Physics
posted by Dave on .
A 3000kg satellite orbits the Earth 400km above the equator. find
a)The speed of the satellite
b) The kinetic energy of the satellite
c) The potential energy of the satellite
d) The total energy of the satellite

a) You need the radius of the Earth,
Re = 6370 km
The orbit radius is then
R = Re + h = 6770 km
The acceleration of gravity at that altitude is
g' = g*(Re/R)^2 = 8.685 m/s^2
Set the gravitational acceleration g' equal to the centripetal acceleration and solve for V
V^2/R = 8.685 m/s^2
V = 7670 m/s
b) K.E. = (1/2) M V^2
c) You need to specify where the potential energy is defined to be zero.
Usually in cases like this it is assumed zero at infinity.