D.1 Gravitational fields | IB DP Physics

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The old site split this across Unit 6 and Unit 10. It is now one topic, with the HL material sitting inside it rather than on a separate page — which is closer to how it is actually examined.

🎯What you need to be able to do

Apply Newton’s law of gravitation.
Use gravitational field strength, and sketch field lines around masses.
Analyse circular orbits, and derive Kepler’s third law.
HLUse gravitational potential and gravitational potential energy, including their signs.
HLCalculate escape speed and the total energy of an orbit; interpret equipotential surfaces.

📚The physics

Newton’s law of gravitation:

\[ F = \frac{GMm}{r^{2}} \qquad G = 6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2} \]

Two points to be careful about. The \(r\) is measured centre to centre, not surface to surface — for a satellite you must add the Earth’s radius to its altitude. And the inverse square is a square: triple the separation and the force falls to a ninth.

Gravitational field strength \( g = F/m = GM/r^{2} \) is the force per unit mass, and it is a vector pointing towards the mass. Note that it does not depend on the mass being placed in the field — which is precisely why all objects fall with the same acceleration, a result that puzzled people for two thousand years and drops out of one line of algebra.

Field lines for a point mass are radial, pointing inwards, and their spacing shows the field strength. They never cross. For two masses the pattern shows a point between them where the fields cancel — nearer the smaller mass.

Orbits are the central application. For a circular orbit, gravity supplies exactly the centripetal force, and setting the two equal is the move that solves most orbit questions:

\[ \frac{GMm}{r^{2}} = \frac{mv^{2}}{r} \qquad\Longrightarrow\qquad v = \sqrt{\frac{GM}{r}} \]

The satellite mass cancels. Higher orbits are slower, which surprises people who expect the opposite.

Substituting \( v = 2\pi r/T \) into that gives

\[ T^{2} = \frac{4\pi^{2}r^{3}}{GM} \qquad\Longrightarrow\qquad T^{2} \propto r^{3} \]

— Kepler’s third law, derived rather than asserted. A geostationary satellite is simply the case \( T = 24 \) hours, which fixes \(r\) at about 42 000 km from the Earth’s centre.

HLPotential and potential energy

Gravitational potential \( V_g = -GM/r \) is the work done per unit mass bringing a small mass from infinity; gravitational potential energy is \( E_p = -GMm/r \). Both are negative, and the reason is a definitional choice: potential is taken as zero at infinity, and since gravity is always attractive, work must be done against the field to escape, so everything closer in sits below zero. A more negative value means more tightly bound.

Do not confuse \( E_p = -GMm/r \) with \( \Delta E_p = mg\Delta h \). The second is the near-surface approximation of the first, valid only where \(g\) is effectively constant.

HLEscape speed

To escape, the kinetic energy must at least cancel the negative potential energy: \( \tfrac{1}{2}mv^{2} = GMm/r \), so

\[ v_{\text{esc}} = \sqrt{\frac{2GM}{r}} \]

The escaping mass cancels, so escape speed is the same for a pebble and a spacecraft. For Earth it is 11.2 km s\(^{-1}\).

HLTotal orbital energy

\( E = E_k + E_p = -\dfrac{GMm}{2r} \), negative because the satellite is bound. A useful consequence: if a satellite in low orbit loses energy to atmospheric drag, \(r\) decreases, and since \( v = \sqrt{GM/r} \) the satellite speeds up while losing energy. Friction makes it faster. That is genuinely counter-intuitive and a favourite of examiners.

HLEquipotentials

Equipotentials are surfaces of constant potential, spheres around a point mass, always perpendicular to field lines. No work is done moving along one.

✏️Worked example

A satellite orbits Earth at an altitude of 800 km. Take \( M_E = 5.97 \times 10^{24} \) kg, \( R_E = 6.37 \times 10^{6} \) m, \( G = 6.67 \times 10^{-11} \).

Get \(r\) right first. \( r = 6.37 \times 10^{6} + 0.80 \times 10^{6} = 7.17 \times 10^{6} \) m. Using the altitude alone here is the error that wrecks the whole question.

(a) Orbital speed. \( v = \sqrt{GM/r} = \sqrt{(6.67 \times 10^{-11} \times 5.97 \times 10^{24})/(7.17 \times 10^{6})} = \sqrt{5.55 \times 10^{5}} = 7.45 \times 10^{3} \) m s\(^{-1}\), about 7.5 km s\(^{-1}\).

(b) Period. \( T = 2\pi r/v = (2\pi \times 7.17 \times 10^{6})/(7.45 \times 10^{3}) = 6045 \) s, about 101 minutes — a plausible low-Earth-orbit period, which is a useful sanity check.

(c) HLEscape speed from that altitude. \( v_{\text{esc}} = \sqrt{2} \times v_{\text{orbit}} = 1.414 \times 7.45 = 10.5 \) km s\(^{-1}\).

A shortcut worth keeping. Escape speed is always \(\sqrt{2}\) times the circular orbital speed at the same radius — useful both as a shortcut and as a check.

🔭See it happen

PhET, Gravity and Orbits. Turn on the velocity and force vectors and watch gravity point permanently at the centre while the velocity stays tangential. Then reduce the orbital speed slightly and watch the orbit decay inwards — and the satellite speed up as it falls.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Find the gravitational field strength at the surface of Mars, given \( M = 6.42 \times 10^{23} \) kg and \( R = 3.39 \times 10^{6} \) m.

\( g = \dfrac{GM}{R^{2}} = \dfrac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{(3.39 \times 10^{6})^{2}} = \dfrac{4.28 \times 10^{13}}{1.15 \times 10^{13}} = 3.7 \) m s\(^{-2}\), about 38% of Earth’s.

2. Find the orbital radius of a geostationary satellite. Take \( M_E = 5.97 \times 10^{24} \) kg and \( T = 24 \) hours.

From \( T^{2} = \dfrac{4\pi^{2}r^{3}}{GM} \), \( r^{3} = \dfrac{GMT^{2}}{4\pi^{2}} = \dfrac{3.98 \times 10^{14} \times (86400)^{2}}{39.5} = 7.53 \times 10^{22} \) m\(^{3}\), so \( r = 4.2 \times 10^{7} \) m from the Earth’s centre.

3. Find the orbital speed of that geostationary satellite.

\( v = \sqrt{GM/r} = \sqrt{\dfrac{3.98 \times 10^{14}}{4.22 \times 10^{7}}} = \sqrt{9.43 \times 10^{6}} = 3.1 \times 10^{3} \) m s\(^{-1}\). Check: \( 2\pi r/T \) gives the same, as it must.

4. Find the escape speed from the surface of the Moon, given \( M = 7.35 \times 10^{22} \) kg and \( R = 1.74 \times 10^{6} \) m.

\( v_{\text{esc}} = \sqrt{\dfrac{2GM}{R}} = \sqrt{\dfrac{2 \times 6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^{6}}} = \sqrt{5.63 \times 10^{6}} = 2.4 \) km s\(^{-1}\) — low enough that the Moon cannot retain an atmosphere.

5. HLFind the gravitational potential at the Earth’s surface, taking \( GM = 3.98 \times 10^{14} \) and \( R = 6.37 \times 10^{6} \) m.

\( V_g = -\dfrac{GM}{R} = -\dfrac{3.98 \times 10^{14}}{6.37 \times 10^{6}} = -6.3 \times 10^{7} \) J kg\(^{-1}\). It is negative because potential is defined as zero at infinity and gravity is always attractive.

6. HLA satellite in low orbit loses energy to atmospheric drag. Explain why it speeds up.

Its total energy \( E = -GMm/2r \) becomes more negative, which requires \(r\) to decrease. But orbital speed is \( v = \sqrt{GM/r} \), which increases as \(r\) falls. So friction makes it faster: the gravitational potential energy lost as it spirals inwards is more than enough to supply both the drag losses and the extra kinetic energy.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

The Physics Classroom — Circular Motion and Satellite Motion
HyperPhysics — gravity, orbits and gravitational potential