(a) 0.448
The gravitational potential energy of a satellite in orbit is given by:
[tex]U=-\frac{GMm}{r}[/tex]
where
G is the gravitational constant
M is the Earth's mass
m is the satellite's mass
r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):
r = R + h
We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as
[tex]\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}[/tex]
and so, substituting:
[tex]R=6370 km\\h_A = 5970 km\\h_B = 21200 km[/tex]
We find
[tex]\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448[/tex]
(b) 0.448
The kinetic energy of a satellite in orbit around the Earth is given by
[tex]K=\frac{1}{2}\frac{GMm}{r}[/tex]
So, the ratio between the two kinetic energies is
[tex]\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}[/tex]
Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.
(c) B
The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:
[tex]E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}[/tex]
For satellite A, we have
[tex]E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J[/tex]
For satellite B, we have
[tex]E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J[/tex]
So, satellite B has the greater total energy (since the energy is negative).
(d) [tex]-2.57\cdot 10^8 J[/tex]
The difference between the energy of the two satellites is:
[tex]E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J[/tex]
