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 130 km above the lunar surface moving with a speed of 1500 m/s .

Answer 1

Assuming there is no dissipative forces, by Conservation of Energy we know that the sum of initial Kinetic Energy and Potential Energy must be equal to the sum of final Kinetic and Potential Energy.

Potential energy for a body with mass m due to a mass M is given by

U = -GMm/r

[Notice that we could have used U = mgh but we can't because it would require the knowledge of Moon's acceleration due to gravity].

therefore our energy expression will be:

$\frac{1}{2}mv_i^2 + (-\frac{GMm}{r_i}) = \frac{1}{2}mv_f^2 + (-\frac{GMm}{r_f})$

or $\frac{1}{2}v_i^2 - \frac{GM}{r_i} = \frac{1}{2}v_f^2 - \frac{GM}{r_f}$

since mass m (lunar module mass) was in all the terms.

v_i = 1500 m/s

Mass of the moon is 7.36x10²²kg and its radius R is 1738 km = 1738000 m

substituting these values in our equation will give:

=> $\frac{1}{2}1500^2 - \frac{6.67\times10^{-11}\times7.36\times10^{22}}{(1738000 + 130000)} = \frac{1}{2}v_f^2 - \frac{6.67\times10^{-11}\times7.36\times10^{22}}{1738000}$

$1321571.412 = \frac{1}{2}v_f^2$

therefore v_f = 1625.77 m/s