_p = −GMm/r, the total energy is E = E_k + E_p = −GMm/2r (derived, not printed). Units: joules. So E_p = −2E_k and E = −E_k = E_p/2. Valid for circular orbits only. To move to a HIGHER orbit the total energy must INCREASE (become less negative): ΔE = GMm/2 × (1/r₁ − 1/r₂). Common misuse: assuming a higher orbit means a faster satellite — v = √(GM/r) falls, yet the energy required rises because E_p increases more than E_k decreases. This also explains orbital decay: drag lowers E, so r falls and v rises.

57. Graph/diagramHL only

    Sketch and describe the graph of gravitational potential V_g against distance r for a planet.

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    Axes: V_g/J kg⁻¹ (vertical, entirely negative) against r/m (horizontal). Outside the planet (r ≥ R) the curve is −GM/r: it starts at its most negative value −GM/R at the surface and rises steeply at first, then flattens, approaching zero asymptotically as r → ∞ — a "potential well". The magnitude of the gradient at any point equals g, since g = −ΔV_g/Δr, so draw a tangent to obtain g. Inside the planet the curve becomes a shallow parabola reaching −3GM/2R at the centre, with zero gradient there because g = 0. Exam tip: the curve must never cross the r-axis and must not touch it at any finite r.

58. Graph/diagramHL only

    Describe how to draw and interpret equipotential surfaces around a spherical planet.

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    Draw concentric circles (spheres in 3-D) centred on the planet, labelled with equal decrements of V_g, for example −10, −20, −30 MJ kg⁻¹. Because V_g = −GM/r, equal potential steps occur at radii that get progressively FURTHER apart as r increases, showing that g weakens. Field lines are drawn radially inwards, crossing every equipotential at 90°. Extract g from the diagram using g ≈ ΔV_g/Δr between adjacent surfaces, and the work to move mass m between two surfaces from W = mΔV_g. Exam tip: no work is done moving along an equipotential; a common error is drawing equally spaced circles, which would represent a uniform field.

59. Graph/diagramHL only

    Describe the graph of gravitational potential along the line joining the Earth and the Moon and how to locate the neutral point.

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    Axes: V_g/J kg⁻¹ (negative throughout) against distance d/m from the Earth's centre. The curve plunges steeply near each body (V_g → −∞ at each centre) and rises to a shallow MAXIMUM between them, since potentials add algebraically: V_g = −GM_E/d − GM_M/(D − d). The maximum is the neutral point, where the two field strengths are equal and opposite so the resultant g is zero and the gradient of the graph is zero. Setting GM_E/d² = GM_M/(D − d)² gives d/(D − d) = √(M_E/M_M) ≈ 9.0, so d ≈ 3.46 × 10⁸ m — about 90 % of the way to the Moon. Exam tip: V_g is a maximum there, NOT zero.

60. Concept/explainHL onlyData booklet: Yes

    Explain why gravitational potential is always negative and is defined as zero at infinity.

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    • Gravitational potential V_g at a point is the work done per unit mass in bringing a small test mass from infinity to that point
    • the zero is chosen at infinity because that is where the gravitational force between masses vanishes, giving a universal reference
    • gravity is attractive, so the field does positive work as the mass approaches; an external agent does negative work
    • hence V_g = −GM/r is negative everywhere and becomes more negative as r decreases
    • potential is a scalar, unit J kg⁻¹, so potentials due to several masses add algebraically with no vector components
    • V_g → 0 as r → ∞, and E_p = mV_g = −GMm/r. Exam tip: dropping the minus sign, or calling potential a vector, are the standard mark losses.
61. Concept/explainHL onlyData booklet: Yes

    Explain what is meant by escape speed and why it does not depend on the mass or launch direction of the projectile.

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    • Escape speed is the minimum speed at which a body must be projected from a planet's surface, with no further propulsion, to reach infinity with zero kinetic energy
    • energy conservation: ½mv² − GMm/r = 0, so v_esc = √(2GM/r)
    • the projectile mass m cancels, so escape speed is the same for a pebble and a spacecraft
    • gravitational potential energy depends only on distance from the centre, not on direction, so ignoring air resistance and rotation any launch direction that avoids the surface requires the same speed
    • for the Earth v_esc ≈ 11.2 km s⁻¹, for the Moon 2.4 km s⁻¹ — which explains why the Moon retains no atmosphere. Exam tip: a rocket with continuous thrust never needs to reach escape speed.
62. Concept/explainHL onlyData booklet: No – derive

    Explain why the total energy of a satellite in a circular orbit is E = −GMm/(2r) and why atmospheric drag makes a satellite speed up.

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    • For a circular orbit GMm/r² = mv²/r gives E_k = ½mv² = GMm/(2r)
    • potential energy is E_p = −GMm/r, so E = E_k + E_p = −GMm/(2r), negative, showing the satellite is bound
    • note E_p = −2E_k and E = −E_k
    • drag does negative work, so the total energy E becomes more negative and r decreases
    • but E_k = GMm/(2r) increases as r falls, so the orbital speed rises even though energy is being lost
    • the loss in potential energy is twice the gain in kinetic energy, the difference being dissipated as heat
    • this is the satellite paradox, ending in re-entry and burn-up. Exam tip: students wrongly argue that friction must slow the satellite.
63. Concept/explainHL onlyData booklet: Yes

    Explain the difference between the point of zero gravitational field and the potential at that point on the line between the Earth and the Moon.

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    • Field strength is a vector: at the null point the fields of Earth and Moon are equal in magnitude and opposite in direction, so g_total = 0, giving GM_E/d² = GM_M/(D − d)²
    • this occurs about 3.5 × 10⁸ m from the Earth's centre, roughly 90% of the way to the Moon, because the Earth is about 81 times more massive
    • potential is a scalar and both contributions are negative, so V_g = −GM_E/d − GM_M/(D − d) is a negative minimum in magnitude but never zero
    • the potential graph has a maximum (least negative point) exactly where the field is zero, since g = −dV_g/dr
    • a spacecraft passing this point coasts downhill to the Moon. Exam tip: potential is not zero where the field is zero.
64. Worked problemHL onlyData booklet: Yes

    Calculate the gravitational potential at the Earth's surface and the energy needed to raise a 2000 kg satellite to an altitude of 1.00 × 10⁶ m. (GM_E = 3.98 × 10¹⁴, R_E = 6.37 × 10⁶ m)

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    Principle: V_g = −GM/r and ΔE_p = mΔV_g. At the surface: V_g = −3.98 × 10¹⁴/6.37 × 10⁶ = −6.25 × 10⁷ J kg⁻¹. At r = 7.37 × 10⁶ m: V_g = −3.98 × 10¹⁴/7.37 × 10⁶ = −5.40 × 10⁷ J kg⁻¹. ΔV_g = −5.40 × 10⁷ − (−6.25 × 10⁷) = +8.5 × 10⁶ J kg⁻¹. ΔE_p = mΔV_g = 2000 × 8.5 × 10⁶ = 1.7 × 10¹⁰ J. Check/Trap: mgh gives 2000 × 9.8 × 10⁶ = 1.96 × 10¹⁰ J, about 15% too large, because g falls with height — mgh is only valid when h ≪ R. Keep the signs: the change in potential is positive even though both potentials are negative.

65. Worked problemHL onlyData booklet: Yes

    Determine the escape speed from the surface of the Moon. (M = 7.35 × 10²² kg, R = 1.74 × 10⁶ m) Comment on why the Moon has no atmosphere.

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    Principle: ½mv² = GMm/r at the surface gives v_esc = √(2GM/r). Substitution: GM = 6.67 × 10⁻¹¹ × 7.35 × 10²² = 4.90 × 10¹² N m² kg⁻¹. 2GM/R = 9.81 × 10¹²/1.74 × 10⁶ = 5.64 × 10⁶ m² s⁻². v_esc = √(5.64 × 10⁶) = 2.37 × 10³ m s⁻¹ ≈ 2.4 km s⁻¹. Comment: typical thermal speeds of gas molecules (v_rms = √(3k_BT/m), a few hundred m s⁻¹ to over 1 km s⁻¹ for light gases) are a significant fraction of this, so over geological time molecules in the high-speed tail escape. Check/Trap: Earth's value is 11.2 km s⁻¹, about 4.7 times larger — factor of 2 inside the root is easily dropped.

66. Worked problemHL onlyData booklet: No – derive

    A 500 kg satellite is moved from a circular orbit of radius 7.00 × 10⁶ m to one of radius 1.40 × 10⁷ m. Determine the energy that must be supplied. (GM_E = 3.98 × 10¹⁴)

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    Principle: total orbital energy E = −GMm/(2r). Initial: E₁ = −(3.98 × 10¹⁴ × 500)/(2 × 7.00 × 10⁶) = −1.99 × 10¹⁷/1.40 × 10⁷ = −1.42 × 10¹⁰ J. Final: E₂ = −1.99 × 10¹⁷/(2.80 × 10⁷) = −7.11 × 10⁹ J. Energy supplied ΔE = E₂ − E₁ = −7.11 × 10⁹ + 1.42 × 10¹⁰ = 7.1 × 10⁹ J. Check/Trap: doubling r halves the magnitude of E, so ΔE = |E₁|/2 — a fast route. Note the satellite ends up SLOWER (v = √(GM/r) falls from 7.5 to 5.3 km s⁻¹) despite energy being added; using only ΔE_p would overestimate the answer by a factor of 2.

67. Worked problemHL onlyData booklet: Yes

    An object is released from rest at a distance of 3R from the Earth's centre. Determine its speed as it reaches the surface, ignoring air resistance. (R = 6.37 × 10⁶ m, GM_E = 3.98 × 10¹⁴)

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    Principle: conservation of energy, ½mv² = mΔV_g, with m cancelling. ΔV_g = GM(1/R − 1/3R) = GM(2/3R) = (2 × 3.98 × 10¹⁴)/(3 × 6.37 × 10⁶) = 7.96 × 10¹⁴/1.91 × 10⁷ = 4.17 × 10⁷ J kg⁻¹. So ½v² = 4.17 × 10⁷, v² = 8.34 × 10⁷ m² s⁻², v = 9.13 × 10³ m s⁻¹ ≈ 9.1 km s⁻¹. Check/Trap: this must be less than the escape speed of 11.2 km s⁻¹ (the value obtained from infinity) — it is. Using v = √(2gh) with a constant g = 9.8 and h = 2R gives 1.6 × 10⁴ m s⁻¹, far too large, because g is much weaker over most of the fall.

68. Worked problemHL onlyData booklet: Yes

    The Earth (5.97 × 10²⁴ kg) and Moon (7.35 × 10²² kg) have centres 3.84 × 10⁸ m apart. Determine the distance from the Earth's centre at which the resultant gravitational field is zero.

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    Principle: fields are vectors and must be equal and opposite: GM_E/d² = GM_M/(D − d)². Take the square root: (D − d)/d = √(M_M/M_E) = √(7.35 × 10²²/5.97 × 10²⁴) = √(1.231 × 10⁻²) = 0.111. So D/d = 1 + 0.111 = 1.111, giving d = 3.84 × 10⁸/1.111 = 3.46 × 10⁸ m from the Earth's centre. Check/Trap: the point is about 90% of the way to the Moon, closer to the smaller mass, as expected. The gravitational potential there is NOT zero: V_g = −(3.98 × 10¹⁴/3.46 × 10⁸) − (4.90 × 10¹²/3.8 × 10⁷) ≈ −1.28 × 10⁶ J kg⁻¹. Remember to take the square root before rearranging.

69. Exam technique/trapHL only

    What sign errors do students make with gravitational potential and potential energy, and how are they avoided?

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    The trap: treating V_g = −GM/r as a magnitude and then adding a second minus sign, or writing ΔE_p = E_p(start) − E_p(final). Why it happens: negative quantities that increase towards zero feel counter-intuitive. Correct approach: always substitute values complete with their signs, use Δ = final − initial, and check that moving further from a mass increases the potential (towards zero). Work done by an external agent is W = mΔV_g and is positive when a mass is raised. For the work done BY the gravitational field the sign reverses. Exam tip: the mark scheme awards the mark for a numerically correct magnitude only if the direction of energy transfer is stated. Quote potential in J kg⁻¹ and potential energy in J — mixing them is a common unit error.

70. Exam technique/trapHL only

    How are V_g against r and g against r graphs for a planet read, and what are the common traps?

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    Outside the planet V_g = −GM/r is a negative curve rising asymptotically towards zero, while g = GM/r² is a positive curve falling more steeply; inside a uniform planet g falls linearly to zero at the centre and V_g reaches its most negative constant-curvature minimum. Key relationships: the gradient of the V_g–r graph equals −g, so where V_g is stationary the field is zero; the area under a g–r graph between two radii gives the magnitude of ΔV_g, and multiplying by m gives the energy required. Traps: confusing the 1/r and 1/r² shapes, reading the potential graph as if it crossed zero at the surface, and forgetting that the area gives potential difference, not force.

71. Exam technique/trapHL only

    What are the classic examination traps involving escape speed and orbital speed?

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    Trap 1: writing v_esc = √(GM/r), omitting the factor 2 — escape speed is √2 times the orbital speed at the same radius (11.2 vs 7.9 km s⁻¹ at the Earth's surface). Trap 2: using the altitude instead of r = R + h. Trap 3: claiming a rocket must reach 11.2 km s⁻¹ to leave the Earth; with continuous thrust any speed suffices, since escape speed applies only to unpowered projectiles. Trap 4: including the projectile's mass, which always cancels. Trap 5: forgetting that the derivation neglects air resistance and the Earth's rotation. Command-term note: suggest invites a plausible physical reason (for example, why the Moon has no atmosphere), while determine demands a numerical answer with working, units and correct significant figures.

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