Recently started book on gravity and there (not only there, it is correct way) the escape velocity was given as $${v_{esc}=\sqrt{\frac{2GM}{R}}}.$$ I wanted to derive this myself and I decided to use centripetal force as key to derivation. My idea is that the gravitational force is equal to the centripetal force of the object. I got: $${F_{grav}=\frac{GMm_{obj}}{R^2};}$$ $${F_{cntr}=m_{obj}\frac{v^2}{R};}$$ $${\frac{GMm_{obj}}{R^2}=m_{obj}\frac{v^2}{R};}$$ $${\frac{GM}{R}={v^2};}$$ $${{v}=\sqrt\frac{GM}{R}}$$ So my question is where is the mistake?

EDIT:I would be very pleased if someone explain the other derivation with potential energy too. Especially the fact that the potential gravitational energy must be equal to the kinetic energy.

  • $\begingroup$ what you derived would just be the tangential velocity of the particle at a certain distance R away from the centre of mass of the earth - particle system. $\endgroup$
    – inya
    Jun 18, 2015 at 15:59
  • $\begingroup$ I agree but is not that the definition of escape velocity-the tangent velocity or it could be velocity perpendicular to the area of sphere at radius R. $\endgroup$
    – Blake
    Jun 18, 2015 at 16:11
  • $\begingroup$ No, if you find the tangential velocity that is related to a centripetal acceleration on a circular path then all you have found is the velocity to maintain that circular path, not escape from the gravity well. $\endgroup$
    – Bill N
    Jun 18, 2015 at 18:11

3 Answers 3


What you have derived is the equation for the orbital velocity for a circular orbit. This is the velocity at which the acceleration required to keep moving in a circle exactly matches the acceleration due to gravity.

To derive the escape velocity you need to know that the potential energy for an object of mass $m$ in the gravitational field of a planet with mass $M$ is given by:

$$ V = -\frac{GMm}{r} $$

The kinetic energy is as usual:

$$ K = \tfrac{1}{2}mv^2 $$

The total energy of a body stationary at infinity is zero, so a body can just escape the gravitational field if its total energy is zero. The total energy is just the sum of the kinetic and potential energies, so we have:

$$ E = K + V = \tfrac{1}{2}mv^2 - \frac{GMm}{r} = 0 $$

and rearranging gives:

$$ \tfrac{1}{2}mv^2 = \frac{GMm}{r} = 0 $$


$$ v = \sqrt{\frac{2GM}{r}} $$

  • $\begingroup$ So we take escape velocity as the velocity at which the object with escape the gravitational field while moving perpendicular to the surface of Earth for example. Is that the definition? $\endgroup$
    – Blake
    Jun 18, 2015 at 16:17
  • $\begingroup$ The object doesn't have to be moving perpendicular to the surface. The kinetic energy is the same no matter which way the object is moving. If it's moving sideways it will still escape but the trajectory will be a spiral instead of a straight line outwards. $\endgroup$ Jun 18, 2015 at 16:19
  • $\begingroup$ Yeah, I meant that. I wanted to say that the perpendicular component of the velocity vector should be greater or equal to the escape velocity (if the case is that we want to shoot rocket slightly on angle but above us). Sorry if it is still confusing.(Maybe it would be simpler if we consider the Earth flat and infinite) $\endgroup$
    – Blake
    Jun 18, 2015 at 16:24
  • 1
    $\begingroup$ @user1163511: no! The perpendicular component of the velocity can be zero or even negative and the object will still escape. Suppose the object starts moving tangential to the surface, so the perpendicular component is zero. Because $v$ is greater than the orbital velocity (twice as great in fact) the object starts accelerating outwards and moves outwards (to infinity) in a spiral trajectory. $\endgroup$ Jun 18, 2015 at 16:29
  • $\begingroup$ Thank you very much for the explanation. One last question-if the Earth is flat and infinite-is now the perpendicular component of the velocity the escape velocity. $\endgroup$
    – Blake
    Jun 18, 2015 at 16:50

When a rocket is fired from Earth with a sudden impulse, its total energy is given by: $$E_k \text{ (kinetic energy)} + E_p \text{ (potential energy)}= \frac{1}{2}mv^2 - \frac{GMm}{r} = constant$$ where the variables have their usual meaning. The potential energy here is taken to be negative because the reference point chosen for potential energy to be zero is when the rocket is unbound in Earth's orbit. Hence, after the rocket is fired (with no propulsion after the initial impulse) it is bound if its: $$E_{total} < 0$$ and unbound if its: $$E_{total} \geq 0$$

Setting $E_{total}$ to zero, you can rearrange and solve for $v_{escape}$.

  • $\begingroup$ Thank you, I appreciate although I have already marked it as answered. It took me time but got it right. $\endgroup$
    – Blake
    Jul 10, 2015 at 10:22

Your derivation is for equality of force at a certain distance from the center. This equality will not hold at other distances as the particle is released. The correct derivation is the one involving energy and is first one involving a factor of 2.


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