Formula & Calculator

Escape Velocity

Minimum speed needed to escape a gravitational field without further propulsion.

AstronomyOrbital MechanicsGravitation

Escape Velocity Calculator vesc = √(2GM / r)

vesc = √(2 · G · M / r)
vesc = escape velocity (m/s)  ·  G = gravitational constant (m³/kg·s²)  ·  M = mass (kg)  ·  r = radius (m)
⟹ Solve vesc, G, M, r
m/s
m³/kg·s²
kg
m
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Escape Velocity (vesc)
vesc: G: M: r:
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Escape Velocity
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vesc = √(2GM/r)  ·  Escape velocity is the minimum speed needed to break free from a gravitational field.

Interpretation

v_esc = √(2GM/r). Minimum speed needed for an object to escape a gravitational field without further propulsion. Used in space mission planning and planetary science.

v_esc = √(2GM / r)
Escape Velocity

Variables

SymbolQuantityUnit
v_escEscape velocitym/s
GGravitational constant
MMass of bodykg
rDistance from centerm

What it means

Escape velocity is the minimum initial speed required for an object to break free from the gravitational influence of a massive body (like a planet or star) without any additional thrust. It is derived from the conservation of energy, balancing kinetic energy against gravitational potential energy. For Earth, v_esc ≈ 11.2 km/s. This concept is essential for launching spacecraft, for understanding planetary atmospheres (gases with molecular speeds exceeding escape velocity are lost), and for studying the formation of celestial bodies. In astrophysics, it determines the size of a black hole’s event horizon (Schwarzschild radius). Understanding escape velocity is critical for astronauts, engineers, and planetary scientists to design missions and to interpret atmospheric evolution.

Worked example

Escape Velocity – Two Detailed Examples

Real‑World
Scenario: A rocket engineer is calculating the minimum speed needed for a spacecraft to escape Earth's gravity without further propulsion. Using Earth's mass (5.972×10²⁴ kg) and radius (6.371×10⁶ m) along with G = 6.67430×10⁻¹¹, they compute the escape velocity. This determines the fuel requirements for the launch vehicle and the trajectory design.
ParameterValue
G (m³/kg·s²)6.67430e-11
M (kg)5.972e24
r (m)6.371e6
1v_esc = √(2 × 6.67430e-11 × 5.972e24 / 6.371e6)
2v_esc ≈ 11,186 m/s ≈ 11.2 km/s
Result 11.2 km/s ✓ Earth escape velocity
Scenario: A planetary scientist is studying the possibility of a moon having an atmosphere. For a moon of mass 7.342×10²² kg and radius 1.737×10⁶ m (similar to our Moon), they compute the escape velocity. If the escape velocity is low, gases can easily escape into space, explaining why the Moon has no atmosphere. This calculation supports their atmospheric loss models.
ParameterValue
G6.67430e-11
M7.342e22
r1.737e6
1v_esc = √(2 × 6.67430e-11 × 7.342e22 / 1.737e6)
2≈ 2.38 km/s (Moon), much lower than Earth
Result 2.38 km/s ✓ Low escape velocity, atmosphere unlikely
Insight: Escape velocity depends on mass and radius. Smaller bodies have lower escape velocities, making it harder to retain an atmosphere.

Common mistakes

  • Escape velocity: The minimum speed to escape a gravitational field.
  • Units: G in m³/(kg·s²), M in kg, r in metres → v in m/s.
  • r: The distance from the centre of the mass – not the surface radius if starting from altitude.
  • Atmospheric effects: Does not account for air resistance – applies to vacuum.
  • Energy conservation: Derived from kinetic = gravitational potential – ensure sign conventions.

Applications

Escape velocity, v_esc = √(2GM/r), is the minimum speed required for an object to break free from the gravitational pull of a celestial body without further propulsion. This concept is critical for space exploration: rockets must reach Earth's escape velocity (about 11.2 km/s) to leave our planet. Astronomers use it to assess whether a planet can retain an atmosphere – if the average molecular speed exceeds escape velocity, the atmosphere will be lost over time. By calculating escape velocity, engineers design launchers and planetary missions. It also helps in understanding the formation of planetary systems. This formula is a cornerstone of astrophysics, relating mass, radius, and gravitational binding.

  • Spacecraft launch and interplanetary mission planning
  • Atmospheric retention and evolution studies (planetary habitability)
  • Determination of black hole event horizons (Schwarzschild radius)
  • Design of gravitational assist manoeuvres
  • Understanding of celestial body mass and size

Frequently Asked Questions

Q01What is escape velocity and what is its formula?
A01

v_esc = √(2GM / r). It is the minimum speed needed for an object to break free from the gravitational pull of a massive body (e.g., a planet, star) without any further propulsion, assuming no atmospheric drag.

Q02What is the escape velocity from Earth's surface, and why is it important?
A02

From Earth's surface, v_esc ≈ 11.2 km/s (about 25,000 mph). This is the speed required for a rocket to leave Earth permanently. It is a fundamental parameter in aerospace engineering and space mission design.

Q03How does escape velocity depend on the mass and radius of the body?
A03

Escape velocity is proportional to √M and inversely proportional to √r. Therefore, more massive or smaller bodies have higher escape velocities. For example, a black hole has such high mass and small size that v_esc exceeds the speed of light.

Q04What is the relationship between escape velocity and orbital velocity?
A04

For a circular orbit at radius r, the orbital velocity is v_orb = √(GM/r). Escape velocity is √2 times larger: v_esc = √2 · v_orb. So to escape, you need about 41% more speed than to stay in a circular orbit.

Q05What happens if an object has a speed less than escape velocity?
A05

If the speed is between orbital speed and escape velocity, the object follows an elliptical orbit. If it is below orbital speed, it may follow a sub‑orbital trajectory (falling back) or, if below the required for a stable orbit, it will crash into the central body.

Q06What is the escape velocity from the Moon, and how does it compare to Earth's?
A06

Moon's mass ≈ 7.35×10²² kg, radius ≈ 1.74×10⁶ m. v_esc = √(2·6.674×10⁻¹¹·7.35×10²² / 1.74×10⁶) ≈ 2.38 km/s. This is much lower than Earth's, making it easier for a spacecraft to leave the Moon.

Q07What is the escape velocity from a black hole?
A07

At the event horizon, the escape velocity equals the speed of light, c. The radius at which this occurs is the Schwarzschild radius: R_s = 2GM/c². Inside this radius, no information or matter can escape.

Q08How is escape velocity related to gravitational binding energy?
A08

The kinetic energy required to escape is ½ m v_esc², which equals the gravitational potential energy: GMm/r. Thus, v_esc² = 2GM/r. This energy is what holds galaxies and star clusters together.

Q09Does atmospheric drag affect the escape velocity requirement?
A09

Escape velocity is defined in vacuum. In a real launch, atmospheric drag and gravity losses increase the required Δv (change in velocity). Rockets must achieve a higher total velocity to overcome these losses, typically around 9–10 km/s for Earth orbit (not escape).

Q10How does escape velocity change for an object starting from a higher altitude (e.g., from orbit)?
A10

If you are already at a distance r (e.g., in orbit), the escape velocity from that altitude is √(2GM/r). For example, at the International Space Station altitude (~400 km), v_esc is about 10.9 km/s, slightly lower than at the surface because r is larger.