# Total energy of elliptical orbit

1. Oct 3, 2009

### E92M3

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?

2. Oct 3, 2009

### Janus

Staff Emeritus
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'.

3. Oct 3, 2009

### E92M3

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.

4. Oct 3, 2009

### Janus

Staff Emeritus
5. Oct 4, 2009

### E92M3

I'm okay up to the part where:
$$2E_{tot}=\frac{L^2}{2m}(\frac{1}{r^2_1}+\frac{1}{r^2_2})-GMm(\frac{1}{r_1}+\frac{1}{r_2})$$
But I'm not sure how to get from there to:
$$E=\frac{-GM}{r_1+r_2}=\frac{-GM}{2a}$$

6. Feb 18, 2010

### shnav34

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.