Velocity of a planet around the sun

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Homework Help Overview

The discussion revolves around the velocity of a planet in an elliptical orbit around the sun, specifically comparing its speed at perihelion and aphelion given known distances. The original poster presents a relationship between the speeds and distances, referencing relevant equations.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the planet's speed at different points in its orbit using mathematical expressions derived from angular momentum and energy conservation principles.

Discussion Status

Some participants have verified the original poster's result through different approaches, noting consistency with conservation laws. There appears to be a productive exchange of ideas, although no explicit consensus is stated.

Contextual Notes

Participants reference known distances and conservation principles, but the discussion does not clarify if all necessary parameters are fully defined or if assumptions are being questioned.

hs764
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1. Asking because the answer I got seems too simple...a planet of mass m moves in an elliptical orbit about the sun. The minimum and maximum distances of the planet from the sun are called the perihelion and aphelion respectively. If the speed of the planet at p is vp, what is its speed at a? Assume the distances rp and ra are known.

Homework Equations



dA/dt = 1/2r2ω, ω=v/r[/B]

The Attempt at a Solution



dA/dt is constant, so 1/2rp2ωp = 1/2ra2ωa. Substitute v = ωr, that gives 1/2rpvp = 1/2rava, va = rpvp/ra. Is this correct?[/B]
 
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Try verifying the result from another rule - like conservation of momentum and energy...
 
So using conservation of momentum I got the same thing...dA/dt = L/2m = mr2ω/2m. Given that ω = v/r, rpvp/2 = rava/2, va = rpvp/ra.
 
hs764 said:
So using conservation of momentum I got the same thing...dA/dt = L/2m = mr2ω/2m. Given that ω = v/r, rpvp/2 = rava/2, va = rpvp/ra.

Yes, it is the same thing as Kepler's law. Good job.
 
Great, thank you!
 

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