5.7 Gravitational fields and orbits
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A2 content for Unit 5 (WPH15), specification points 154–160. Centripetal force (4.2) and electric fields (4.3) are used here, not re-derived; stars and distances are on the next page (5.8).
What this page covers
6 points
- What a gravitational field is, and gravitational field strength \(g=F/m\)
- Newton's law of gravitation, \(F=Gm_1m_2/r^2\), and the inverse square law
- Deriving and using \(g=Gm/r^2\) for a radial field
- Gravitational potential \(V_{grav}=-Gm/r\), equipotentials, and the work done moving a mass between two points
- Similarities and differences between gravitational and electric fields
- Circular orbits: orbital speed, period, geostationary satellites and the derivation of \(T^2\propto r^3\)
In one minute
6 points
- A gravitational field is a region where a mass experiences a force, and its strength \(g=F/m\) is the force per unit mass, in N kg⁻¹.
- Two point masses attract with \(F=Gm_1m_2/r^2\), where \(r\) is the distance between their centres: double \(r\) and the force falls to one quarter.
- Around a planet of mass \(M\), \(g=GM/r^2\) (a radial field); close to the surface the field is almost uniform.
- Potential \(V_{grav}=-GM/r\) is zero at infinity and negative everywhere else. The work to move a mass \(m\) is \(m\Delta V\); no work is done along an equipotential.
- Gravity and the electric force are both inverse square laws; gravity only attracts and is far weaker between particles.
- In a circular orbit gravity provides the centripetal force: \(GMm/r^2=mv^2/r\), so \(v=\sqrt{GM/r}\) and \(T^2=(4\pi^2/GM)\,r^3\).
Contents
12 sections
- What is a gravitational field?
- What is Newton's law of gravitation?
- How does g change with distance from a planet?
- What is gravitational potential?
- How do gravitational and electric fields compare?
- How do Newton's laws explain orbits?
- Exam tips for 5.7
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is a gravitational field?
This whole topic grows from one force law. Before meeting it, we need a way to describe what a mass does to the space around it.
- Field strength is a vector: its direction is the direction of the force on a mass. 1 N kg⁻¹ is the same as 1 m s⁻², so \(g\) is also the acceleration of a freely falling object.
- The field is drawn with field lines. The arrow shows the direction of the force on a mass; the closer the lines, the stronger the field.
- Around a spherical planet the field is radial: the lines point in towards the centre and spread out with distance, so the field weakens.
- Over a region that is small compared with the planet, the lines are almost parallel and equally spaced: the field is uniform. That is why \(g=9.81\ \text{N kg}^{-1}\) can be used everywhere in a laboratory.
- \(g=F/m\) defines the field at a point whatever is producing it; the test mass must be small enough not to disturb that field.
A lander of mass 420 kg rests on the surface of Titan, Saturn's largest moon. The gravitational force on it is 567 N. (a) Calculate the gravitational field strength at Titan's surface. (b) An astronaut with a spacesuit has a total mass of 115 kg. Calculate the gravitational force on the astronaut on Titan, and compare it with the force on Earth (\(g=9.81\ \text{N kg}^{-1}\)).
(a) \(g=F/m=567\div420=\) 1.35 N kg⁻¹
(b) Same field, different mass: \(F=mg=115\times1.35=\) 155 N. On Earth, \(115\times9.81=1130\ \text{N}\), so the force on Titan is about 0.14 of the Earth value.
What is Newton's law of gravitation?
Newton's insight was that the force that makes an apple fall also holds the Moon in orbit, and that it weakens with distance in a precise way.
\[F=\frac{Gm_1m_2}{r^2}\]\(F\) is the force each mass exerts on the other (N), \(m_1\) and \(m_2\) are the two masses (kg), \(r\) is the distance between their centres (m) and \(G\) is the gravitational constant, \(6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}\). The law is in the formulae list. It is valid for point masses, and for uniform spheres if \(r\) is measured between their centres. The force is always attractive.
- The force is proportional to each mass and to \(1/r^2\): an inverse square law. Doubling \(r\) divides the force by 4; tripling it divides it by 9.
- The forces on the two masses form a Newton's third law pair: equal in size and opposite in direction, even if one is a planet and the other a pebble. The pebble accelerates far more because its mass is so much smaller.
- A uniform sphere attracts as if all its mass were at its centre. So for a satellite, \(r\) = radius of the planet + height above the surface.
- \(G\) is tiny, so gravitational forces between everyday objects are minute; gravity only becomes important when at least one mass is astronomical.
In a laboratory, a lead sphere of mass 12.0 kg and a small lead sphere of mass 0.800 kg are placed with their centres 0.150 m apart. (a) Calculate the gravitational force between them. (b) Compare it with the weight of the small sphere. (c) Calculate the gravitational field strength due to the large sphere at the centre of the small one.
(a) \(F=\dfrac{6.67\times10^{-11}\times12.0\times0.800}{0.150^2}=\) 2.85 × 10⁻⁸ N
(b) \(W=0.800\times9.81=7.85\ \text{N}\), so \(F/W=\) 3.6 × 10⁻⁹, far too small to notice without very sensitive apparatus.
(c) \(g=F/m=2.85\times10^{-8}\div0.800=\) 3.56 × 10⁻⁸ N kg⁻¹ (the same as \(GM/r^2\)).
How does g change with distance from a planet?
Place a small test mass \(m\) at distance \(r\) from the centre of a planet of mass \(M\). By Newton's law the force on it is \(F=GMm/r^2\). By definition \(g=F/m\), so the test mass cancels:
\[g=\frac{F}{m}=\frac{GMm/r^2}{m}\quad\Rightarrow\quad g=\frac{Gm}{r^2}\](In the formulae list the producing mass is written \(m\); here it is \(M\) to keep it apart from the test mass.) The equation is in the formulae list, and you must also be able to derive it in two lines, as above. It applies outside a spherical mass (\(r\) ≥ its radius), and the field points towards the centre.
- \(g\) does not depend on the mass placed in the field: a feather and a hammer fall with the same acceleration in a vacuum.
- Outside the planet, \(g\propto1/r^2\). The graph is a steep curve that never reaches zero.
- Rearranged, \(M=gr^2/G\): measuring \(g\) at the surface and knowing the radius gives the planet's mass. For the Earth, \(9.81\times(6.37\times10^6)^2\div6.67\times10^{-11}=5.97\times10^{24}\ \text{kg}\).
A space station orbits 420 km above the Earth's surface. The Earth has mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m. (a) Calculate the gravitational field strength at the station. (b) Express it as a percentage of the surface value, and explain why astronauts inside still seem weightless.
(a) Distance from the centre, not from the surface: \(r=6.37\times10^6+0.42\times10^6=6.79\times10^6\ \text{m}\). Then \(g=\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{(6.79\times10^6)^2}=\) 8.64 N kg⁻¹
(b) \(8.64\div9.81=\) 88% of the surface value. Gravity is still strong; the astronauts and the station are both in free fall with acceleration \(g\), so nothing presses them against the floor.
Using the height above the surface as \(r\). With \(r=4.2\times10^5\ \text{m}\) the answer would be over 2000 N kg⁻¹, a sign that something is wrong. Always add the planet's radius.
What is gravitational potential?
Field strength tells you the force. Potential tells you about energy: how much work it takes to move a mass around in the field.
- Why negative? Zero is defined at infinity. Gravity attracts, so a mass moving in from infinity has work done on it by the field, and the work done by an external agent is negative. Every point near a planet is therefore at a potential below zero.
- As \(r\) increases, \(V\) increases (becomes less negative) towards zero. Moving away from a planet means moving to a higher potential.
- Work done moving a mass \(m\) between two points: \(W=m\Delta V=m(V_{\text{final}}-V_{\text{initial}})\). This is the change in gravitational potential energy, and it depends only on the start and end points, not on the path.
- Near the surface, where \(g\) is almost constant, \(m\Delta V\) reduces to the familiar \(mg\Delta h\). Over large distances you must use \(m\Delta V\), because \(g\) falls with height.
- The field strength is the potential gradient: \(g=-\Delta V/\Delta r\). The tangent to a \(V\)–\(r\) graph gives the size of \(g\); the minus sign means the field points towards decreasing potential.
Equipotentials
- An equipotential is a surface on which every point has the same potential. Around a spherical planet they are concentric spheres (circles in a diagram).
- \(\Delta V=0\) along an equipotential, so no work is done moving a mass along it. A satellite in a circular orbit stays on one equipotential.
- Field lines always cross equipotentials at right angles.
- Drawn at equal steps of potential, equipotentials get further apart with distance, because the field (the potential gradient) gets weaker. In a uniform field they are equally spaced, parallel planes.
Mars has mass 6.42 × 10²³ kg and radius 3.39 × 10⁶ m. A probe of mass 250 kg is lifted from the surface to a point 3.39 × 10⁶ m above the surface. (a) Calculate the gravitational potential at the surface and at the final position. (b) Calculate the minimum work done on the probe. (c) Explain why using \(W=mg\Delta h\) with the surface value of \(g\) gives too large an answer.
(a) The final point is at \(r=2R\). \(V_{\text{surface}}=-\dfrac{6.67\times10^{-11}\times6.42\times10^{23}}{3.39\times10^6}=\) −1.263 × 10⁷ J kg⁻¹; \(V_{2R}=-1.263\times10^7\div2=\) −6.32 × 10⁶ J kg⁻¹
(b) \(W=m\Delta V=250\times\left(-6.32\times10^6-(-1.263\times10^7)\right)=250\times6.32\times10^6=\) 1.58 × 10⁹ J
(c) \(g_{\text{surface}}=GM/R^2=3.73\ \text{N kg}^{-1}\), so \(mg\Delta h=250\times3.73\times3.39\times10^6=3.16\times10^9\ \text{J}\), twice the true value, because \(g\) falls to one quarter of its surface value at 2R. Over large heights, use potential.
Taking the probe all the way to infinity (\(V=0\)) needs \(250\times1.263\times10^7=3.16\times10^9\ \text{J}\). A probe launched from the surface with at least this much kinetic energy would never fall back: the idea of escape from a planet.
Write both potentials with their signs, then subtract: \(\Delta V=V_{\text{final}}-V_{\text{initial}}\). Moving away from a planet always gives a positive \(\Delta V\) and positive work done on the mass; if your answer is negative, check the order of subtraction.
How do gravitational and electric fields compare?
The field of a point mass and the field of a point charge have the same mathematical shape. Learn the pairs side by side:
| Feature | Gravitational field (point mass \(m\)) | Electric field (point charge \(Q\)) |
|---|---|---|
| Acts on | mass | charge |
| Field strength defined as | \(g=F/m\) (N kg⁻¹) | \(E=F/Q\) (N C⁻¹) |
| Force law | \(F=Gm_1m_2/r^2\) | \(F=Q_1Q_2/4\pi\varepsilon_0r^2\) |
| Radial field strength | \(g=Gm/r^2\) | \(E=Q/4\pi\varepsilon_0r^2\) |
| Potential | \(V=-Gm/r\) | \(V=Q/4\pi\varepsilon_0r\) |
| Dependence on distance | inverse square (force, field); \(1/r\) (potential) | inverse square (force, field); \(1/r\) (potential) |
| Attract or repel? | always attracts (mass is never negative) | like charges repel, unlike charges attract |
| Sign of potential | always negative | positive near \(+Q\), negative near \(-Q\) |
| Strength | very weak (\(G=6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}\)) | very strong (\(k=8.99\times10^9\ \text{N m}^2\,\text{C}^{-2}\)) |
| Can it be shielded or cancelled? | no, there is no negative mass | yes, positive and negative charges cancel |
Between individual particles the electric force is overwhelmingly stronger. Yet gravity dominates on the scale of planets and galaxies, because large bodies are almost exactly electrically neutral (their positive and negative charges cancel) while every bit of mass adds to the gravitational pull.
In an atom, an electron and a proton are 1.0 × 10⁻¹⁰ m apart. Calculate the electric force and the gravitational force between them, and the ratio of the two.
Electric: \(F_E=\dfrac{8.99\times10^9\times(1.60\times10^{-19})^2}{(1.0\times10^{-10})^2}=\) 2.30 × 10⁻⁸ N
Gravitational: \(F_G=\dfrac{6.67\times10^{-11}\times9.11\times10^{-31}\times1.67\times10^{-27}}{(1.0\times10^{-10})^2}=\) 1.01 × 10⁻⁴⁷ N
Ratio: \(F_E/F_G=\) 2.27 × 10³⁹. Both forces follow \(1/r^2\), so this ratio is the same at any separation: gravity is negligible inside atoms.
How do Newton's laws explain orbits?
A satellite in a circular orbit moves at constant speed, but its velocity changes direction continuously, so it is accelerating towards the centre. By Newton's second law there must be a resultant force towards the centre: the centripetal force. The only force on it is gravity, which points towards the centre of the planet, at right angles to the velocity. Gravity therefore does no work, and the speed stays constant. Setting the gravitational force equal to the centripetal force gives everything else:
\[\frac{GMm}{r^2}=\frac{mv^2}{r}\ \Rightarrow\ v=\sqrt{\frac{GM}{r}}\qquad v=\frac{2\pi r}{T}\ \Rightarrow\ \frac{4\pi^2r^2}{T^2}=\frac{GM}{r}\ \Rightarrow\ T^2=\frac{4\pi^2}{GM}\,r^3\]\(M\) is the mass of the central body, \(m\) the mass of the satellite, \(r\) the orbital radius from the centre, \(v\) the orbital speed and \(T\) the period. These results are not in the formulae list: derive them from \(F=Gm_1m_2/r^2\) and \(F=mv^2/r\), both of which are in the formulae list.
- The satellite's mass \(m\) cancels: every object at the same orbital radius has the same speed and period. Since \(4\pi^2/GM\) is a constant for one central body, \(T^2\propto r^3\).
- Further out means slower and longer: \(v\propto1/\sqrt r\) and \(T\propto r^{3/2}\). A satellite at 4 times the radius moves at half the speed and takes 8 times as long to go round.
- Rearranged, \(M=4\pi^2r^3/GT^2\): timing any moon or satellite gives the mass of the body it orbits.
Geostationary orbits
A geostationary satellite stays above the same point on the Earth's surface, so a fixed dish can point at it. It must:
- have a period equal to the Earth's rotation period (about 24 hours);
- orbit above the equator (its orbit must be centred on the Earth's centre and lie in the plane in which the surface rotates);
- move in the same direction as the Earth rotates (west to east).
Because \(T\) fixes \(r\) through \(T^2=(4\pi^2/GM)\,r^3\), there is only one possible radius for a geostationary orbit.
The Earth rotates once every 8.62 × 10⁴ s. Its mass is 5.97 × 10²⁴ kg and its radius is 6.37 × 10⁶ m. (a) Show that the radius of a geostationary orbit is about 4.2 × 10⁷ m. (b) Calculate the height of the orbit above the surface and the orbital speed.
(a) The gravitational force provides the centripetal force, so \(T^2=4\pi^2r^3/GM\). Rearranged: \(r^3=\dfrac{GMT^2}{4\pi^2}=\dfrac{6.67\times10^{-11}\times5.97\times10^{24}\times(8.62\times10^4)^2}{4\pi^2}=7.49\times10^{22}\ \text{m}^3\), so \(r=\) 4.22 × 10⁷ m ≈ 4.2 × 10⁷ m
(b) \(h=4.22\times10^7-6.37\times10^6=\) 3.58 × 10⁷ m; \(v=2\pi\times4.22\times10^7\div8.62\times10^4=\) 3.07 × 10³ m s⁻¹. Check: \(\sqrt{GM/r}\) gives the same 3.07 × 10³ m s⁻¹.
Testing T² ∝ r³ with real orbits
If \(T=kr^n\), then \(\lg T=n\lg r+\lg k\), so a graph of \(\lg T\) against \(\lg r\) is a straight line of gradient \(n\). For moons of one planet the gradient should be 1.5, confirming \(T^2\propto r^3\). Equally, a graph of \(T^2\) against \(r^3\) is a straight line through the origin of gradient \(4\pi^2/GM\).
Europa orbits Jupiter in a circle of radius 6.71 × 10⁸ m with a period of 3.07 × 10⁵ s (3.55 days). Calculate the mass of Jupiter.
Gravity provides the centripetal force, so \(T^2=4\pi^2r^3/GM\); make \(M\) the subject: \(M=\dfrac{4\pi^2r^3}{GT^2}\).
Substitute, with the period in seconds: \(M=\dfrac{4\pi^2\times3.02\times10^{26}}{6.67\times10^{-11}\times9.42\times10^{10}}=\) 1.90 × 10²⁷ kg
The mass of Europa does not appear: it cancelled in the derivation. Any of the other three moons gives the same answer.
Leaving the period in hours or days. \(G\) is in SI units, so \(T\) must be in seconds and \(r\) in metres. A period of 3.55 days left unconverted would make the mass about 7 × 10⁹ times too large (86 400²).
Exam tips for 5.7
In every equation on this page, \(r\) is the distance from the centre of the planet or star. For a satellite, add the planet's radius to the height above the surface before substituting.
If the distance is multiplied by \(k\): force and field strength are divided by \(k^2\), potential by \(k\), orbital speed by \(\sqrt k\), and the period is multiplied by \(k^{3/2}\). Ratios are quicker and avoid rounding errors.
For "show that" or "derive" questions on orbits, start with the physics, not the answer: "the gravitational force provides the centripetal force", then write \(GMm/r^2=mv^2/r\), substitute \(v=2\pi r/T\), and rearrange step by step.
Gravitational potentials are negative. Write them with their signs, find \(\Delta V\) as final minus initial, then multiply by the mass. The work done in moving away from a planet is always positive.
When asked to compare the two fields, give matching pairs: both inverse square, both have field strength = force per unit (mass or charge); gravity only attracts while electric forces can attract or repel; gravity is far weaker between particles.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Astronauts in orbit float because there is no gravity in space. | At a low orbit \(g\) is still about 90% of its surface value. | The astronauts and their craft are in free fall together, so there is no contact force. |
| A heavier satellite needs to go faster to stay in the same orbit. | The satellite's mass cancels in \(GMm/r^2=mv^2/r\). | Orbital speed and period depend only on \(M\) and \(r\). |
| Gravitational potential is zero at the Earth's surface. | That is a convenient choice for \(mg\Delta h\), not the definition used for radial fields. | \(V=0\) at infinity, so \(V\) is negative everywhere near a planet. |
| Field strength halves when the distance doubles. | The field follows an inverse square law. | Doubling \(r\) gives one quarter of \(g\); it is the potential that halves. |
| An orbiting satellite needs a forward force to keep it moving. | No force is needed to maintain speed (Newton's first law). | Only a force towards the centre is needed, to change the direction of the velocity. |
| A geostationary satellite can hover above any city. | Its orbit must be centred on the Earth's centre and match the surface's rotation. | It must orbit above the equator, west to east, at one particular radius. |
How this topic connects
4.2 Circular motion and 4.3 Electric fields
\(F=mv^2/r\) and \(v=\omega r\) are what turn the law of gravitation into orbital speed and period. Coulomb's law, \(E=Q/4\pi\varepsilon_0r^2\) and electric potential have exactly the same structure as the equations here; learn the two topics as a pair.
Work and energy (Unit 1)
\(mg\Delta h\) is the special case of \(m\Delta V\) for a uniform field; this page explains when that approximation fails.
5.8 Stars and 5.9 The universe
Gravity holds stars and galaxies together; orbital speeds of stars in galaxies give evidence for dark matter, and the fate of the universe depends on its mean density.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(g=F/m\) | defines gravitational field strength (N kg⁻¹) | yes |
| \(F=Gm_1m_2/r^2\) | force between two point masses or spheres; \(G=6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}\) | yes |
| \(g=Gm/r^2\) | field strength in a radial field | yes (and derive it from the two above) |
| \(V_{grav}=-Gm/r\) | potential in a radial field (J kg⁻¹) | yes |
| \(W=m\Delta V\) | work done moving a mass between two points | no — learn it |
| \(g=-\Delta V/\Delta r\) | field strength from the gradient of a \(V\)–\(r\) graph | no — learn it |
| \(v=\sqrt{GM/r}\) | speed in a circular orbit (m s⁻¹) | no — derive it from \(GMm/r^2=mv^2/r\) |
| \(T^2=(4\pi^2/GM)\,r^3\) | period and radius of a circular orbit; mass of the central body | no — derive it, using \(v=2\pi r/T\) |
Check your understanding
1. Two asteroids attract each other with a force of 800 N. What is the force when the distance between their centres is doubled?
200 N. The force is proportional to \(1/r^2\), so doubling \(r\) divides it by 4.
2. Explain why gravitational potential is negative near a planet.
Potential is zero at infinity. Because gravity attracts, the field does work on a mass as it moves in from infinity, so the work done per unit mass by an external agent is negative; every nearby point is below zero.
3. At what height above the Earth's surface is the field strength one quarter of its surface value? (Earth radius 6.37 × 10⁶ m)
\(g\propto1/r^2\), so one quarter needs \(r=2R\). The height is \(2R-R=\) 6.37 × 10⁶ m.
4. Satellite B orbits the Earth at four times the orbital radius of satellite A. How do their periods compare?
\(T^2\propto r^3\), so \(T\propto r^{3/2}\): \(4^{3/2}=8\). Satellite B's period is 8 times A's.
5. A satellite moves in a circular orbit. How much work does gravity do on it during one orbit, and why?
None. The orbit lies on an equipotential (\(\Delta V=0\)), and the gravitational force is always at right angles to the velocity, so the speed and kinetic energy stay constant.
Specification checklist
Numbers refer to the specification points.
- (154) I can describe a gravitational field as a region where a mass feels a force, and draw field lines for a radial field and for a uniform field near a surface.
- (155) I can use \(g=F/m\) and state that field strength is the force per unit mass, a vector in N kg⁻¹.
- (156) I can use \(F=Gm_1m_2/r^2\), measuring \(r\) between centres, and explain the inverse square law.
- (157) I can derive \(g=Gm/r^2\) from the law of gravitation and use it, including to find the mass of a planet.
- (158) I can use \(V_{grav}=-Gm/r\), explain why it is negative, calculate the work done moving a mass between two points, and describe equipotentials.
- (159) I can compare gravitational and electric fields, giving similarities and differences.
- (160) I can use Newton's laws of motion and gravitation for circular orbits: derive \(v=\sqrt{GM/r}\) and \(T^2\propto r^3\), calculate orbital speeds, periods and masses, and explain the conditions for a geostationary orbit.
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