Understanding the Work Needed to Escape a Gravitational Pull

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Discussion Overview

The discussion revolves around the integration of Newton's law of universal gravitation and its implications for calculating work needed to escape a gravitational pull. Participants explore the relationship between force, work, and potential energy in the context of gravitational fields.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the meaning of integrating Newton's law of universal gravitation, suggesting that the result may represent work needed to escape gravitational pull.
  • Another participant asserts that the integration was performed incorrectly and invites the original poster to try again.
  • A different participant references the work-energy theorem and states that integrating the gravitational force yields potential energy rather than work.
  • There is a correction regarding the expression for work, with a participant indicating that the correct expression is related to potential energy differences.

Areas of Agreement / Disagreement

Participants express disagreement regarding the integration results and interpretations of work versus potential energy. No consensus is reached on the correct approach or interpretation.

Contextual Notes

Some participants highlight potential misunderstandings in the integration process and the definitions of work and potential energy, but these remain unresolved.

Mr-R
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Hello everyone,
What does it mean if I integrate Newton's law of universal gravitation with respect to r.
F= GMm/r^2 become 3GMm/r^3 . Is this the work needed to escape a gravitational pull ?

Thank you
 
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Mr-R said:
Hello everyone,
What does it mean if I integrate Newton's law of universal gravitation with respect to r.
F= GMm/r^2 become 3GMm/r^3 . Is this the work needed to escape a gravitational pull ?

Thank you
Dear Mr-R. Welcome to Physics Forums.

You integrated incorrectly. Please try again.
 
This is the work-energy theorem.

$$\int ^b _a \vec{F} \cdot d \vec{r} = \Delta KE = W $$

If you integrate ##\frac{GMm}{r^2}## you get the potential energy.

Try integrating it again correctly.
 
Chestermiller said:
Dear Mr-R. Welcome to Physics Forums.

You integrated incorrectly. Please try again.

Oh that's quiet embarrassing. It should be = GMm/r which is the work. Thank you very much.
 
Mr-R said:
Oh that's quiet embarrassing. It should be = GMm/r which is the work. Thank you very much.

No, that's potential energy. In this case, ## W = \Delta U = \frac{GMm}{r_2}- \frac{GMm}{r_1}##
 

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