Question

An object of mass $m$is dropped from height $h$above a planet of mass $M$and radius $R.$Find an expression for the object’s speed as it hits the ground.

Short answer

the expression for object's speed is$v=\sqrt{2GM\left(\frac{1}{R+h}-\frac{1}{R}\right)}$

Step-by-step solution

Given information

mass of an object is$m$

height of an object to be dropped is $h$

mass of the planet $M$

radius of the planet$R$

Explanation

The initial mechanical energy of the object, when it is located at height h above the the planet, is just gravitational potential energy:

$E=U=\frac{GMm}{(R+h)}$

where

$G$is the gravitational constant

$M$is the mass of the planet

$m$is the mass of the object

$R$is the radius of the planet

$h$is the altitude of the object

When the object hits the ground, its mechanical energy will sum of potential energy and kinetic energy:

$E=\frac{GMm}{R}+\frac{1}{2}m{v}^2$

where

$v$is the speed of the object at the ground

Since the mechanical energy is conserved, we can write

$\frac{GMm}{R}+\frac{1}{2}m{v}^2=\frac{GMm}{R+h}$

and solving for $v$, we find

$$\begin{gathered} v^2=2GM\left(\frac{1}{R+h}-\frac{1}{R}\right) \\ v=\sqrt{2GM\left(\frac{1}{R+h}-\frac{1}{R}\right)} \end{gathered}$$