Escape Velocity Calculator: Formula & Examples

This escape velocity calculator determines the minimum speed an object needs to break free from a celestial body’s gravitational pull without further propulsion. You enter the mass and radius of the planet, moon, or star, and the calculator returns the escape velocity in kilometers per second, miles per second, and meters per second. It uses the standard physics formula derived from the conservation of energy.

Formula

The escape velocity is given by the formula:

v_esc = √(2GM / r)

Where:

• v_esc = escape velocity (m/s or km/s)
• G = gravitational constant = 6.674 × 10^−11 N·m²/kg²
• M = mass of the celestial body (kg)
• r = radius from the center of mass to the object (m)

Note that the formula uses the radius in meters, but the calculator accepts kilometers for convenience and converts internally.

Worked Example

Using the default values for Earth (M = 5.972 × 10²⁴ kg, r = 6371 km):

1. Convert radius to meters: 6371 km × 1000 = 6,371,000 m
2. Compute the product 2GM: 2 × 6.674 × 10^−11 × 5.972 × 10²⁴ = 7.973 × 10¹⁴ m³/s²
3. Divide by radius: 7.973 × 10¹⁴ / 6,371,000 = 1.251 × 10⁸ m²/s²
4. Take the square root: √(1.251 × 10⁸) ≈ 11,186 m/s = 11.19 km/s

The calculator shows 11.19 km/s, 11186 m/s, and 6.95 mi/s for Earth—matching the well-known value.

What Escape Velocity Really Means

Escape velocity is the speed needed to overcome a body’s gravitational pull without any additional thrust. It is independent of the object’s mass—a feather and a rocket need the same speed to escape from the same altitude. However, it does depend on the starting distance from the center of the body. The higher you start, the lower the required speed.

For example, from Earth’s surface, escape velocity is 11.2 km/s. But from the International Space Station’s altitude (~400 km), it drops to about 10.8 km/s. This is why rockets launch from the surface with most of their fuel—they must achieve a large fraction of this speed while still fighting gravity.

How to Use the Calculator Correctly

To get accurate results, you need the mass and radius of the celestial body. These values are well-known for planets and moons, but you can also use it for asteroids, comets, or even artificial objects if you know their mass and size.

• Enter the mass in kilograms (kg). Use scientific notation for large numbers, e.g., 5.972e24 for Earth.
• Enter the radius in kilometers (km). For an object like a rocket at altitude, add the height to the body’s radius.
• Read the escape velocity in km/s (the headline result), m/s, or mi/s.

If you are calculating from a point above the surface, add the altitude to the radius. For example, to escape from a 500 km orbit around Earth, use r = 6371 + 500 = 6871 km.

Common Mistakes and Misconceptions

A frequent error is mixing up radius and diameter. Always use the radius (distance from center to surface), not the diameter. Another mistake is forgetting to convert kilometers to meters in the formula—the calculator does this for you, but if you try the math manually, remember that 1 km = 1000 m.

Misconception: Escape velocity is the speed you need to reach orbit. Actually, orbital speed is lower—about 7.9 km/s for low Earth orbit. Escape velocity is about 40% higher. Another misconception: escape velocity is a single number for a planet. It varies with altitude, so it’s only constant at a given distance from the center.

Also, escape velocity ignores air resistance. On Earth, a projectile would need to be launched faster than 11.2 km/s because atmospheric drag slows it down. In space, where there is no air, the calculated value is exact.

Escape Velocity for Different Bodies

Here are some example values to give you a sense of scale (using known masses and radii):

• Moon: 2.38 km/s (0.238 × 10⁴ m/s) – much easier to escape than Earth.
• Mars: 5.03 km/s – about half of Earth’s.
• Jupiter: 59.5 km/s – huge due to its enormous mass.
• Sun: 617.7 km/s – escaping the Sun from its surface is extremely difficult.
• Black hole (event horizon): equals the speed of light (299,792 km/s) by definition.

These values illustrate why launching from smaller bodies is cheaper, and why interstellar travel from Earth is so challenging.

Limitations of the Formula

The formula assumes a perfectly spherical, non-rotating body with uniform density. Real planets are slightly oblate and rotate, which reduces the effective escape velocity at the equator (because rotation gives you a head start). For most calculations, the difference is small—less than 0.5% for Earth—but it matters for precise mission planning.

Also, the formula gives the speed relative to the body’s center. If the body is moving (like a planet orbiting the Sun), you must consider that motion for interplanetary trajectories. In practice, spacecraft use gravity assists and multiple burns, so the actual required delta-v is different from the theoretical escape velocity.

Frequently Asked Questions

What is escape velocity in simple terms?

It’s the speed an object needs to travel to break free from a planet or moon’s gravity without any additional push. Think of throwing a ball upward—if you throw it fast enough, it never comes back down.

Why is escape velocity independent of the object’s mass?

Because both the kinetic energy (½mv²) and gravitational potential energy (GMm/r) are proportional to the object’s mass. The mass cancels out when you equate the two, leaving a formula that depends only on the body’s mass and radius.

Does escape velocity change with altitude?

Yes. The farther you are from the center of the body, the weaker the gravity, so the lower the escape velocity. For example, at Earth’s surface it’s 11.2 km/s, but at 10,000 km altitude it’s about 7.1 km/s.

Is escape velocity the same as orbital velocity?

No. Orbital velocity is the speed needed to stay in a circular orbit (about 7.9 km/s for low Earth orbit). Escape velocity is √2 times larger (about 11.2 km/s). You need more speed to leave than to circle.

What happens if you launch at less than escape velocity?

The object will rise to a maximum height and then fall back down, unless it achieves orbit. If it reaches orbital velocity, it will circle the planet; below that, it follows a ballistic arc and returns to the surface (ignoring air resistance).

Can escape velocity be used for black holes?

Yes. For a black hole, the escape velocity at the event horizon equals the speed of light. That’s why nothing, not even light, can escape. The formula still works if you plug in the mass and the Schwarzschild radius.

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in m | Mass of celestial body | kg | 5.972e24 | 0
in r | Radius of celestial body | km | 6371 | 0
out v_km | Escape velocity | km/s | sqrt(2 * 6.674e-11 * m / (r * 1000)) | 2
out v_ms | Escape velocity | m/s | sqrt(2 * 6.674e-11 * m / (r * 1000)) * 1000 | 0
out v_mps | Escape velocity | mi/s | sqrt(2 * 6.674e-11 * m / (r * 1000)) * 0.621371 | 2