Key Idea: Topic 4.1 is about gravity as a field — how a mass fills the space around it with a pull, and what that pull does to planets, satellites and rockets. It ties together four ideas: how strong the pull is (Newton's law F = GMm/r² and the field strength g = GM/r²), how things orbit (Kepler's laws, with T² ∝ r³), why a satellite stays up (gravity is the centripetal force, giving v = √(GM/r)), and the energy of being in a gravity well (the negative potential energy, and the escape speed). It is examined on Paper 1A (quick MCQs — inverse-square reasoning, recognising T² ∝ r³, the direction of a satellite's acceleration, why escape speed ignores the rocket's mass) and on Paper 2 (substitute into g = GM/r², compare two orbits with T²/r³, 'weigh' a central body with M = 4π²r³/(GT²), or use energy conservation for escape and impact speeds).
📐 Key formulas
Three of these are given in the data booklet (D.1) — you pick the right one rather than memorising it. The orbit and energy results below them are built from the given equations, so know how each one comes about.
- gravitational force between the masses (N)
- gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
- the two masses (kg)
- distance between their centres (m)
- gravitational field strength (N kg⁻¹), also the free-fall acceleration
- gravitational force on the small mass (N)
- the small mass placed in the field (kg)
- mass of the planet or star making the field (kg)
- distance from the centre of M (m)
- gravitational potential — energy per kilogram (J kg⁻¹)
- gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
- mass of the planet or star (kg)
- distance from the centre of M (m)
- orbital speed (m s⁻¹)
- orbit radius (m)
- orbital period — time for one full orbit (s)
- orbital speed (m s⁻¹)
- gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
- mass of the central body (kg)
- orbit radius from the centre of the central body (m)
- orbital period — time for one full orbit (s)
- orbital radius — distance from the central body's centre (m)
- gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
- mass of the central body being orbited (kg)
- gravitational potential energy of the object (J)
- gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
- mass of the planet or star (kg)
- mass of the object placed in the field (kg)
- distance from the centre of M (m)
- escape speed — the launch speed needed to escape (m s⁻¹)
- gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻²
- mass of the planet or star (kg)
- distance from the centre of M at launch — usually its radius (m)
🧭 Which equation, and when?
The most-tested decision in this topic: is the question about the field at a point, an orbit's timing or speed, or the energy of being in the well?
| What the question wants | Equation to reach for | Watch out for |
|---|---|---|
| Field strength g (or force F) at a distance r | g = GM/r² (or F = GMm/r²) | Inverse-square: r ×n → g ÷n² |
| Compare two orbits round the same body | TA²/rA³ = TB²/rB³ | G and M cancel — you never need them |
| Orbital speed (no period given) | v = √(GM/r) | The orbiting mass cancels — it is not in it |
| Link a period to a radius, or weigh a body | T² = 4π²r³/(GM) | Rearrange to M = 4π²r³/(GT²) to find M |
| Potential energy, or a launch/impact speed | Eₚ = -GMm/r, vₑₛc = √(2GM/r) | Eₚ is negative; vₑₛc ignores the object's mass |
🛰️ The four big results — what each depends on
| Quantity | Formula | Grows with… | Does NOT depend on |
|---|---|---|---|
| Field strength g | g = GM/r² | central mass M | the falling object's mass — so all masses fall equally |
| Orbital speed v | v = √(GM/r) | central mass M (smaller for big r) | the orbiting body's mass |
| Orbital period T | T² = 4π²r³/(GM) | orbit radius r (T² ∝ r³) | the orbiting body's mass |
| Escape speed vₑₛc | vₑₛc = √(2GM/r) | central mass M (∝ √M) | the escaping object's mass |
🪐 Kepler's three laws at a glance
| Law | What it says | Exam use |
|---|---|---|
| 1st (shape) | Orbits are ellipses, with the Sun at one focus | State the law; the Sun is offset from the centre |
| 2nd (speed) | Faster near the Sun, slower far away (equal areas in equal times) | Explain why a planet's speed / kinetic energy changes |
| 3rd (timing) | T² ∝ r³ — bigger orbit, longer period | Ratio two orbits with TA²/rA³ = TB²/rB³ |
🔄 The inverse-square shortcut
| Move the distance to… | Field strength g = GM/r² becomes… | Why |
|---|---|---|
| 2r (twice as far) | g ÷ 4 | g ∝ 1/r², so ÷2² = ÷4 |
| 3r (three times as far) | g ÷ 9 | ÷3² = ÷9 — the classic trap is dividing by 3 |
| ½r (half as far) | g × 4 | ×(½)⁻² = ×4 — closer in is much stronger |
✍️ IB-style worked examples
A planet has mass 5.4 × 10²⁴ kg and radius 6.0 × 10⁶ m. (a) Find the gravitational field strength at its surface. (b) State the field strength at a point three times as far from the centre. Take G = 6.67 × 10⁻¹¹ N m² kg⁻².
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
Two moons, P and Q, orbit the same planet. Moon Q's orbital radius is 4.0 times that of moon P, and moon P's period is 1.5 days. Find moon Q's orbital period.
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
A planet orbits a star in a circle of radius 1.8 × 10¹¹ m with a period of 1.6 × 10⁷ s. Determine the mass of the star. Take G = 6.67 × 10⁻¹¹ N m² kg⁻².
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
A moon has mass 8.0 × 10²² kg and radius 1.8 × 10⁶ m. (a) Find the escape speed from its surface. (b) State whether a 2000 kg lander or a 500 kg probe needs the greater launch speed to escape. Take G = 6.67 × 10⁻¹¹ N m² kg⁻².
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
✅ Quick self-check
Tap each card to reveal the answer.
Move three times farther from a planet — what happens to g? It is divided by 3² = 9 (inverse-square law). Dividing by 3 is the classic mistake.
Two different masses are dropped at the same point — do they accelerate the same? Yes — the acceleration is g = GM/r², which does not depend on the falling mass.
Plot T² against r³ for moons of one planet — what shape? A straight line through the origin, because T²/r³ is a constant (T² ∝ r³).
Does a heavier satellite at the same radius orbit faster? No — v = √(GM/r) has no satellite mass in it, so both orbit at the same speed.
Why is gravitational potential energy negative? We set it to zero at infinity; anywhere closer in, gravity has already pulled the object 'downhill', so it sits in a well.
Does escape speed depend on the rocket's mass? No — the mass cancels in vₑₛc = √(2GM/r). It depends only on the planet's M and r.
🎯 Highest-yield exam reminders
Exam Tips
- g = GM/r² and F = GMm/r² are both inverse-SQUARE: multiply the distance by n and you divide g (or F) by n². Three times farther → one ninth, not one third.
- Field strength g is also the free-fall acceleration, and it does not depend on the falling mass — so all masses dropped at one place accelerate equally. N kg⁻¹ and m s⁻² are the same unit for g.
- For two orbits round the SAME body, use TA²/rA³ = TB²/rB³ — the constant 4π²/(GM) cancels, so you never need G or M. Watch the powers: T is squared, r is cubed.
- Gravity is the centripetal force for an orbit (GMm/r² = mv²/r). No period given for a speed? Use v = √(GM/r). Linking T and r, or finding the central mass? Use T² = 4π²r³/(GM), i.e. M = 4π²r³/(GT²).
- Orbit radius r is measured from the CENTRE of the planet — a satellite's height above the surface is r minus the planet's radius. A geostationary satellite has a 24 h period over the equator, so it stays above one fixed point.
- Eₚ = -GMm/r and V = -GM/r are always negative and zero at infinity — never drop the minus sign. Many 'launch or impact speed' questions are energy conservation: kinetic energy gained = depth of the well climbed.
- Escape speed vₑₛc = √(2GM/r) depends only on the planet (M and r), not on the escaping object — and vₑₛc ∝ √M, so quadrupling the mass at fixed radius doubles it.