(a) 0.473
The potential energy of a satellite orbiting around Earth is given by
[tex]U=-\frac{GMm}{R+h}[/tex]
where
G is the gravitational constant
M is the Earth's mass
m is the satellite's mass
R is the Earth's radius
h is the altitude of the satellite above the Earth's surface
So the potential energy of satellite A is
[tex]U_A=-\frac{GMm}{R+h_A}[/tex]
while potential energy of satellite B is
[tex]U_B=-\frac{GMm}{R+h_B}[/tex]
Therefore the ratio of the potential energy of satellite B to that of satellite A is
[tex]\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}[/tex]
and using
hA = 5920 km
hB = 19600 km
R = 6370 km
we find
[tex]\frac{U_B}{U_A}=\frac{6370+5920}{6370+19600}=0.473[/tex]
(b) 0.473
The kinetic energy of a satellite orbiting around Earth instead is given by
[tex]K=\frac{GMm}{2(R+h)}[/tex]
So the kinetic energy of satellite A is
[tex]K_A=\frac{GMm}{2(R+h_A)}[/tex]
while kinetic energy of satellite B is
[tex]K_B=\frac{GMm}{2(R+h_B)}[/tex]
Therefore the ratio of the kinetic energy of satellite B to that of satellite A is
[tex]\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}[/tex]
which is identical to before, so it gives
[tex]\frac{K_B}{K_A}=\frac{6370+5920}{6370+19600}=0.473[/tex]
(c) Satellite B
The total energy of a satellite in orbit is given by
[tex]E=U+K = -\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}[/tex]
We see that the total energy is:
1) negative (because the satellite is on a bound orbit)
2) inversely proportional to the distance of the satellite from the Earth's center, R+h
So the magnitude of the fraction in the equation is larger for the satellite which is closer to the Earth's surface (satellite A), but since the energy is negative, this means that the total energy of this satellite is smaller than that of satellite B. So, satellite B has a greater total energy.
(d) [tex]1.03\cdot 10^7 J[/tex]
We have to calculate the total energy of each satellite.
Given:
[tex]G=6.67\cdot 10^{-11}[/tex]
[tex]M=5.98\cdot 10^{24} kg[/tex]
m = 12.0 kg
[tex]R+h_A = 6370 km+5920 km=12290 km = 12.3 \cdot 10^6 m[/tex]
[tex]R+h_B = 6370 km+19600 km=25970 km = 26.0 \cdot 10^6 m[/tex]
We find:
[tex]E_A = - \frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(12.0)}{2(12.3\cdot 10^6)}=-1.95\cdot 10^{7} J[/tex]
[tex]E_B = - \frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(12.0)}{2(26.0\cdot 10^6)}=-9.2\cdot 10^{6} J[/tex]
So the difference in total energy is
[tex]E_B-E_A = -9.2\cdot 10^6 - (-1.95\cdot 10^7) =1.03\cdot 10^7 J[/tex]
