Total energy of elliptical orbit

Click For Summary
SUMMARY

The total energy of an elliptical orbit is defined by the equation E_{tot} = -GMm / (2a), where 'a' represents the semi-major axis. This relationship is derived from the principles of gravitational force and kinetic energy, paralleling the derivation for circular orbits. The justification for substituting 'a' for 'r' in elliptical orbits is based on the constancy of total energy throughout the orbit. The discussion also highlights the derivation of the vis-viva equation, which relates orbital speed to distance from the central body.

PREREQUISITES
  • Understanding of gravitational force and potential energy
  • Familiarity with kinetic energy equations
  • Knowledge of elliptical orbits and semi-major axis
  • Proficiency in algebraic manipulation of equations
NEXT STEPS
  • Study the derivation of the vis-viva equation in detail
  • Explore the implications of Kepler's laws on elliptical orbits
  • Investigate the relationship between orbital mechanics and energy conservation
  • Learn about the applications of elliptical orbits in celestial mechanics
USEFUL FOR

Astronomy students, physicists, and anyone interested in orbital mechanics and the dynamics of celestial bodies.

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
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 13 ·
Replies
13
Views
989
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
4K
  • · 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