1. Sep 10, 2009

### F.Turner

1.
{This paradox denotes the fact that a satellite in a near circular orbit suffers an increase in velocity when subject to a drag force.}

The Specific energy of the satellite is K + U = E where K =v^2/2 is specific kinetic energy. U= -u/r is specific energy and E is specific total energy. If satellite in circular orbit then u=(v^2)*r.

If orbital radius changes by a small amount dr, what is the resulting change dE of the total energy?

3. The attempt at a solution

What I figured to do is take derivative with respect to r but if I do that it will cancel out the distance completely when I plug in the u into the E equation. I i'm not sure if that's the approach I should take. Or should I take derivative with respect to time, because as time changes distance also changes. Not to sure...

2. Sep 10, 2009

### kuruman

Can you show what you mean by this?

3. Sep 10, 2009

### F.Turner

So if I take the derivative of the specific total energy E with respect to the distance value r something like this:

dE/dr = -$$\mu$$/2r --> i plug in the value of $$\mu$$ = (v^2)*r

it turns into -(v^2)*r/2r the r will cancel in this case...(or maybe i should leave one alone and continue to eval the derivative....

The other case would be to take dE/dt to see the change in dr.

4. Sep 10, 2009

### D H

Staff Emeritus
This is paradoxically true -- assuming the drag force is small, of course.

What is the total energy for a circular orbit as a function of orbital radius?

5. Sep 10, 2009

### F.Turner

That is what I show up there its E=-$$\mu$$/2r where $$\mu$$ = (v^2)r

6. Sep 10, 2009

### kuruman

If -μ/r has dimensions of energy, then how can dE/dr = -μ/2r? The dimensions don't match.

7. Sep 10, 2009

### F.Turner

Ok then that would mean dE/dr wouldn't work so, it has to be dE/dt in order to get my answer; so in order to take the derivative with respect to time I would have to plug in v=d/t into the equation right...

8. Sep 10, 2009

### D H

Staff Emeritus
dE/dr is the correct approach.

9. Sep 10, 2009

### F.Turner

Okay thanks I got it