# Newtonian Gravity Problem

1. Nov 16, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data
A spacecraft is in a circular orbit around the moon, observing the lunar surface from an altitude of 50 km. The on board thrusters fire, decreasing the speed of the spacecraft by 20 m/s, What is the speed (in km/h) in which the spacecraft crashes into the moon?

2. Relevant equations
velocity of circular orbit = (Gm/r)^(1/2)
conservation of energy
v^2/r = a_c
radius of moon = 1.74E6 meters
Mass of moon = 7.35E24 kg

3. The attempt at a solution
so you can solve the the velocity of the circular orbit and it comes out to 1678. If you decrease this by 20 as the problem suggests it becomes 1658. From here I took v = (2gh)^(1/2) to see how much velocity would be added from the decreasing potential energy as it falls towards the moon. This calculation led me to it would gain 989.9 m/s in the direction towards the moon. It would still have 1658 m/s in the direction tangential to it's initial centripetal acceleration. adding the 1658 m/s + 989.9 m/s = 2647.9 m/s which comes out to 9532.44 km/h which his incorrect.

If I take (1658^2 + 989.9^2)^(1/2) I get 1931 m/s which comes out to 6951.7 km / hour which is also wrong.

The back of the book is looking for 6060 km/h, i'm having problems getting this number.

2. Nov 17, 2013

### voko

g = 9.8 m/s is the acceleration due to gravity at the surface of the Earth. It has nothing to do with the acceleration due to gravity at 50 km off the Moon.