Formulas & tables/Physics/Gravitation/Second cosmic velocity (escape velocity)

Second cosmic velocity (escape velocity)

vII=2GMRv_{II} = \sqrt{\dfrac{2GM}{R}}

Result unit: m/s (often given in km/s)

What it describes

The second cosmic velocity (escape velocity) is the smallest speed that must be given to a body at a planet's surface for it to escape the planet's gravitational field completely and move away without limit — so that it never returns to the surface.

Symbols

vIIv_{II}
second cosmic velocity (escape velocity)
[m/s]
GG
gravitational constant (≈6.67⋅10−11\approx 6.67\cdot10^{-11})
[N·m²/kg²]
MM
mass of the central body (the planet)
[kg]
RR
radius of the planet (distance from its centre)
[m]

Where it comes from

We derive it from the conservation of energy. For a body to "escape", its kinetic energy must be enough to overcome the gravitational attraction: 12mv2=GMmR\dfrac{1}{2}mv^2=\dfrac{GMm}{R}. The body's mass mm cancels, leaving v2=2GMRv^2=\dfrac{2GM}{R}, that is vII=2GMRv_{II}=\sqrt{\dfrac{2GM}{R}}. Comparing with the first cosmic velocity gives a simple relation: vII=2⋅vIv_{II}=\sqrt{2}\cdot v_I — the escape velocity is 2\sqrt{2} (≈1.41\approx1.41) times the first.

When to use it

When asking what speed lets a body leave the planet for good (escape the gravitational field), not merely enter orbit. For Earth vII≈11.2v_{II}\approx11.2 km/s. The relation vII=2⋅vIv_{II}=\sqrt{2}\cdot v_I always holds.

Example

Find the second cosmic velocity for Earth.
SubstitutionM=5.97⋅1024M=5.97\cdot10^{24} kg, R=6.37⋅106R=6.37\cdot10^{6} m, G=6.67⋅10−11G=6.67\cdot10^{-11}.
ResultvII=2⋅6.67⋅10−11⋅5.97⋅10246.37⋅106≈1.12⋅104v_{II}=\sqrt{\dfrac{2\cdot6.67\cdot10^{-11}\cdot5.97\cdot10^{24}}{6.37\cdot10^{6}}}\approx1.12\cdot10^{4} m/s ≈11.2\approx11.2 km/s. (Check: 2⋅7.9\sqrt2\cdot7.9 km/s ≈11.2\approx11.2 km/s.)

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