Potential Energy of a Three-Particle System

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SUMMARY

The discussion focuses on calculating the gravitational potential energy of a three-particle system after removing one sphere. The masses of the spheres are mA = 200 kg, mB = 100 kg, mC = 1900 kg, and mD = 250 kg, with specific coordinates provided. The formula used for potential energy is -G(m1m2)/r, leading to a calculated value of -4.6433E-7 J. The participant realizes the importance of converting centimeters to meters for accurate calculations.

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  • Understanding of gravitational potential energy calculations
  • Familiarity with the universal gravitational constant (G)
  • Knowledge of converting units from centimeters to meters
  • Basic principles of coordinate geometry in physics
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  • Practice unit conversion techniques in physics problems
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sizzler
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I am working on this problem:

Four uniform spheres, with masses mA = 200 kg, mB = 100 kg, mC = 1900 kg, and mD = 250 kg, have (x, y) coordinates of (0, 50 cm), (0, 0), (-80 cm, 0), and (40 cm, 0), respectively. Sphere A is then removed. Calculate the gravitational potential energy of the remaining three-particle system

And I do not understand why I cannot get the correct answer. Potential energy equals...

= -G(m1m2)/r

Or in my case...
= -G(100*1900/80 + 100*250/40 + 1900*250/120)
= -4.6433E-7 J

Do I need to account for the loss of sphere A? Or is my error not accounting for the opposite direction of sphere C? Thanks for any advice.
 
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Hm..did you convert cm's into m's?
 
Wow, I can't believe I'm that dumb. Thanks! :)
 

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