# Work done in moving a body out of Gravitational influence

By "out of Gravitational influence ", I suppose it's meant that Gravity of a planet has no effect whatsoever on the body. Am I correct?

Could we do it by making the object reach Escape Velocity by doing Work = GMm/2D? {assuming circular orbit}

(m, mass of object;M, mass of planet; D= distance from center)

It could also be done by taking the object to infinity (hypothetically) This would require a work of GMm/D

In first case total mechanical energy is zero but in second it is equal to Kinetic Energy.

In which of these is the body actually out of Gravitational influence of the Planet?

Or is there some other way?

What should the total work done be in putting it out of Gravitational influence of a planet?

By "out of Gravitational influence ", I suppose it's meant that Gravity of a planet has no effect whatsoever on the body. Am I correct?

Yes, that's correct.

An object that is subjected to a central gravitational field has potential energy $U$, given by:

$$U(r)=-\frac{GMm}{r},$$

where $G$ is the universal constant of gravity, $M$ the mass of the central object, $m$ the mass of the object and $r$ the distance between the centres of gravity of the objects.

For $r \to \infty$, $U \to 0$ and the object is no longer under the influence of the gravitational field. Due to conservation of energy the work $W$ needed to move the object from $r$ to $\infty$ is:

$$W=U(\infty)-U(r),$$

$$W=0-(-\frac{GMm}{r}),$$

Or:

$$W=\frac{GMm}{r}.$$

If the object was launched from position $r$ with an initial velocity $v$ (and with no means of propulsion/thrust) then if:

$$\frac{mv^2}{2}=\frac{GMm}{r},$$

$$v_e=\sqrt{\frac{2GM}{r}},$$

would be the escape velocity.