Escape Velocity and Orbital Mechanics — Worked Examples
This guide is deliberately calculation-focused rather than a restatement of Newton's law of gravitation itself — if you need the full step-by-step derivation of g from G, M and R, see our Gravitation guide first. Here we work through four numericals that guide alone does not cover: the Moon's escape velocity, the exact relationship between orbital speed and escape speed, the radius of a geostationary orbit, and the (negative) total energy that keeps a satellite bound to its orbit.
Escape velocity, derived from energy conservation
A projectile "escapes" a planet when it has just enough kinetic energy to overcome the planet's gravitational pull all the way to infinity, arriving there with zero speed. Setting kinetic energy equal to the magnitude of gravitational potential energy at the surface:
The escaping mass m cancels out completely — every object, from a dust grain to a rocket, needs the same escape speed at a given R.
Orbital velocity, and how it relates to escape velocity
This √2 factor is worth remembering directly: whatever speed keeps a satellite in a circular orbit at some radius, escaping from that same radius always needs exactly √2 ≈ 1.414 times that speed.
Total mechanical energy of an orbiting satellite
The total energy of a bound circular orbit is always negative — that negative value is the "binding energy," the amount of energy that would need to be supplied to lift the satellite from that orbit all the way to infinity (E = 0). A positive total energy means the object is not in a closed orbit at all; it escapes.
Worked example 1 — escape velocity of the Moon
Moon: M = 7.342 × 10²² kg, R = 1.737 × 10⁶ m, G = 6.674 × 10⁻¹¹ N m² kg⁻².
2GM = 2 × 6.674 × 10⁻¹¹ × 7.342 × 10²²
6.674 × 7.342 = 49.00, so 2GM = 2 × 49.00 × 10¹¹ = 98.00 × 10¹¹ = 9.800 × 10¹²
2GM ÷ R = 9.800 × 10¹² ÷ 1.737 × 10⁶ = (9.800 ÷ 1.737) × 10⁶ = 5.642 × 10⁶ m²/s²
ve = √(5.642 × 10⁶) = 2.375 × 10³ m/s ≈ 2.38 km/s
Compare that with Earth's 11.2 km/s (derived in the Gravitation guide linked above) — the Moon's much smaller mass and radius make it roughly 4.7 times easier to escape from, which is exactly why a lunar-ascent rocket needs far less fuel than one launching from Earth.
Worked example 2 — escape velocity from orbital velocity, cross-checked two ways
A satellite orbits Earth at height h = 2000 km (r = R + h = 6371 + 2000 = 8371 km = 8.371 × 10⁶ m). Using the standard Earth value GM = 3.986 × 10¹⁴ m³/s², find its orbital speed, then find the escape speed at that same radius two different ways.
Orbital speed: vo = √(GM/r) = √(3.986×10¹⁴ ÷ 8.371×10⁶)
3.986 ÷ 8.371 = 0.4762, so GM/r = 4.762 × 10⁷ m²/s²
vo = √(4.762 × 10⁷) = 6.90 × 10³ m/s = 6.90 km/s
Route A — multiply by √2: ve = 1.4142 × 6.90 = 9.758 km/s
Route B — direct formula: ve = √(2GM/r) = √(2 × 4.762×10⁷) = √(9.524×10⁷) = 9.758 km/s
Both routes agree to four significant figures, confirming ve = √2 × vo numerically as well as algebraically.
Worked example 3 — radius of a geostationary orbit
A geostationary satellite has the same period as Earth's rotation. Using the school-level approximation T = 24 h = 86400 s (a more precise treatment uses the sidereal day, 23 h 56 min ≈ 86164 s — state clearly which one you are using), find the orbital radius from Kepler's third law, T² = 4π²r³/(GM).
r³ = GM·T² / (4π²)
T² = 86400² = 7.465 × 10⁹ s²
GM·T² = 3.986×10¹⁴ × 7.465×10⁹ = 3.986 × 7.465 × 10²³ = 29.75 × 10²³ = 2.975 × 10²⁴
4π² = 39.48
r³ = 2.975×10²⁴ ÷ 39.48 = 7.537 × 10²² m³
Taking the cube root: r ≈ 4.224 × 10⁷ m ≈ 42,240 km
Height above the surface: h = r − R = 42,240 − 6,371 = ≈ 35,870 km — close to the commonly quoted value of about 35,800 km; the small difference comes from using the 24-hour approximation instead of the sidereal day.
Worked example 4 — binding energy of an orbiting satellite
Find the total mechanical energy of a 500 kg satellite in the r = 8.371 × 10⁶ m orbit from example 2, using E = −GMm/(2r).
GMm = 3.986×10¹⁴ × 500 = 1.993×10¹⁷
2r = 2 × 8.371×10⁶ = 1.6742×10⁷
E = − (1.993×10¹⁷ ÷ 1.6742×10⁷) = −1.190 × 10¹⁰ J
The negative sign confirms the satellite is genuinely bound to Earth. Its binding energy — the energy that would have to be added to send it to infinity with zero leftover speed — is the positive magnitude of that same number, about 11.9 gigajoules.
Common mistakes that cost marks
- Forgetting the √2 factor between orbital and escape velocity — a very common JEE trap when a question asks for one given the other.
- Ignoring the sidereal-day subtlety in geostationary orbit questions. Using T = 24 h is the standard school-level approximation and is acceptable at that level, but state which value you used — the two give slightly different radii.
- Dropping the negative sign in the total-energy formula. A bound orbit always has E < 0; a positive result means an arithmetic slip, not a genuinely unbound trajectory.
- Thinking escape velocity depends on the planet the same way everywhere. It depends on that specific body's M and R — the Moon's 2.38 km/s and Earth's 11.2 km/s are not interchangeable constants.
- Using diameter instead of radius anywhere in these formulas — R must always be the radius, not the diameter, of the central body.
Where orbital mechanics appears in exams
| Exam | Typical use |
|---|---|
| CBSE Class 11 | Gravitation — escape speed, orbital speed, energy of a satellite, geostationary orbits |
| JEE Main & Advanced | Escape-vs-orbital-velocity comparisons, binding energy, energy required to change orbits |
| NEET | Direct ve = √(2GM/R) substitution and escape-vs-orbital ratio questions |
| GATE (Aerospace-adjacent papers) | Orbital radius and period calculations from Kepler's third law |
These numericals are mostly powers-of-ten and cube-root arithmetic. The Scientific Calculator's exponent and root keys make a step like cube-rooting 7.537 × 10²² fast while you check your working.
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