**Problem : **
What is the gravitational potential energy of the moon with respect to the
earth? The mass of the moon is 7.35×10^{22} kilograms and the mass of
the earth is 5.98×10^{24} kilograms. The earth moon distance is 384
400 kilometers.

Plugging into the formula,

*U* = - = - = - 7.63×10^{22} Megajoules.

**Problem : **
What is the gravitational potential with respect to the sun at the position
of the earth? The mass of the sun is 1.99×10^{30} kilograms and the
mass of the earth is 5.98×10^{24} kilograms. The mean earth-sun
distance is 150×10^{6} kilometers.

We can just use the formula:

*Φ*_{g} = = = 8.85×10^{8} J/kg.

**Problem : **
What is the total energy of a 90 kilogram satellite with a perigee distance 595
kilometers and apogee distance 752 kilometers, above the surface of the earth?
The mass of the earth is 5.98×10^{24} kilograms and its radius is 6.38×10^{6} m.

The total energy of a satellite in orbit is given by

*E* = , where

*a* is the semi-major axis length of orbit. The perigee distance from the
center of the earth is

595000 + 6.38×10^{6} m and the apogee distance
is

752000 + 6.38×10^{6}. The semi-major axis length is given by

(595000 + 752000 + 2×6.38×10^{6})/2 = 7.05×10^{6} m.
The energy is therefore:

= 2.55×10^{9} Joules.

**Problem : **
Calculate the orbital energy and orbital speed of a rocket of mass 4.0×10^{3} kilograms and radius 7.6×10^{3} kilometers above the center of the
earth. Assume the orbit is circular. (*M*_{e} = 5.98×10^{24} kilograms).

The total orbital energy of a circular orbit is given by:

*E* = - = - 1.05×10^{11} Joules. The kinetic contribution is

*T* = = 1.05×10^{11} Joules This is also equal to

1/2*mv*^{2} so we can find the
orbital speed as

*v* = = = 7.2×10^{4} m/s.

**Problem : **
A satellite of mass 1000 kilograms is launched with a speed of 10 km/sec. It
settles into a circular orbit of radius 8.68×10^{3} km above the center
of the earth. What is its speed in this orbit? (*M*_{e} = 5.98×10^{24} and
*r*_{e} = 6.38×10^{6} m).

This problem involves the conservation of energy. The initial kinetic energy is
given by

1/2*mv*^{2} = 1/2×1000×(10000)^{2} = 5×10^{1}0 Joules.
It also has some initial gravitational potential energy associated with its
position on the surface

*U*_{i} = - = - 6.25×10^{1}0
Joules. The total energy is then given by

*E* = *T* + *U*_{i} = - 1.25×10^{10}
Joules. In its new orbit the satellite now has a potential energy

*U* = - = - 4.6×10^{10} Joules. The kinetic energy is given by

*T* = *E*–*U* = (- 1.25 + 4.6)×10^{10} = 3.35×10^{10} Joules. We can
easily now find the velocity:

*v* = = 8.1×10^{3} m/s.