Respuesta :
(a) 0.439
The potential energy of a satellite in orbit is given by
[tex]U=-\frac{GmM}{R+h}[/tex]
where
G is the gravitational constant
m is the mass of the satellite
M is the mass of the Earth
R is the Earth's radius
h is the altitude of the satellite
If we call
[tex]U_A=-\frac{GmM}{R+h_A}[/tex]
the potential energy of satellite A, with
[tex]h_A = 6380 km = 6.38\cdot 10^6 m[/tex]
being its altitude, and
[tex]U_B=-\frac{GmM}{R+h_B}[/tex]
the potential energy of satellite B, with
[tex]h_B = 22700 km = 22.7\cdot 10^6 m[/tex]
being the altitude of satellite B
and
[tex]R=6370 km = 6.37 \cdot 10^6 m[/tex] being the Earth's radius
The ratio between the potential energy of satellite B to that of satellite A will be
[tex]\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}=\frac{6.37\cdot 10^6 m+6.38\cdot 10^6 m}{6.37\cdot 10^6 m+22.7\cdot 10^6 m}=0.439[/tex]
(b) 0.439
The kinetic energy of a satellite in orbit has a similar expression to the potential energy
[tex]K=\frac{1}{2} \frac{GmM}{R+h}[/tex]
As before, if we call
[tex]K_A=\frac{1}{2} \frac{GmM}{R+h_A}[/tex]
the kinetic energy of satellite A, with
[tex]h_A = 6380 km = 6.38\cdot 10^6 m[/tex]
being its altitude, and
[tex]K_B=\frac{1}{2} \frac{GmM}{R+h_B}[/tex]
the kinetic energy of satellite B, with
[tex]h_B = 22700 km = 22.7\cdot 10^6 m[/tex]
being the altitude of satellite B,
the ratio between 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}=\frac{6.37\cdot 10^6 m+6.38\cdot 10^6 m}{6.37\cdot 10^6 m+22.7\cdot 10^6 m}=0.439[/tex]
(c) Satellite B
The total energy of each 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}[/tex]
While for satellite B we have
[tex]E_B = -\frac{1}{2}\frac{GMm}{R+h_B}[/tex]
We see that the total energy is inversely proportional to the altitude of the satellite: therefore, the higher the satellite, the smaller the energy. So, satellite A will have the greater total energy (in magnitude), since [tex]h_A < h_B[/tex]; however, the value of the total energy is negative, so actually satellite B will have a greater energy than satellite A.
(d) [tex]3.07\cdot 10^8 J[/tex]
The total energy of satellite A is
[tex]E_A = -\frac{1}{2}\frac{GMm}{R+h_A}[/tex]
with
[tex]h_A = 6380 km = 6.38\cdot 10^6 m[/tex]
while the total energy of satellite B is
[tex]E_B = -\frac{1}{2}\frac{GMm}{R+h_B}[/tex]
with
[tex]h_B = 22700 km = 22.7\cdot 10^6 m[/tex]
So the difference between the two energies is
[tex]E_B - E_A = -\frac{1}{2}\frac{(6.67\cdot 10^{-11}(35 kg)(5.98\cdot 10^{24} kg)}{6.37\cdot 10^6 m +22.7\cdot 10^6 m}-(-\frac{1}{2}\frac{(6.67\cdot 10^{-11}(35 kg)(5.98\cdot 10^{24} kg)}{6.37\cdot 10^6 m +6.38\cdot 10^6 m})=3.07\cdot 10^8 J[/tex]
