Answer :
Answer:
[tex]\displaystyle \frac{R}{R + 2\, h}[/tex],
Where [tex]R[/tex] is the radius of the Earth, assuming that gravitational potential energy is measured relative to when the satellite is on the surface of the Earth and stationary.
Explanation:
Let [tex]R[/tex] denote the radius of the Earth. At a height of [tex]h[/tex] from the surface of the Earth, the satellite would be at a distance of [tex](R + h)[/tex] from the center of mass of the Earth. The gravitational potential energy of the satellite would be:
Where [tex]G[/tex] is the gravitational constant, [tex]M[/tex] is the mass of the Earth, and [tex]m[/tex] is the mass of the satellite.
[tex]\displaystyle - \frac{G\, M\, m}{R + h}[/tex].
In comparison, when the satellite is on the surface of the Earth, gravitational potential energy would be:
[tex]\displaystyle - \frac{G\, M\, m}{R}[/tex].
Hence, the gravitational potential energy of the satellite in orbit, relative to when the satellite was on the surface, would be:
[tex]\begin{aligned} & - \frac{G\, M\, m}{R + h} - \left(- \frac{G\, M\, m}{R}\right) \\ =\; & \frac{G\, M\, m}{R} - \frac{G\, M\, m}{R + h} \\ =\; & \left(\frac{1}{R} - \frac{1}{R + h}\right)\, G\, M\, m\end{aligned}[/tex].
Let [tex]v[/tex] denote the tangential velocity of the satellite. Since the satellite is in a centripetal motion of radius [tex](R + h)[/tex], the net force on the satellite would be:
[tex]\displaystyle \frac{m\, v^{2}}{R + h}[/tex].
Assume that gravitational attraction [tex](G\, M\, m) / ((R + h)^{2})[/tex] is the only force on the satellite in the orbit:
[tex]\displaystyle \frac{m\, v^{2}}{R + h} = (\text{net force}) = \frac{G\, M\, m}{(R + h)^{2}}[/tex].
The kinetic energy of the satellite would be:
[tex]\displaystyle (\text{KE}) = \frac{1}{2}\, m\, v^{2} = \frac{1}{2}\, \left(\frac{G\, M\, m}{R + h}\right)[/tex].
Hence, relative to when the satellite was stationary on the surface of the Earth, the total energy of the satellite while in orbit would be:
[tex]\begin{aligned} & (\text{KE}) + (\text{GPE}) \\ = \; & \frac{1}{2}\, \left(\frac{G\, M\, m}{R + h}\right) + \left(\frac{1}{R} - \frac{1}{R + h}\right)\, G\, M\, m \\ =\; & (G\, M\, m)\, \left(\frac{(1/2)}{R + h} + \frac{1}{R} - \frac{1}{R + h}\right) \\ =\; & (G\, M\, m)\, \left(\frac{1}{R} - \frac{(1/2)}{R + h}\right) \\ =\; & (G\, M\, m)\, \left(\frac{R + h - (1/2)\, R}{R\, (R + h)}\right) \\ =\; & (G\, M\, m)\, \left(\frac{(1/2)\, R + h}{R\, (R + h)}\right)\end{aligned}[/tex].
The ratio between kinetic energy and total energy (again, relative to when the satellite was on the surface of the Earth and stationary) would be:
[tex]\begin{aligned}& \frac{\displaystyle (G\, M\, m)\, \left(\frac{(1/2)}{R + h}\right)}{\displaystyle (G\, M\, m)\, \left(\frac{(1/2)\, R + h}{R\, (R + h)}\right)} \\ =\; & \frac{(1/2)\, R}{(1/2)\, R + h} \\ =\; & \frac{R}{R + 2\, h}\end{aligned}[/tex].
The ratio of kinetic energy to total energy for a satellite in a circular orbit is 1.
To find the ratio of kinetic energy to total energy for a satellite moving around the earth at a height h from the surface, we begin by understanding the mechanics of satellite motion in a circular orbit.
The gravitational potential energy (U) of a satellite at a distance r from the center of the Earth (which is the sum of the Earth's radius and the satellite's altitude h) is given by the formula:
U = - G M m / r,
where G is the gravitational constant, M is the mass of the Earth, and m is the mass of the satellite.
The kinetic energy (K) of the satellite is:
K = 1/2 (m v²)
but since the satellite is in orbit, its velocity can be found from the gravitational force equating to the centripetal force required for circular motion, resulting in:
K = G M m / (2 r)
The total mechanical energy (E) of the satellite is the sum of its kinetic and potential energy:
E = K + U = - G M m / (2 r)
The ratio of kinetic energy to the total energy is then calculated as:
R = K / E = [- G M m / (2 r)] /[- G M m / (2 r)] = 1
