An orbit is an energy account: A satellite in orbit owns two kinds of energy at once:
- 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)
The total energy is their sum. 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
- 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 fraction smaller, so the total energy becomes less negative — it rises toward zero. 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):
Final answer
ΔE = +5×10⁹ J must be supplied — the higher orbit has the less negative (higher) energy.
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). You find it by setting the launch kinetic energy equal to the depth of the gravity well:
½mv² = GMm/r ⇒ v_esc = √(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):
- 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.