Escape speed is the minimum speed an object must have at a given distance from a massive body in order to reach infinity with zero final speed, assuming no air resistance and no further propulsion.
1. Energy condition for escape
The total mechanical energy of an object of mass \(m\) moving near a spherical body of mass \(M\) is
\[
E=K+U.
\]
The kinetic energy is
\[
K=\frac12mv^2.
\]
With zero gravitational potential energy at infinity, the gravitational potential energy is
\[
U=-\frac{GMm}{r}.
\]
Therefore,
\[
E=\frac12mv^2-\frac{GMm}{r}.
\]
2. Deriving the escape-speed formula
For the object to just escape, it reaches infinity with zero final speed. At infinity,
\[
K_{\infty}=0,
\qquad
U_{\infty}=0.
\]
Therefore the total energy for just escaping is
\[
E=0.
\]
Set the initial total energy equal to zero:
\[
\frac12mv_{\mathrm{esc}}^2-\frac{GMm}{r}=0.
\]
Move the potential-energy term to the other side:
\[
\frac12mv_{\mathrm{esc}}^2=\frac{GMm}{r}.
\]
The object mass \(m\) cancels:
\[
\frac12v_{\mathrm{esc}}^2=\frac{GM}{r}.
\]
Solving for \(v_{\mathrm{esc}}\) gives
\[
\boxed{
v_{\mathrm{esc}}=\sqrt{\frac{2GM}{r}}
}.
\]
3. Launch from the surface or from height
If the launch point is at height \(h\) above a body of radius \(R\), then the distance from the center is
\[
r=R+h.
\]
The escape speed from height \(h\) is
\[
\boxed{
v_{\mathrm{esc}}(R+h)=\sqrt{\frac{2GM}{R+h}}
}.
\]
At the surface, \(h=0\), so \(r=R\):
\[
\boxed{
v_{\mathrm{esc,surface}}=\sqrt{\frac{2GM}{R}}
}.
\]
4. Relationship with local gravity
The local gravitational field strength at radius \(r\) is
\[
g(r)=\frac{GM}{r^2}.
\]
Since \(GM=g(r)r^2\), escape speed can also be written as
\[
v_{\mathrm{esc}}
=
\sqrt{\frac{2g(r)r^2}{r}}
=
\boxed{\sqrt{2g(r)r}}.
\]
This form is useful when local \(g\) and radius \(r\) are known.
5. Why the rocket mass cancels
In the escape condition,
\[
\frac12mv_{\mathrm{esc}}^2=\frac{GMm}{r},
\]
both sides contain the escaping object’s mass \(m\). Dividing both sides by \(m\) removes it.
Therefore escape speed depends on the central mass \(M\) and launch radius \(r\), not on rocket mass.
However, the energy required to give a rocket that speed does depend on the rocket mass:
\[
K_{\mathrm{required}}=\frac12mv_{\mathrm{esc}}^2.
\]
6. Worked example: Earth surface
Use
\[
G=6.67430\times10^{-11}\ \mathrm{N\,m^2/kg^2},
\qquad
M_{\oplus}=5.9722\times10^{24}\ \mathrm{kg},
\qquad
R_{\oplus}=6.371\times10^6\ \mathrm{m}.
\]
At Earth’s surface,
\[
r=R_{\oplus}.
\]
Substitute:
\[
v_{\mathrm{esc}}
=
\sqrt{
\frac{2GM_{\oplus}}{R_{\oplus}}
}.
\]
\[
v_{\mathrm{esc}}
=
\sqrt{
\frac{2(6.67430\times10^{-11})(5.9722\times10^{24})}
{6.371\times10^6}
}.
\]
\[
v_{\mathrm{esc}}
\approx
1.12\times10^4\ \mathrm{m/s}.
\]
Therefore,
\[
\boxed{
v_{\mathrm{esc}}\approx11.2\ \mathrm{km/s}
}.
\]
7. Escape speed from orbit or altitude
Escape speed is smaller at larger radius. For example, at altitude \(h\),
\[
r=R+h.
\]
Then
\[
v_{\mathrm{esc}}(R+h)
=
v_{\mathrm{esc}}(R)
\sqrt{\frac{R}{R+h}}.
\]
This is why escape speed from low Earth orbit is slightly smaller than escape speed from Earth’s surface.
8. Black-hole event-horizon teaser
If Newtonian escape speed is set equal to the speed of light \(c\), then
\[
c=\sqrt{\frac{2GM}{r}}.
\]
Squaring both sides gives
\[
c^2=\frac{2GM}{r}.
\]
Solving for \(r\) gives the Schwarzschild radius:
\[
\boxed{
r_s=\frac{2GM}{c^2}
}.
\]
In general relativity, this radius is the event horizon of a non-rotating, uncharged black hole.
The calculator includes this as a conceptual preview, not as a full relativity model.
9. Typical surface escape speeds
| Body |
Approximate surface escape speed |
Comment |
| Moon |
2.38 km/s |
Much easier to escape than Earth |
| Earth |
11.2 km/s |
Standard reference value |
| Mars |
5.03 km/s |
Lower than Earth because Mars is less massive |
| Jupiter |
About 59.5 km/s |
Very high because Jupiter is extremely massive |
| Sun |
About 618 km/s |
Very high because the Sun is massive and compact compared with orbital distances |
10. Assumptions and limits
- The formula assumes a spherical or point-like central mass.
- The launch point must be outside the body for the simple outside-body formula to apply directly.
- Air resistance is ignored.
- No additional propulsion is included after the initial launch speed.
- The black-hole preview uses the Schwarzschild-radius idea and is not a full relativistic trajectory calculation.
Key idea: escape speed is found by giving the object enough kinetic energy to make its total mechanical energy zero.
