Question: On Apollo Moon missions, the lunar module would blast off from the Moo

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Question: On Apollo Moon missions, the lunar module would blast off from the Moon's surface and dock with the command module in lunar orbit. After docking, the lunar module would be jettisoned and allowed to crash back onto the lunar surface. Seismometers placed on the Moon's surface by the astronauts would then pick up the resulting seismic waves.

Find the impact speed of the lunar module, given that it is jettisoned from an orbit 110 km above the lunar surface moving with a speed of 1600 m/s .

My Approach:
Ei = Ef

1/2*m*vi2 - (G*m*ME)/(radius of moon + orbital distance) = 1/2*m*vf2 - (G*m*ME)/r

=> (0.5 * m * 16002) - (6.67 * 10-11 * 7.35*1022 * m/(1737.4*103 +180*103) = (0.5*m*v^2) - (6.67*10-11 * 7.35*1022 * m/(1737.4*103 )

The m's cancel and I got v = 2421.62 m/s, which is incorrect!

I would really appreciate any help in trying to solve this question. Thank you!

Explanation / Answer

We can use conservation of energy to find the crash speed ...Let K = kinetic energy = 1/2*m*v^2 & U is potential energy = -GMm/r The mass of the moon is 7.36x10^22kg and its radius is 1738 km

So (K + U) orbit = (K + U)crash

So 1/2*m*vo^2 - GMm/(1738000 + 110000) = 1/2*m*vc^2 - GMm/1738000. Note mass drops out leaving

vo^2/2 - GM/1848000 = vc^2/2 - GM/1738000

solving for vc we get vc = sqrt(vo^2 +2*GM*(1/1738000 - 1/1848000)) =

vc = sqrt(1600^2 +2*6.67x10^-11*7.36x10^22*(1/1738000- 1/1848000)) =

So vc = 1701 m/s

your approach is correct but wrong substituion of 180 km in place of 110 km

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