Total energy of elliptical orbit

Click For Summary

Discussion Overview

The discussion revolves around the total energy of an elliptical orbit, specifically the expression E_{tot}=\frac{-GMm}{2a}, where 'a' is the semi-major axis. Participants explore the justification for this expression, its relation to circular orbits, and the derivation of related equations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions the justification for using the semi-major axis 'a' in the energy expression for elliptical orbits, comparing it to the circular orbit case.
  • Another participant suggests that the total energy remains constant throughout the orbit and relates it to the orbital speed at any point in an elliptical orbit.
  • A participant expresses the need for an alternative explanation to avoid circular logic when deriving the vis-visa equation from total energy.
  • One participant references a source to support their understanding of the relationship between total energy and the semi-major axis.
  • A later post discusses a specific problem involving a comet's elliptical orbit, asking for equations and a solution to determine its speed at a given distance from the Sun.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the justification for the energy expression in elliptical orbits, and multiple viewpoints regarding the derivation and application of related equations are present.

Contextual Notes

Some participants express uncertainty about the steps needed to derive the total energy from the given equations, indicating potential gaps in assumptions or mathematical clarity.

E92M3
Messages
64
Reaction score
0
Why is the total energy of an elliptical orbit given by:
E_{tot}=\frac{-GMm}{2a}
Where a=semi major axis.
I agree for a circular orbit I can do the following:
F_c=F_g
ma_c=\frac{GMm}{r^2}
\frac{v^2}{r}=\frac{GM}{r^2}
v^2=\frac{GM}{r}
Since the total energy also equal to the kinetic plus potential energy we have:
E_{tot}=\frac{1}{2}mv^2-\frac{GMm}{r}=\frac{1}{2}m\frac{GM}{r}-\frac{GMm}{r}=\frac{-GMm}{2r}
Ok this is a similar form for circular orbit. But how can we just put a in instead of r for elliptical orbit? What is the justification?
 
Physics news on Phys.org
Consider the fact that the total energy of the orbiting object is constant for any point of its orbit. Then take the visa-vis equation for finding the orbital speed at any point of an elliptical orbit:

\sqrt{GM \left (\frac{2}{r}-\frac{1}{a} \right )}

Note that when r=a, you get

v=\sqrt{\frac{GM}{a}}

or

v^2=\frac{GM}{a}

Thus the total energy of a elliptical orbit with a semi-major axis of 'a' is the same as a that for a circular orbit with a radius of 'a'.
 
Well actually I was trying to derive the vis-visa equation through the total energy. So I must have an alternate way to explain it or i'll be circular logic.
 
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 5e10 m (inside the orbit of Mercury), at which point its speed is 9e4 m/s. Its farthest distance from the Sun is far beyond the orbit of Pluto. What is its speed when it is 6e12 m from the Sun?

please show the equations needed, if you feel generous work the problem out as well.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K