# What is Escape Velocity – Step by Step Derivation

Two planets of radii (R) each, but mass (M) and (4M) have a center to center separation of (6R), as shown. A rocket of mass (m) is projected from the surface of planet of mass (M) directly towards the center of second planet. Obtain an expression for the minimum speed (v) of the rocket so that it reaches the surface of the second planet.

### Step by step derivation

The rocket is acted upon by two mutually opposing gravitational forces of the two planets. The neutral point (N) is defined as the position on O-C line where the two forces cancel each other exactly.

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Once the rocket is projected to neutral point (N), its velocity would be zero and then the gravitational pull of second planet would pull it to its surface. If ON = r, we have,

$\frac {G M m} {r^2} ~=~ \frac {4 G M m} {(6 R ~-~ r)^2}$

$(6 R ~-~ r)^2 ~=~ 4 r^2$

$(6 R ~-~ r) ~=~ \pm 2 r$

$r~=~2 R ~~"or" ~~ - 6 R$

The neutral point r = -6R does not concern us. Thus ON = r = 2R

It is sufficient to project the rocket with a speed that would enable it to reach N. The mechanical energy at surface of planet M is

$E_M ~=~ \frac {1}{2} m v^2 ~-~ \frac {G M m} {R} ~-~ \frac {4 G M m} {5 R }$

At neutral point N, the speed approaches zero. The mechanical energy at N is purely potential energy.

$E_N ~=~ ~-~ \frac {G M m} {2 R} ~-~ \frac {4 G M m} {4 R }$

Now applying the principle of conservation of energy,

$\frac {1}{2} v^2 ~-~ \frac {G M} {R} ~-~ \frac {4 G M} {5 R } ~=~ ~-~ \frac {G M} {2 R} ~-~ \frac {G M} {R}$

or

$v^2 ~=~ \frac {2 G M} {R} ( \frac {4} {5} ~-~ \frac {1} {2} )$

or

$v ~=~ ( \frac {3 G M} {5 R} )^ {1/2}$

At neutral point speed is zero, but when it strikes the second planet, it is not zero.

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