A satellite is put into an elliptical orbit around the Earth. When the satellite is at its perigee, its nearest point to the Earth, its height above the ground is
h
p
=
215.0
km,
and it is moving with a speed of
v
p
=
8.450
km/s.
The gravitational constant
G
equals
6.67
×
10
−
11
m
3
⋅kg
−
1
⋅s
−
2
and the mass of Earth equals
5.972
×
10
24
kg.
When the satellite reaches its apogee, at its farthest point from the Earth, what is its height
h
a
above the ground?

For this problem, choose gravitational potential energy of the satellite to be
0
at an infinite distance from Earth.

Amass m = 74 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius r = 18.4 m and finally a flat straight section at the same height as the center of the loop (18.4 m off the ground). since the mass would not make it around the loop if released from the height of the top of the loop (do you know why? ) it must be released above the top of the loop-the-loop height. (assume the mass never leaves the smooth track at any point on its path.) 1. what is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track? 2. what height above the ground must the mass begin to make it around the loop-the-loop? 3. if the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop? 4. if the mass has just enough speed to make it around the loop without leaving the track, what is its speed at the final flat level (18.4 m off the ground)? 5. now a spring with spring constant k = 15600 n/m is used on the final flat surface to stop the mass. how far does the spring compress?