Object In Orbit pushed by a velocity and crashes onto moon.

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Homework Help Overview

The discussion revolves around a physics problem related to an object in orbit, specifically focusing on the Lunar Module (LEM) and its velocity as it approaches docking with the command module and subsequently drops towards the moon's surface. The subject area includes concepts of energy conservation, angular momentum, and orbital mechanics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the LEM's velocity and the angle of descent, with attempts to apply conservation of energy and angular momentum equations. There are questions about the correct values to use for velocity and how the LEM's speed changes as it approaches its docking point.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and attempting to clarify the relationships between various velocities. Some guidance has been offered regarding the need for the LEM to match the command module's speed before being cast off, but no consensus has been reached on the correct approach or values to use.

Contextual Notes

Participants express frustration with the complexity of the problem compared to the textbook material, indicating a potential gap in understanding how to apply theoretical concepts to practical scenarios. There is also mention of a deadline for submission, adding urgency to the discussion.

uhohitzluke
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Homework Statement



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.

Homework Equations



U + KE
rvsinθ = rvsinθ

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


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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uhohitzluke said:
I found the relationship between velocity c and the angle by using

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

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.
 
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?
 
uhohitzluke said:
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?

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.
 
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 ):
 
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?
 
(1585 - 30 + 200) m/s?
 
Scratch that. (1585-200+30) m/s?!
 
BY GOD I THINK I FIGURED IT OUT. thank you so much!
 
  • #10
uhohitzluke said:
BY GOD I THINK I FIGURED IT OUT. thank you so much!

I love a good eureka :smile:
 

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