Question

An object of mass m is dropped from height h above a planet of mass M and radius R .

Answer 1

Answer:

$v = \sqrt{2GM(\frac{1}{R+h)}-\frac{1}{R})}$

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:

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

where

v is the speed of the object at the ground

Since the mechanical energy is conserved, we can write

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

and solving for v, we find

$\frac{GMm}{R}+\frac{1}{2}mv^2=\frac{GMm}{R+h}\\v = \sqrt{2GM(\frac{1}{R+h)}-\frac{1}{R})}$