2 VMain Question The Command-Service Module (CSM) orbited the moon while Apollo

Question

2 VMain Question The Command-Service Module (CSM) orbited the moon while Apollo astronauts took the Lunar Mod- ule (LM) down to the surface below for a moon walk. After walking on the moon, the LM took off from the surface, reaching orbit and coming to rest. Meanwhile, the CSM was moving towards it at a velocity of u = 1575 m/s from a distance d away. The LM needs to accelerate (in the same direction the CSM is moving) in order to dock with the CSM- a successful docking requires both the LM and CSM to be at approximately the same position, moving with approximately the same velocity. The LM can accelerate at a constant rate of 7.5 m/s?, and the initial situation (when the LM first begins accelerating) is shown in the figure below. Hint: Which way should the LM accelerate to match velocity with the CSM? a) How much time will it take the LM to match ve- locity with the CSM for docking (assuming the CSM doesn't change its velocity)? b) How far will the CSM travel trom the time that the LM begins its burn to the time that they have the same velocity? CSM LM c) How far will the LM travel in this time? d) What is the inital distance between the LM and Give all distances in km. the CSM (d in the figure)? Distance for CSM: Time to match: d: Distance for LM:

Explanation / Answer

Given,

Vcsm = 1575 m/s ; aLM = 7.5 m/s^2

a)We know from eqn of motion

v = u + at

final velocity should match that of CSM

1575 = 0 + 7.5 t => t = 1575/7.5 = 210 s

Hence, t = 210 s

b)CSM travels, distance equal to

d = Vcsm x t

d = 1575 x 210 = 330750 m

Hence, d = 330750 m = 330.75 km

c)from eqn of motion

v^2 = u^2 + 2 a S

S = (v^2 - u^2)/2a

S = (1575^2 - 0)/2 x 7.5 = 165375 m

Hence, S = 165375 m = 165.38 km

d)Intial distance is:

d = 330.75 - 165.38 = 165.37 km

Hence, d = 165.37 km

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