# What is the impact speed of the lunar module?

• MozAngeles
In summary: According to the equation, the gravitational potential energy at h=0 is 0. So your equation reduces to KE_{initial}+PE_{initial}=KE_{final}. So solving for vfinal i get vf=√(vinitial2-2G*M(moon mass)/r(orbit i.e. 100*103m)) However, if you use the value of g = GM/r2 where M is the mass of the moon, r is the distance the lunar module is from the core, then vfinal would be equal to √(2gh(initial=100km)+v(initial)^2).
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)

Using KEfinal + PEfinal = KEinitial + PEinitial,

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

mghi +0.5mv2i = mghf + 0.5mv2f, and notice that the mass of the lunar module is irrelevant as it can be canceled out from the equation.

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

Epot = 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, Epot = 0. And for any r smaller than infinity, we take the convention and thus it becomes negative.

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$$_{final}$$.

That helps. So solving for vfinal i get vf=√(vinitial2-2G*M(moon mass)/r(orbit i.e. 100*103m))
am i using the right value for r and M?

Another thing to note is that
U = -GMm/r is GRAVITATIONAL POTENTIAL, which is with respect to infinity.
Epot = 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 Epot = mgr, since the "floor" is given as the surface of moon.

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)

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.

## 1. What is the "gravity energy problem"?

The "gravity energy problem" refers to the challenge of finding a way to harness and utilize the potential energy generated by gravity, which is constantly exerted by the Earth on all objects and can be seen in phenomena such as falling objects and ocean tides.

## 2. Why is it important to address the gravity energy problem?

Addressing the gravity energy problem is crucial in our efforts towards finding sustainable and renewable energy sources. The potential energy of gravity is vast and can be harnessed to produce electricity, potentially reducing our dependence on non-renewable energy sources.

## 3. What are some potential solutions to the gravity energy problem?

One potential solution is the use of hydropower, which involves converting the kinetic energy of falling water into electricity. Another solution is the development of gravity-based energy storage systems, where energy is stored by lifting heavy objects against the force of gravity and then releasing them to generate electricity when needed.

## 4. What are the challenges in implementing gravity-based energy solutions?

One of the main challenges is the high cost of building and maintaining infrastructure for gravity-based energy systems. Additionally, the efficiency of these systems is limited by factors such as friction and resistance, making it difficult to fully harness the potential energy of gravity.

## 5. Are there any current applications of gravity energy solutions?

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