Question

A satellite of mass mS orbits the Earth (with mass mE and radius RE) with a velocity v and an
altitude h. The gravitational force FG and the centripetal force FC are given by:
FG = G
mS · mE
(RE + h)
2
, FC =
mS · v
2
RE + h
, G = const.
(a) Find an equation for the kinetic energy Ekin(h) of a satellite with an altitude h.
(b) Based on the kinetic energy, how much liquid hydrogen (energy density: 106
J/Litre) is at least
needed to bring a small 1 kg satellite in an orbit of 400 km. (Use literature to find mE, RE, G.)

Answer 1

If the satellite is at height "h" then the Kinetic energy will be K.E = GMm/r + h

Explanation:

Question statement:

Find an equation for the kinetic energy Ekin(h) of a satellite with an altitude h.?

Solution:

The kinetic energy of an object in orbit can easily be found from the following equations:

Centripetal force on a satellite of mass m moving at velocity v in an orbit of radius r = mv^2/r

F =GMm/r^2 and so mv^2 = GMm/r

But kinetic energy K.E = mv^2 and so:

Kinetic energy of the satellite K.E = GMm/r

If the satellite is at height "h" then the Kinetic energy will be K.E = GMm/r + h

Answer 2

Answer:

the kinetic energy of the satellite = 1/2*G * ME*Ms / (RE+h)

Explanation:

(a) Let's assume RE + h = L

As Fc = Ms * v^2 / L and Fg = G * Ms*ME / L^2 and Kinetic energy = 1/2 M*v^2

As the centripetal force (in this situation) is the same as the gravitational force, we'll conclude the

kinetic energy by saying that Fc = Fg which is Ms * v^2 / L = G * Ms*ME / L^2

This will end with Ms*v^2 = G * ME*Ms / L

Finally, by dividing both sides by 2, 1/2 M*v^2 = 1/2*G * ME*Ms / L = 1/2*G * ME*Ms / (RE+h)

So, the kinetic energy of the satellite = 1/2*G * ME*Ms / (RE+h)