Is Calculating Work Done by Gravity as Simple as It Seems?

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Discussion Overview

The discussion revolves around the calculation of work done by gravity, particularly in the context of an object moving towards the center of the Earth. Participants explore the implications of the Universal Law of Gravitation and the behavior of gravitational force as the radial distance changes.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant proposes using the formula for gravitational force, F=(G*m*m)/(r^2), and suggests that the work done by gravity can be calculated using the integral of this force over the distance to the center of the Earth.
  • Another participant questions the validity of taking the radial distance to be zero, suggesting instead that the limits of integration should reflect the radius of the Earth or use initial and final radial distances for a more general solution.
  • A later reply acknowledges the confusion about the concept of zero radial distance, humorously noting the implications of occupying the same space as the Earth.
  • One participant challenges the initial claim about gravitational force increasing as radial distance decreases, stating that only the mass inside the radial distance contributes to the gravitational force, referencing the shell theorem.

Areas of Agreement / Disagreement

Participants express differing views on the behavior of gravitational force as an object approaches the center of the Earth, indicating that there is no consensus on this aspect of the discussion.

Contextual Notes

Limitations include the assumption about the behavior of gravitational force inside the Earth and the implications of integrating to zero radial distance, which may not accurately reflect the physical situation.

Martin23
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So I was thinking about the Universal Law of Gravitation, and the force of an object depends on the radial distance from Earth for simplicity. As an object travels to the center of the Earth the Fg increases as the radial distance decreases.F=(G*m*m*1)/(r^2). Knowing that the Force is not constant, the work done by gravity would be ∫F*dr from lower limit of r to upper limit of 0. Now, if you put the F in terms of r you get ∫[(G*m*m)/(r^2)]*dr from r to 0. As I tried to solve for Work, I arrived at 1/0 and now I am stuck. Was all this valid or just nonsense? If you think about this in vector form 3 dimensions, it is more understandable to me at least. Here is a picture of my work. Help please.
 

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Think about what you mean by r being 0. Are you really taking the limit of integration as the center of Earth? I would think you would want to substitute 0 for the radius of Earth. Or even better, just use r_initial and r_final as your limits, it gives you a more general answer. Then just plug-n-chug for your solution.
 
oh your right haha. 0 radial distance would be like super imposed on Earth hahaa
they would occupy the same space right? haha ty.
 
Martin23 said:
As an object travels to the center of the Earth the Fg increases as the radial distance decreases.

No, it doesn't. When the radial distance is less than the Earth's radius, only the mass inside the radial distance "counts" in calculating the gravitational force. Google "shell theorem" for details.
 

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