How Do You Calculate Gravitational Field Strength Between Two Planets?

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SUMMARY

The discussion focuses on calculating the gravitational field strength (E) between two planets with masses M and m, separated by a distance of 2a. The correct expression for E is derived using vector addition and the cosine rule, resulting in E = (2GMr) / (a^2 + r^2)^(3/2). The participant initially attempted to use the parallelogram method for vector addition but faced confusion regarding its application. Ultimately, the solution emphasizes the cancellation of y components due to symmetry and the addition of x components.

PREREQUISITES
  • Understanding of gravitational field strength and Newton's law of universal gravitation
  • Familiarity with vector addition techniques, specifically the parallelogram method
  • Knowledge of trigonometric functions, particularly cosine
  • Basic algebra and manipulation of equations
NEXT STEPS
  • Study the parallelogram method for vector addition in physics
  • Learn about gravitational field strength calculations in multi-body systems
  • Explore trigonometric identities and their applications in physics problems
  • Investigate the implications of symmetry in gravitational fields
USEFUL FOR

Students studying physics, particularly those focusing on gravitational forces and field strength calculations, as well as educators seeking to clarify concepts related to vector addition in gravitational contexts.

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Homework Statement



Write an expression for the gravitational field strength E at a point distance r from a planet of mass M. (In my diagram, there are two planets with mass, m and distance 2a apart)

Homework Equations





The Attempt at a Solution



The resultant gravitational field strength is towards the left. Using the parallelogram method to solve for the resultant vector and by the cosine rule,

|E|^2=(\frac{GM}{R^2})^2+(\frac{GM}{R^2})^2-2(\frac{GM}{R^2})(\frac{GM}{R^2})\cos 120

|E|=\frac{GM\sqrt{3}}{(r^2+a^2)^2}

But the answer given is

E=\frac{2GMr}{(a^2+r^2)^{\frac{3}{2}}}
 

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I don't know what a parallelogram method is. You know from symmetry that the y components will cancel, so all you have to do is look at the two x components, which will both add to give

2cosØ(GM)/(a^2+r^2)^(1/2), and cosØ=r/(r^2+a^2)
 
Last edited:
Mindscrape said:
I don't know what a parallelogram method is. You know from symmetry that the y components will cancel, so all you have to do is look at the two x components, which will both add to give

2cosØ(GM/R^2)/(a^2+r^2)^(1/2), and cosØ=r/(r^2+a^2)

Thanks for your help, Mindscrape!
 

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