Calculating Earth Displacement from Free Fall of 1kg Object

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Homework Help Overview

The discussion revolves around calculating the displacement of the Earth when a 1 kg object is released from a height of 1 meter. Participants are exploring the implications of conservation laws in this context.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum and energy, questioning whether the scenario describes an elastic collision. There are attempts to derive the displacement of the Earth based on these principles, with some participants suggesting limits to test the validity of the approach.

Discussion Status

The discussion is ongoing, with participants raising questions about the assumptions made in the calculations. Some guidance has been offered regarding the nature of the collision and the relevance of the center of mass, but no consensus has been reached on the correct approach.

Contextual Notes

There is uncertainty regarding the nature of the collision (elastic vs. inelastic) and how it affects the calculations. Participants are also considering the implications of different mass ratios between the object and the Earth.

gasar8
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Homework Statement


We release m_o = 1 kg object from h=1m height. How much does the Earth move (x)? I just need the comfirmation if I did correctly?

Homework Equations


Conservation of momentum:
m_o v_o + m_E v_E = 0

The Attempt at a Solution


I wrote the energy conservation law as (beginning = end):
m_o g h = \frac{1}{2} m_o v_o^2 + \frac{1}{2} m_E v_E^2,
and from momentum conservation got:
g h= \frac{1}{2} v_o^2 \bigg(1 + \frac{m_o}{m_E} \bigg),
and now, I insert the free fall relation v_o = \sqrt{2g(h-x)} from where I extract x as:
x = h\frac{\frac{m_o}{m_E}}{1+\frac{m_o}{m_E}} .
It seems reasonable, because it is something small.
 
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gasar8 said:
I wrote the energy conservation law
Is your scenario describing a fully elastic collision ? If not, you'll need some other relationship...
 
Doesn't say anything about that, so I would assume that it is elastic?
 
then the object bounces back up 1 m
 
gasar8 said:

Homework Statement


We release m_o = 1 kg object from h=1m height. How much does the Earth move (x)? I just need the comfirmation if I did correctly?

Homework Equations


Conservation of momentum:
m_o v_o + m_E v_E = 0

The Attempt at a Solution


I wrote the energy conservation law as (beginning = end):
m_o g h = \frac{1}{2} m_o v_o^2 + \frac{1}{2} m_E v_E^2,
and from momentum conservation got:
g h= \frac{1}{2} v_o^2 \bigg(1 + \frac{m_o}{m_E} \bigg),
and now, I insert the free fall relation v_o = \sqrt{2g(h-x)} from where I extract x as:
x = h\frac{\frac{m_o}{m_E}}{1+\frac{m_o}{m_E}} .
It seems reasonable, because it is something small.

To assure yourself further what limits can you look at? What if the two masses are equal? What if the object is much more massive than earth? Those limits give you reasonable answers so you can do this without any "expert" advice!
 
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Ok, that does not seem right. :) So I must change the momentum conservation also? I am not sure what exactly changes?
 
BvU said:
then the object bounces back up 1 m
I think you misunderstand the question
 
Enlighten me
 
The question is how far the Earth moves during the fall, I believe. This is independent of how elastic the colision.
 
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  • #10
I agree. What happens to the center of mass, my dear gasar ?
 
  • #11
During the fall, I supose it should stay at rest?
 

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