Calculating Final Velocity of a Spacecraft Crashing into the Moon

AI Thread Summary
A spacecraft in a circular orbit around the Moon at 50 km altitude decreases its speed by 20 m/s, leading to a new speed of 1658 m/s. The calculation of the additional velocity gained from falling towards the Moon was determined to be 989.9 m/s. Attempts to combine these velocities using both direct addition and the Pythagorean theorem yielded incorrect results, with the expected final speed being 6060 km/h. The discussion highlights the importance of correctly applying gravitational acceleration relevant to the Moon rather than Earth. The calculations reveal the complexities involved in determining the final velocity during a lunar descent.
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Homework Statement


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?



Homework 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



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.
 
Physics news on Phys.org
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.
 
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