# Homework Help: Object In Orbit pushed by a velocity and crashes onto moon.

1. Dec 4, 2011

### uhohitzluke

1. The problem statement, all variables and given/known data

I attached the problem to this post. I need help on the second question and I need to turn it in tomorrow. I'm not sure if I'm supposed to use anything from the first as they are both related.

2. Relevant equations

U + KE
rvsinθ = rvsinθ

These are all the equations I can think of that apply to this problem.

3. The attempt at a solution

I found the relationship between velocity c and the angle by using

(1880km)*(200m/s)sin(90°) = (1740km)*(v)sin(θ)

I initially thought this was correct and went on to substitute values into the energy conservation formula. The conservation of energy formula was not working and I realized that the answers given did not fit my original relationship between the angle and velocity C.

I'm totally lost right now and have no clue what I'm doing. I read the corresponding chapter to these problems in my book, but I'm still lost. I feel like what the book teaches is fairly easy to grasp but the problems that are given to solve are so difficult. Please help!

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2. Dec 5, 2011

### Staff: Mentor

The velocity you want to use on the L.H.S. is the velocity of the LEM, not the casting-off relative velocity between the LEM and the Command Module.

3. Dec 5, 2011

### uhohitzluke

I changed 200 m/s to 1704 m/s as that is the speed of the LEM at engine shut off. It's still wrong. Does the LEM gain speed as it travels closer to point B?

4. Dec 5, 2011

### Staff: Mentor

It has to dock with the command module in order for the excursion team to transfer to it from the LEM. So assume that it has the same velocity as the command module before it is cast away. Also think about which direction the cast-away speed change has to be in order for the LEM to drop towards the surface.

5. Dec 6, 2011

### uhohitzluke

I found that the initial velocity of the LEM is 1585 m/s from the first problem. I do not understand how to incorporate the 200 m/s cast away speed into the problem. I've tried every combination of numbers into the angular momentum and energy conservation formulas but no luck. I'm desperate and I need to turn this in soon. I've been working on it forever and i'm losing it. I need a big big hint ):

6. Dec 6, 2011

### Staff: Mentor

The LEM docks with the command module. In order to do so it must achieve the same velocity as the command module and be in the same orbit as the command module. Therefore it has the same speed as the command module before it is cast off. So what is its speed when cast off?

7. Dec 6, 2011

### uhohitzluke

(1585 - 30 + 200) m/s?

8. Dec 6, 2011

### uhohitzluke

Scratch that. (1585-200+30) m/s?!

9. Dec 6, 2011

### uhohitzluke

BY GOD I THINK I FIGURED IT OUT. thank you so much!

10. Dec 6, 2011

### Staff: Mentor

I love a good eureka