# Find eccentricity of orbit after star has lost mass

1. May 19, 2013

### refractor

1. The problem statement, all variables and given/known data

Initially, a planet with mass m moves on a circular orbit (r = R) around a star with mass M. Now M is instantaneously decreased to M'. Find the eccentricity e of the elliptical orbit the planet now follows.

2. Relevant equations

specific angular momentum l = L/m

$\frac{\mbox{total energy}~ E}{m} = \frac{1}{2}v_r^2 + \frac{1}{2}\frac{l^2}{r^2} - \frac{γM}{r}$

The effective potential is:

$V_{eff}=\frac{1}{2}\frac{l^2}{r^2} - \frac{γM}{r}$

click

3. The attempt at a solution

The mimimum of the effective potential is at R. But then as M->M' the minimum is shifted upwards and to the right. Energy is not conserved: the potential becomes less negative and kinetic energy also increases

The radius of the circular orbit R provides us with the semi-axis b of the ellipse, because at the moment of mass loss M -> M' the planet will leave it's circular orbit and enter the elliptical orbit at perihelion. The circle lies exactly inside the ellipse.

The mass loss does not effect angular momentum because the gradient of the potential is parallel to r, so the force responsible for the change of orbit is also only acting parallel to r.

$l_{circle} = R^2 \omega$

Now I have to find a relation between l and e, but I'm stuck on how to calculate the angular momentum of the elliptical orbit. I don't even know it's period.

Michael

2. May 19, 2013

### Thaakisfox

Check "Kepler problem" on wikipedia.

See here the relation between the energy, eccentricity and angular momentum of the motion.

3. May 20, 2013

### refractor

sorry, this relation is known but it doesn't solve the problem. Could anybody help?

4. May 20, 2013

### Thaakisfox

That is precisely the relation you need.... just look at the derivation a little longer.

For the angular momentum is conserved, so you just need to calculate the angular momentum in the first case with the circular orbit. (suppose the star is fixed ...)

This is simply:

$$L=mR^2\omega$$

Where the angular velocity can simply be calculated (circular motion provided by gravity):

$$\omega = \sqrt{\frac{\gamma M}{R^2}}$$

Use this to find the angular momentum ... and then use the relation...

5. May 20, 2013

### refractor

I think I still haven't understood. I'd say it's

$$\omega = \sqrt{\frac{\gamma M}{R^3}}$$

and thus

$$L = m\sqrt{\gamma M R}$$

In my first post I wrote angular momentum is conserved, but why does it depend on M?
$$e = \sqrt{1 + \frac{2EL^{2}}{k^{2}m}}$$