1
$\begingroup$

A body is projected vertically upwards from the surface of the earth with a velocity equal to half the escape velocity. If $R$ is the radius of the earth, then find the maximum height attained by the body.

If we use Law of Conservation of Energy,

$$\frac{-GMm}{R}+\frac12mu^2=\frac{-GMm}{R+h}+0$$

Here, $\displaystyle u=\sqrt\frac{GM}{2R}$.

On solving this, we get $\displaystyle h=\frac R3$

If we use kinematical equations,

$$v^2=u^2-2gh$$

Here, $v=0,\;\displaystyle u=\sqrt\frac{GM}{2R}=\sqrt\frac{gR}{2}$

$$\therefore \frac{gR}{2}=2gh\implies h=\frac R4$$

Why do we get different answers?

$\endgroup$
8
$\begingroup$

You cannot use the second kinematical equation because it is valid only when the acceleration due to gravity, $g$ , is constant. This is incorrect for distances comparable to the radius of the earth, and velocities comparable to the escape velocity. The first correctly assumes a $\frac{1}{R^2}$ fall-off of the gravitational attraction on the body due to Earth's pull.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.