What is the impact speed of the lunar module?

[MozAngeles]

Homework Statement

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 100 km above the lunar surface moving with a speed of 1550 m/s.

v(final) =--------------m/s

Homework Equations

KEfinal + PEfinal = KEinital + PEinital

It is the gravitational potential energy = -GMm/r
KE=1/2mv²
G is a universal constant = 6.67 x 10^-11 Nm²/kg²
M = mass of the moon
r = distance of the module from the center of the moon.

M = 7.349 x 10²²kg
radius of the moon = 1737.1 km
rinitial = 100*10³ m?? don't know if this is right..
rfinal = 100*10³ m

The Attempt at a Solution

No idea where to start except that E(inital)=E(final)

 

[Ambidext]

Using KE_final + PE_final = KE_initial + PE_initial,

mass of the lunar module is assumed to be unchanged, so perhaps you can start with equating them to:

mgh_i +0.5mv²_i = mgh_f + 0.5mv²_f, and notice that the mass of the lunar module is irrelevant as it can be canceled out from the equation.

 

[MozAngeles]

I thought that when dealing with gravity problems the potential energy is = -GMm/r

 

[Ambidext]

E_pot = mgr = m (GM/r²) r = -GMm/r

The units on both sides agree, and the negative sign is due to the convention that when r = infinity, E_pot = 0. And for any r smaller than infinity, we take the convention and thus it becomes negative.

 

[Bhumble]

One other thing that might help is that the potential energy at h=0 is 0. So your equation reduces to KE_{initial}+PE_{initial}=KE_{final}.

 

[MozAngeles]

That helps. So solving for v_final i get v_f=√(v_initial²-2G*M(moon mass)/r(orbit i.e. 100*10³m))
am i using the right value for r and M?

 

[Ambidext]

Another thing to note is that
U = -GMm/r is GRAVITATIONAL POTENTIAL, which is with respect to infinity.
E_pot = mgr is GRAVITATIONAL POTENTIAL ENERGY, which is with respective to the "floor" stated in the question.

For this question, it's more appropriate to use E_pot = mgr, since the "floor" is given as the surface of moon.

 

[MozAngeles]

So using mgr, what do I use for g...
so since the m cancels out.. gh(initial)+.5v(initial)^2=.5v(final)^2+ gh(final, where final h is =0)
so gh(initial)+.5v(initial)^2=.5v(final)^2
solve for v(final)
would the equation for v(final)=√(2gh(initial=100km)+v(initial)^2)

 

[Ambidext]

That one, I haven't checked if the g can entirely cancel out. Safer to use g = GM/r² where M is mass of the moon, r is distance the lunar module is from the core.

Looking at your equations, I think you're on the right track. This method I suggested, the tricky part is calculating the right value of g.