Escape velocity calculator

What is escape velocity?

Escape velocity is the minimum speed an object needs in order to break free from the gravitational pull of a massive body without any further propulsion. Once a projectile reaches this speed, its kinetic energy is large enough to overcome the gravitational potential energy binding it to the body, allowing it to travel away indefinitely and never fall back. The concept applies to rockets leaving Earth, probes departing the Moon, and even particles escaping the surface of a star.

A useful way to picture it: as an object climbs away from a planet, gravity steadily slows it down. If the object starts slower than escape velocity, it will eventually stop rising and fall back. If it starts at exactly escape velocity, it will keep moving outward, getting ever slower but never quite stopping. Any faster, and it leaves with energy to spare.

How the escape velocity calculator works

Enter the mass of the central body and the distance from its center (usually its radius if you are launching from the surface), and the calculator returns the escape velocity. You can supply the radius in meters, kilometers, or miles, and read the result in meters per second, kilometers per second, kilometers per hour, feet per second, or miles per hour. This makes it easy to compare the escape speeds of different worlds or to convert a result into whichever unit your problem requires.

Because escape velocity depends only on mass and distance, it is independent of the mass of the escaping object. A pebble and a spaceship launched from the same point need the same speed to escape, which is why this single formula is so widely used in orbital mechanics and astrophysics.

Formula

The escape velocity ($v$) from a body of mass $M$ at a distance $r$ from its center is:

where:

• $G$ is the gravitational constant, $6.6743 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2$,
• $M$ is the mass of the body (in kilograms),
• $r$ is the distance from the center of the body (in meters).

The formula comes from setting the kinetic energy of the object equal to the gravitational potential energy needed to reach an infinite distance. Notice that the mass of the escaping object cancels out, and that escape velocity grows with the square root of the central mass and falls with the square root of the distance.

Examples

1. Generic body: Take a body with mass $M = 1 \times 10^{24} \, \text{kg}$ and radius $r = 1 \times 10^{6} \, \text{m}$. Applying the formula:

   $$v = \sqrt{\frac{2 \times 6.6743 \times 10^{-11} \times 10^{24}}{10^{6}}} \approx 11{,}553.6 \, \text{m/s}$$

   So an object must reach roughly 11.55 km/s to escape this body’s gravity.

2. Planet Earth: Earth has a mass of about $M = 5.972 \times 10^{24} \, \text{kg}$ and a mean radius of $r = 6.371 \times 10^{6} \, \text{m}$. The escape velocity is:

   $$v = \sqrt{\frac{2 \times 6.6743 \times 10^{-11} \times 5.972 \times 10^{24}}{6.371 \times 10^{6}}} \approx 11{,}185.6 \, \text{m/s}$$

   That is about 11.19 km/s, the familiar figure quoted for leaving Earth from its surface.

Notes

• Escape velocity ignores atmospheric drag and assumes a single, non-rotating central body. A real launch needs more energy to fight air resistance.
• It is a scalar speed, not a direction; an object can escape along any path as long as it does not later strike the surface.
• Reaching escape velocity sends an object onto an open (parabolic or hyperbolic) trajectory, whereas staying below it leaves the object on a bound, returning path.

FAQs

Does escape velocity depend on the mass of the object that is escaping?

No. The mass of the escaping object cancels out of the equation, so a small satellite and a large rocket require the same escape velocity from the same point. Heavier objects simply need more total energy, because energy scales with their own mass.

Why is Earth’s escape velocity about 11.2 km/s?

Plugging Earth’s mass and radius into the formula gives roughly 11.19 km/s. This is the speed needed at the surface, neglecting air resistance and the boost from Earth’s rotation, to reach infinity with zero leftover speed.

What is the difference between escape velocity and orbital velocity?

Orbital velocity is the speed required to maintain a stable circular orbit at a given altitude, while escape velocity is the speed required to leave entirely. Escape velocity is exactly $\sqrt{2}$ times the circular orbital velocity at the same distance.

Does escape velocity change with altitude?

Yes. Because the distance $r$ appears in the denominator, escape velocity decreases as you move farther from the body’s center. It is highest at the surface and smaller from a high orbit.

Can anything exceed escape velocity and still return?

If an object reaches or exceeds escape velocity and is not acted on by other forces, it will not return to the body. Objects only fall back when their speed stays below the local escape velocity.

How does escape velocity relate to black holes?

A black hole is a region where the escape velocity equals or exceeds the speed of light, so not even light can escape. The same formula, treated relativistically, leads to the definition of the event horizon.

Related tools: see the velocity, gravitational force, and kinetic energy calculators. You can bookmark this page at https://www.mega-calculator.com/physics/escape-velocity/.