# A satellite of mass 2500 kg is orbiting the Earth i n an elliptical orbit. At the farthest point from the Earth, its altitude is 3600 km, while at t he nearest point, it is 1100 km. Calculate the...

A satellite of mass 2500 kg is orbiting the Earth i
n an elliptical orbit. At the farthest point
from the Earth, its altitude is 3600 km, while at t
he nearest point, it is 1100 km. Calculate
the energy and angular momentum of the satellite an
d its speed at the aphelion and
perihelion.

Charlie Hooper | Certified Educator

Gravitational potential energy, U:

`U= -(G*M*m)/r`

where: G = gravitational constant, m = satellite's mass, and M = Earth's mass

Aphelion is where the satellite is furthest away:

`r_1= R_E + 3600 km=9.967*10^6 m`

Perihelion is where the satellite is the closest:

`r_2= R_E + 1100 km=7.467*10^6 m`

Conservation of Energy:

`K_1+U_1 = K_2+U_2`

`1/2*m*v_1^2-G*m*M/r_1=1/2*m*v_2^2-G*m*M/r_2`

Conservation of Angular Momentum:

`mv_1 r_1=mv_2 r_2`

Solve system of equations for v2.

`1/2*m*(v_2*r_2/r_1)^2-G*m*M/r_1=1/2*m*v_2^2-G*m*M/r_2`

`v_2^2 = 2GM*(1/r_1-1/r_2)/((r_2/r_1)^2-1)`

Plug in numerical values to and solve to get the velocities.

`v_1 = 4387.3 m/s` at aphelion, and `5856.2 m/s` at perihelion.

Next, plug in either position 1 or position 2 values to get E at BOTH aphelion and perihelion of `E=K+U=1/2*m*v_1^2-G*m*M/r_1=7.603*10^10 J`

Your angular momentum will also be the same for both aphelion and perihelion

`J=m*r_1*v_1=1.09x10^14 (kgm^2)/s`

Therefore this problem was completly solved by the conservation of energy and angular momentum.

I included a link that attempts to solve this same problem in more detail. There numerical values are incorrect because they are missing a factor of two when they solve for the velocity.

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