Calculating Earth Displacement from Free Fall of 1kg Object

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SUMMARY

The discussion focuses on calculating the displacement of the Earth when a 1 kg object is released from a height of 1 meter. The conservation of momentum and energy equations are utilized to derive the formula for Earth's movement, expressed as x = h(m_o/m_E)/(1+m_o/m_E). The participants confirm that the scenario does not depend on the elasticity of the collision, emphasizing the importance of the center of mass remaining at rest during the fall. The calculations yield a small displacement, consistent with the expected physical behavior.

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  • Understanding of conservation of momentum principles
  • Familiarity with energy conservation laws in physics
  • Basic knowledge of gravitational acceleration (g)
  • Concept of center of mass in mechanics
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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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