Deriving escape velocity using centripetal force 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.
 A: 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:
$$ \tfrac{1}{2}mv^2 = \frac{GMm}{r} = 0 $$
or:
$$ v = \sqrt{\frac{2GM}{r}} $$
A: 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 $v_{escape}$.
A: 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.
