Circular orbit change after gaining mass ejected from sun

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Homework Statement


A satellite of mass m is in a circular orbit of radius R about a star of mass M. The star ejects 1% of its mass by means of a spherically symmetric wind which removes the mass to a large distance. What are the new nearest and furthest distances of the satellite’s orbit around the star?

Homework Equations



Angular momentum $$\vec L=\vec r \times \vec p$$

Total energy of orbit $$E= \frac {1}{2}m\dot r^2+\frac{L^2}{2mr^2} -\frac{GMm}{r}$$

surface area of a sphere $$S=4\pi r^2$$

The Attempt at a Solution



If m<<M then it can treated as a point source. So the satellite gains mass ##\frac{M}{400\pi R^2}##

The angular momentum of the circular orbit of the satellite ##\vec L = mr \times \dot r = mr \times (\vec\omega \times \vec r)=mr^2\omega##

For circular motion ##E=\frac{1}{2}mv_0^2 - \frac{GMm}{R}##

I am not entirely sure how to approach this question, as when I considered it I thought that angular momentum and energy should both increase, as the satellite gains mass?

Either way, I calculated the initial angular momentum and energy for the circular orbit of the satellite:

##L=mr^2\omega## and ##\omega=\frac{v_0}{r}## so ##L_0=\frac{GMm}{v_0}##

Now as ##\dot r## is zero for a circular orbit, ##E_0=\frac{1}{2}mv_0^2 - \frac{GMm}{r}##, and equating gravitational force with that of circular motion, ##R=\frac{GM}{v_0^2}## so ##E_0=\frac{1}{2}mv_0^2##.

Now I guess it could be argued that angular momentum is constant as it is conserved for radial forces, but even so, I don't really see how this would help me equate energies or something, as if energy does change, I'm not sure by which factor? Equally, I suppose potentially energy could be conserved if its seen as the total energy of the sun/satellite system, but I really am at a bit of a loss.

I'd really appreciate a bit of a steer in a more productive direction, thanks!
 

Answers and Replies

  • #2
Orodruin
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You know the initial velocity and position of the satellite. This should be sufficient to find its orbit. There is no need to look at energies.
 

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