Escape Velocity
Minimum speed to escape a gravitational field.
Formula
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Example
Earth: ~11.2 km/s
How it works
Escape velocity is the launch speed needed so that an object can move infinitely far from a massive body without any further propulsion, assuming no air resistance and only gravity acts. It comes from equating the object’s initial kinetic energy to the gravitational potential energy required to climb out of the body’s gravity well, with the final speed at infinity taken as zero. Because the formula depends on \(M\) and \(r\), it applies to any spherical mass distribution outside its surface, not just planets.
When to use it
Use this when comparing how hard it is to leave planets, moons, stars, or other spherical astronomical bodies. It is also useful in orbital mechanics and astrophysics whenever you need the minimum launch speed needed to avoid falling back.
Worked example, step by step
Find the escape velocity from Mars. Given: G = 6.674 \times 10^{-11} m^3 kg^{-1} s^{-2}, M = 6.42 \times 10^{23} kg, r = 3.39 \times 10^6 m Use v_e = \sqrt{2GM/r} Substitute: v_e = \sqrt{(2)(6.674 \times 10^{-11})(6.42 \times 10^{23}) / (3.39 \times 10^6)} Compute the numerator: 2GM \approx 8.57 \times 10^{13} Divide by r: (8.57 \times 10^{13}) / (3.39 \times 10^6) \approx 2.53 \times 10^7 Take the square root: v_e \approx \sqrt{2.53 \times 10^7} \approx 5.03 \times 10^3 m/s Final answer: v_e \approx 5.0 km/s
Common mistakes to avoid
- Using the planet’s orbital speed instead of escape speed
- Forgetting that r is measured from the center of the body, not from the surface
- Mixing units and plugging mass or radius into kilometres instead of SI units
Did You Know?
Earth's escape velocity is about 11.2 km/s, but a black hole's exceeds the speed of light — which is why nothing escapes it.
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