What is Escape Velocity?
Escape velocity is the minimum speed an object must have to permanently break free from a planet's or star's gravitational field without any further propulsion. It is derived by equating kinetic energy and gravitational potential energy. For Earth, escape velocity is approximately 11.2 km/s. Notably, it is completely independent of the object's mass.
Core Gravitation Formulas
| Symbol | Quantity | Value / Unit | Notes |
|---|
| G | Gravitational constant | 6.674 × 10−11 N m2 kg−2 | Universal constant |
| M | Mass of central body | kg | e.g. Earth = 5.972 × 1024 kg |
| R | Radius of central body | m | Earth = 6.371 × 106 m |
| r | Orbital radius | m | R + altitude above surface |
| ve | Escape velocity | m/s or km/s | Independent of escaping mass |
| vo | Orbital speed | m/s or km/s | vo = ve / sqrt(2) |
| T | Orbital period | seconds | Time for one complete revolution |
| g | Surface gravitational acceleration | m/s2 | Earth g = 9.8 m/s2 |
| rs | Schwarzschild radius | m | Radius at which escape velocity = c |
Key Concepts in Gravitation
Derivation of Escape Velocity
Escape velocity is derived by applying conservation of energy. At the surface, the object has kinetic energy (1/2)mve2 and gravitational potential energy −GMm/R. At infinity, both energies are zero. Setting total energy = 0:
(1/2)mve2 − GMm/R = 0
ve2 = 2GM/R
ve = sqrt(2GM/R)
Crucially, the mass m of the escaping object cancels out, proving escape velocity is the same for a feather and a rocket — one of the most elegant results in classical mechanics.
Relationship Between Orbital Speed and Escape Velocity
ve = sqrt(2) × vo ≈ 1.414 × vo
A spacecraft in a circular orbit travels at orbital speed vo. To escape the gravitational field entirely, it must reach escape velocity, which is exactly sqrt(2) times the orbital speed. This is why rockets execute a trans-lunar injection burn — a speed increase from orbital velocity to escape velocity.
Kepler's Third Law
The square of the orbital period (T2) is proportional to the cube of the orbital radius (r3): T2 ∝ r3. This applies to all planets orbiting the same central body.
Geostationary Orbit
A geostationary satellite orbits at ~35,786 km above Earth's equator with T = 24 hours. Used for weather satellites, TV broadcasting, and GPS augmentation.
Black Holes & Schwarzschild Radius
If a body is compressed inside its Schwarzschild radius rs = 2GM/c2, its escape velocity exceeds the speed of light and it becomes a black hole. For Earth, rs ≈ 8.87 mm.
Variation of g with Altitude
g decreases with altitude as g(h) = g0 × R2 / (R + h)2. At the International Space Station (400 km altitude), g is still about 88.5% of surface value.
Escape Velocities Across the Solar System
| Body | Mass (kg) | Radius (km) | Escape Velocity | Surface g (m/s2) |
|---|
| Mercury | 3.301 × 1023 | 2,440 | 4.25 km/s | 3.70 |
| Venus | 4.867 × 1024 | 6,051 | 10.36 km/s | 8.87 |
| Earth | 5.972 × 1024 | 6,371 | 11.19 km/s | 9.81 |
| Moon | 7.342 × 1022 | 1,737 | 2.38 km/s | 1.62 |
| Mars | 6.39 × 1023 | 3,390 | 5.03 km/s | 3.72 |
| Jupiter | 1.898 × 1027 | 69,911 | 59.5 km/s | 24.79 |
| Saturn | 5.683 × 1026 | 58,232 | 35.5 km/s | 10.44 |
| Sun | 1.989 × 1030 | 695,700 | 617.5 km/s | 274.0 |
Solved Examples (NCERT / JEE Level)
Example 01 — Escape Velocity from Earth
Calculate the escape velocity from the surface of Earth. Given: M = 5.972 × 1024 kg, R = 6.371 × 106 m, G = 6.674 × 10−11 N m2 kg−2.
Formula: ve = sqrt(2GM / R)
2GM = 2 × 6.674 × 10−11 × 5.972 × 1024
= 2 × 3.986 × 1014
= 7.972 × 1014 m3 s−2
2GM / R = 7.972 × 1014 / 6.371 × 106
= 1.251 × 108 m2 s−2
ve = sqrt(1.251 × 108) = 11,187 m/s
ve = 11.19 km/s | A rocket must reach this speed to escape Earth's gravity
Example 02 — Orbital Speed of ISS
The International Space Station orbits at an altitude of 408 km above Earth's surface. Find its orbital speed and orbital period. (MEarth = 5.972 × 1024 kg, REarth = 6.371 × 106 m)
Orbital radius r = R + h = 6.371 × 106 + 408 × 103
r = 6.779 × 106 m
vo = sqrt(GM / r)
GM = 6.674 × 10−11 × 5.972 × 1024 = 3.986 × 1014
vo = sqrt(3.986 × 1014 / 6.779 × 106)
vo = sqrt(5.880 × 107) = 7,668 m/s = 7.67 km/s
T = 2πr / vo = 2 × 3.14159 × 6.779 × 106 / 7668
T = 5,552 s = 92.5 minutes
vo = 7.67 km/s | T = 92.5 min | ISS completes ~15.5 orbits per day
Example 03 — Escape Velocity on the Moon
Calculate the escape velocity from the Moon's surface and explain why the Moon has no atmosphere. (MMoon = 7.342 × 1022 kg, RMoon = 1.737 × 106 m)
ve = sqrt(2GM / R)
2GM = 2 × 6.674 × 10−11 × 7.342 × 1022
= 9.796 × 1012
ve = sqrt(9.796 × 1012 / 1.737 × 106)
ve = sqrt(5.641 × 106)
ve = 2,375 m/s = 2.37 km/s
The average speed of N2 molecules at 300 K is ~517 m/s, but thermal
extremes on Moon surface reach 400 K, giving some molecules speeds
exceeding the Moon's escape velocity → atmosphere gradually lost.
ve = 2.37 km/s | Moon's low escape velocity causes atmospheric gas molecules to escape into space
Frequently Asked Questions (FAQs)
Why is escape velocity independent of the object's mass?
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When deriving escape velocity using conservation of energy, the equation (1/2)mve2 = GMm/R has the escaping mass m on both sides, so it cancels out. This means escape velocity depends only on the mass M and radius R of the central body, not on what is trying to escape. A tennis ball and a rocket ship both need exactly 11.2 km/s to escape Earth's gravity (ignoring air resistance).
What is the relationship between orbital speed and escape velocity?
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Escape velocity ve = sqrt(2GM/R) is exactly sqrt(2) times the circular orbital speed vo = sqrt(GM/r) at the same radius. So ve = sqrt(2) × vo ≈ 1.414 × vo. For Earth at surface, orbital speed would be ~7.9 km/s and escape velocity is ~11.2 km/s. To leave orbit and escape Earth, a spacecraft must increase speed by a factor of sqrt(2).
What is a geostationary orbit and at what altitude does it exist?
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A geostationary orbit is a circular orbit above Earth's equator where the orbital period is exactly equal to Earth's rotation period (24 hours). A satellite in this orbit appears stationary relative to the ground. Using T2 = 4π2r3/GM with T = 86,400 s, the geostationary orbital radius works out to approximately 42,164 km from Earth's centre (about 35,786 km altitude above the surface). Most TV broadcast and weather satellites are placed here.
What is the Schwarzschild radius and what is Earth's Schwarzschild radius?
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The Schwarzschild radius (rs = 2GM/c2) is the critical radius below which a body becomes a black hole — the point where escape velocity equals the speed of light. If Earth were somehow compressed into a sphere smaller than its Schwarzschild radius, it would collapse into a black hole. Earth's Schwarzschild radius is approximately 8.87 mm (about the size of a marble). The Sun's Schwarzschild radius is about 2.95 km.
How does escape velocity explain why the Moon has no atmosphere?
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The Moon's escape velocity is only about 2.38 km/s, compared to Earth's 11.2 km/s. Gas molecules in an atmosphere have a range of speeds based on temperature (Maxwell-Boltzmann distribution). On the Moon, extreme surface temperatures (up to 127°C on the dayside) give enough thermal energy to many gas molecules — especially lighter ones like hydrogen and helium — to reach or exceed the low escape velocity. Over billions of years, essentially all atmospheric gases have been lost to space through this process, called Jeans escape.