Calculating Final Velocity of a Spacecraft Crashing into the Moon

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SUMMARY

The discussion focuses on calculating the final velocity of a spacecraft crashing into the Moon after reducing its speed by 20 m/s from a circular orbit at an altitude of 50 km. The initial orbital velocity is determined using the formula for circular orbit velocity, yielding 1678 m/s. After accounting for the decrease in speed and the potential energy conversion during the fall, the calculations yield incorrect results compared to the expected answer of 6060 km/h. The key equations utilized include the conservation of energy and the gravitational parameters of the Moon.

PREREQUISITES
  • Understanding of gravitational physics, specifically the equations of motion in a gravitational field.
  • Familiarity with the formula for circular orbital velocity, specifically (Gm/r)^(1/2).
  • Knowledge of energy conservation principles in physics.
  • Ability to perform vector addition of velocities in a two-dimensional motion context.
NEXT STEPS
  • Study the derivation and application of the circular orbital velocity formula (Gm/r)^(1/2).
  • Learn about gravitational acceleration variations at different altitudes, particularly around celestial bodies like the Moon.
  • Explore the concept of energy conservation in gravitational fields and its implications for falling objects.
  • Investigate vector addition techniques for calculating resultant velocities in two-dimensional motion scenarios.
USEFUL FOR

Students and educators in physics, aerospace engineers, and anyone interested in orbital mechanics and gravitational effects on spacecraft motion.

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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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