An orbit is an energy account: A satellite in orbit owns two kinds of energy at once. Its total energy is their sum.
The orbit's energy account
- Kinetic energy KE — because it is moving (always positive).
- Gravitational potential energy PE — because it sits inside a gravity well (always negative, taken as zero at infinity).
- For a body bound in orbit the total comes out negative — that minus sign is the signature of a trapped object.
At HL you don't just quote the orbital speed — you account for the energy. Starting from the circular-orbit condition (gravity supplies the centripetal force), the kinetic energy works out to exactly half the size of the potential energy, and with opposite sign. Adding them gives a compact result for the total.
Each piece of the energy
- KE = +GMm / 2r (positive — it is moving)
- PE = −GMm / r (negative — it is bound)
- PE is exactly twice the size of KE
Added together
- E = KE + PE = +GMm/2r − GMm/r
- E = −GMm / 2r
- Total is negative ⇒ the body is bound
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- total mechanical energy of the orbiting body (J)
- gravitational constant, 6.67×10⁻¹¹ N m² kg⁻²
- mass of the central body, e.g. the Earth (kg)
- mass of the orbiting satellite (kg)
- orbital radius, measured from the centre of M (m)
KE and PE separately: If a question asks for the pieces, the kinetic energy is KE = +GMm/2r and the potential energy is PE = −GMm/r. Their sum, E = −GMm/2r, is the total — and note |PE| = 2 × KE.
Worked example — total energy of a satellite
A 500 kg satellite orbits the Earth at radius r = 7.0×10⁶ m. Take M = 6.0×10²⁴ kg and G = 6.67×10⁻¹¹ N m² kg⁻². Find its total orbital energy.
Solution
- Write the given formula first:
- Substitute the numbers:
- Evaluate — keep the unit and the minus sign:
Final answer
E = −1.4×10¹⁰ J. The energy is negative because the satellite is bound to the Earth.
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Higher orbit → less negative E: Because E = −GMm/2r, raising the orbit (bigger r) makes the total energy less negative — it rises toward zero.
What that means
- A bigger r makes the fraction smaller.
- A higher orbit is a higher-energy, more weakly bound state.
- Reaching E = 0 means the body is no longer bound at all — it can just escape to infinity.
| Orbit radius r | Total energy E = −GMm/2r | How bound? |
|---|---|---|
| small r (low orbit) | large negative E | tightly bound |
| larger r (high orbit) | less negative E | weakly bound |
| r → ∞ | E → 0 | free (just escaped) |
Worked example — energy to raise an orbit
The same 500 kg satellite (E = −1.4×10¹⁰ J at r = 7.0×10⁶ m) is moved to a much higher orbit where its total energy is −0.9×10¹⁰ J. How much energy must be supplied?
Solution
- Energy supplied = change in total energy:
- Substitute (both negative):
- Evaluate:
Final answer
ΔE = +5×10⁹ J must be supplied — the higher orbit has the less negative (higher) energy.
Into orbit, and out of it: From rest on the surface into orbit (non-rotating planet of radius R): energy needed ΔE = Eorbit − Esurface = −GMm/2r − (−GMm/R) = GMm(1/R − 1/2r).
From the orbit to escape: raise the total energy from −GMm/2r to 0, so add GMm/2r.
A little air drag: the satellite spirals in and speeds up
The surprise: Drag slows things down on Earth, yet it makes an orbiting satellite go faster: it loses height, and gravity speeds it up by more than drag slows it.
Breaking free of the well: The escape velocity is the minimum launch speed that lets an object coast away to infinity, ending with zero total energy (just barely free).
Finding it
- Set the launch kinetic energy equal to the depth of the gravity well: ½mv² = GMm/r.
- Solve for v: vesc = √(2GM/r).
- The object's mass m cancels — escape velocity does not depend on what you launch.
- escape velocity (m s⁻¹)
- gravitational constant, 6.67×10⁻¹¹ N m² kg⁻²
- mass of the body being escaped, e.g. the Earth (kg)
- distance from the centre of M to the launch point (m)
Worked example — escape velocity from Earth
Find the escape velocity from the Earth's surface. Take M = 6.0×10²⁴ kg, r = 6.4×10⁶ m and G = 6.67×10⁻¹¹ N m² kg⁻².
Solution
- Write the given formula first:
- Substitute the numbers:
- Work out the bracket, then the root — keep the unit:
Final answer
vesc = 1.1×10⁴ m s⁻¹ ≈ 11 km s⁻¹. Notice the satellite's mass never entered — escape velocity is the same for a pebble or a rocket.
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Where it shows up: Orbital energetics and escape velocity are HL only (Theme D). They come up in Paper 1A and Paper 2.
| Paper | What they ask |
|---|---|
| Paper 1A | A one-step 'is the total energy positive or negative?', 'does escape velocity depend on the satellite's mass?', or 'higher orbit → larger or smaller |E|?' |
| Paper 2 | Determine the total/kinetic/potential energy of a satellite, the energy needed to change orbit, or compute an escape velocity from M and r |
Three easy marks: (1) Total orbital energy is negative — keep the minus sign. (2) In vesc = √(2GM/r) the launched mass cancels, so never plug it in. (3) A higher orbit means a less negative (higher) total energy.
IB-style question — kinetic energy in orbit
A 1200 kg spacecraft orbits a planet of mass M = 4.0×10²⁴ kg at radius r = 8.0×10⁶ m (G = 6.67×10⁻¹¹ N m² kg⁻²). Determine its kinetic energy in orbit.
Solution
- The kinetic energy of a circular orbit is KE = +GMm/2r:
- Substitute the numbers:
- Evaluate — keep the unit:
Final answer
KE = 2.0×10¹⁰ J (positive). The total energy would be the negative of this, E = −2.0×10¹⁰ J.