Speed of orbiting planet given eccentricity of orbit

Click For Summary

Homework Help Overview

The discussion revolves around determining the speed of a planet in an elliptical orbit given its eccentricity, specifically focusing on the ratios of the lengths of the axes and the speeds at the ends of the major axis.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the ratio of the lengths of the major to minor axes using the eccentricity and seeks a more elegant method for finding the ratio of speeds at the ends of the major axis. Some participants suggest using the formula for orbital speed and conservation of angular momentum.

Discussion Status

Participants are exploring different methods to approach the problem, with some finding success in using conservation laws while others are still grappling with the algebra involved. There is no explicit consensus on a single method, but various strategies are being discussed.

Contextual Notes

The problem involves specific constraints related to the eccentricity of the orbit and the definitions of the semi-major and semi-minor axes, which are under discussion. The original poster expresses difficulty in simplifying expressions related to kinetic energy and total energy.

ephedyn
Messages
169
Reaction score
1

Homework Statement


If the eccentricity of a planet's orbit about the sun is 0.4, find (a) the ratio of the lengths of the major to minor axes of the planet's orbit, and (b) the ratio of the speeds of the planet when it is at the ends of the major axis of its elliptical orbit.


Homework Equations



[itex]E=\dfrac{1}{2} mv^2 + \dfrac{-GMm}{r} + \dfrac{L^2}{2mr^2}[/itex]

The Attempt at a Solution



The first part is rather short and sweet:

[itex]\dfrac{a}{b}=\dfrac{1}{\sqrt{1-\epsilon^{2}}}=\dfrac{1}{\sqrt{1-0.4^{2}}}\approx1.0911[/itex]

But I have no idea how to proceed on the second part. If I take the ratio of the kinetic energies I get a very messy expression and have problem eliminating the total energy E. Is there a more elegant way to do this?
 
Physics news on Phys.org
hi

orbital speed is given by

[tex]v=\sqrt{\mu\left(\frac{2}{r}-\frac{1}{a}\right)}[/tex]

where [tex]\inline{\mu =G(M+m)}[/tex] is the standard gravitational parameter. and r is the distance from the sun(focus). at the farthest point in the orbit from the sun
[tex]\inline{r=(1+e)a}[/tex] and the closest point we have [tex]\inline{r=(1-e)a}[/tex]
where a is semi major axis length. use this info

Newton
 
Last edited:
Hey thanks for your help. I tried using your approach but I couldn't simplify it so I tried using conservation of angular momentum and got the correct answer. Simpler than I expected.
 
well algebra is pretty straight forward I guess.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
3
Views
10K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
15
Views
2K