Question

How much gravitational potential energy does a system comprising a 102-kg object and Earth have if the object is one Earth radius above the ground?
Answer: U = −3.19×109 J
How fast would a 102-kg object have to be moving at this height to have zero energy?

Answer 1

1) $-3.19\cdot 10^9 J$

2) 7909 m/s

Explanation:

1)

The potential energy of a system consisting of the Earth and an object in orbit around the Earth is given by

$U=-\frac{GMm}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the mass of the object

r is the distance between the object and the Earth's centre

Here we have:

$M=5.98\cdot 10^{24} kg$

m = 102 kg is the mass of the object

r = 2R is the distance of the object from the Earth's centre, where

$R=6.37\cdot 10^6 m$ is the Earth's radius

Substituting,

$U=-\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(102)}{2(6.37\cdot 10^6)}=-3.19\cdot 10^9 J$

2)

The total mechanical energy of the object is the sum of its potential energy and its kinetic energy:

$E=U+K$

where

E is the mechanical energy

U is the potential energy

K is the kinetic energy

Here we want the object to have zero energy, so

E = 0

This means that the kinetic energy is

$K=-U=3.19\cdot 10^9 J$

The kinetic energy of the object can be rewritten as

$K=\frac{1}{2}mv^2$

where

m = 102 kg is the mass

v is the speed of the object

And solving for v, we find:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.19\cdot 10^9)}{102}}=7909 m/s$