Why does centripetal force not apply in gravitation problem?

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SUMMARY

The discussion clarifies that centripetal force does not apply to elliptical orbits, specifically in the context of a comet's motion around the Sun. The ratio of the comet's speeds at two positions in its orbit is determined to be 1/10, while the ratio of the gravitational forces acting on the comet at these positions is 1/100. The gravitational force formula, Fg = Gm1m2/r^2, is essential for calculating these ratios, as centripetal force is only applicable to circular motion. This distinction is crucial for understanding the dynamics of elliptical orbits.

PREREQUISITES
  • Understanding of gravitational force (Fg = Gm1m2/r^2)
  • Knowledge of centripetal force (Fc = mv^2/r)
  • Familiarity with elliptical orbits in celestial mechanics
  • Basic algebra for manipulating equations and ratios
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  • Research the properties of elliptical orbits in celestial mechanics
  • Learn about the conservation of angular momentum in orbital dynamics
  • Study the differences between circular and elliptical motion
  • Explore the implications of Kepler's laws of planetary motion
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Students studying physics, particularly those focusing on orbital mechanics, as well as educators and anyone interested in the principles governing celestial bodies' movements.

Dennis Heerlein
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Homework Statement


The elliptical orbit of a comet is shown above (hidden document I apologize but easy to picture). Positions 1 and 2 are, respectively, the farthest and nearest positions to the Sun, and at position 1 the distance from the comet to the Sun is 10 times that at position 2.
What is the ratio v1/v2 of the speed of the comet at position 1 to the speed at position 2? What is the ratio F1/F2 of the force on the comet at position 1 to the force on the comet at position 2?

Homework Equations


Fc = mv^2/r
Fg = Gm1m2/r^2

The Attempt at a Solution


I used IW = IW to solve that the ratio v1/v2 is 1/10. Then, for the force equations, I divided centripetal force of v1 by centripetal force of v2, making substitutions, like this (the line of dashes representing division)
[m(v1)^2]/R2(10)
----------------------- = 1/10
[m(v1x10)^2]/R2

The answer is 1/100, which is found if Fg is used, as the only difference in Fg is the radius substitution of (10xR2)^2
 
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The centripetal force formula you are using only applies to circular motion, not to ellipses. Indeed, you could say that because the gravity at closest approach is less then the centripetal force requirement, this is the reason the orbit is swings out into a wide ellipse instead of staying in a tight circle.
 
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Ken G said:
The centripetal force formula you are using only applies to circular motion, not to ellipses. Indeed, you could say that because the gravity at closest approach is less then the centripetal force requirement, this is the reason the orbit is swings out into a wide ellipse instead of staying in a tight circle.
That makes sense. Thanks a bunch, I appreciate it.
 

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