What is the effect of friction on the orbit of a satellite?

[bodensee9]

Homework Statement

A satellite of mass m is orbiting the earth. The radius of the orbit is r_{0} and the mass of Earth is M_{e}.

a. Find total mechanical energy.
b. Now suppose that satellite encounters constant frictional force f which retards its motion. The satellite will spiral to the earth. Assume the radius changes so slowly that you can treat the satellite as being in circular orbit of average radius r. Find approximate change in radius per revolution of satellite.

I found the first a. I think it's -\frac{GmM_{e}}{r_{0}}.
But then for part b, I assume that the frictional force is in the direction of its velocity. And so, I think I have \frac{\partial^2\theta}{dt^2} = f. But then I am not sure what to do after that? I also know that the work done by f is equal to the change in energy? Thanks.

Homework Equations

The Attempt at a Solution

 

[bodensee9]

Um, I meant d^2theta/dt^2 = f. BUt somehow Latex came out the other way.

 

[alphysicist]

Hi bodensee9,

Homework Statement

A satellite of mass m is orbiting the earth. The radius of the orbit is r_{0} and the mass of Earth is M_{e}.

a. Find total mechanical energy.
b. Now suppose that satellite encounters constant frictional force f which retards its motion. The satellite will spiral to the earth. Assume the radius changes so slowly that you can treat the satellite as being in circular orbit of average radius r. Find approximate change in radius per revolution of satellite.

I found the first a. I think it's -\frac{GmM_{e}}{r_{0}}.

No, I don't think that is correct. Remember that they ask for the total mechanical energy.

But then for part b, I assume that the frictional force is in the direction of its velocity. And so, I think I have \frac{\partial^2\theta}{dt^2} = f.

That cannot be the right relation, since it does not have the same units on each side of the equation.

Instead, I would suggest thinking about how much work is done by the frictional force every time the satellite goes around once. How will that work affect the total energy that you found in part a?

 

[bodensee9]

Hi

Oh, right. So the total mechanical energy E would be then \frac{l^2}{mr^2} - \frac{GmM_{e}}{r}, where l is the angular momentum of the satellite. And you can find l because for a circular orbit, \frac{dE}{dr}= 0.

Okay, I think the amount of work done by friction as the satellite goes around once would be 2fr\pi? Hm .. as the satellite gets closer, the speed increases but potential energy decreases. So then the change in energy is then 2fr\pi? Then would I solve for r using the energy equation above?

Thanks.

 

[bodensee9]

sorry, i meant dE/dr = 0, but somehow the latex came out the other way.

 

[alphysicist]

Hi

Oh, right. So the total mechanical energy E would be then \frac{l^2}{mr^2} - \frac{GmM_{e}}{r},

I don't think that's right; it looks like you're missing a numerical factor.

But in the end I think you want to find a single term for part a. Just use translational kinetic energy:

<br /> E_{\rm total}=\frac{1}{2}mv^2-\frac{GMm}{r}<br />

Since it is a circular orbit, you can also find v in terms of r. Once you plug that back into this energy expression, you'll get a much simpler form that makes solving part b easier.

where l is the angular momentum of the satellite. And you can find l because for a circular orbit, \frac{dE}{dr}= 0.

Okay, I think the amount of work done by friction as the satellite goes around once would be 2fr\pi? Hm .. as the satellite gets closer, the speed increases but potential energy decreases. So then the change in energy is then 2fr\pi?

Okay, but don't forget the sign. Friction is causing the satellite to lose energy.

Then would I solve for r using the energy equation above?

Yes, I think you would want to use the work-energy formula for this, since energy is not conserved.

 

[bodensee9]

Thanks!
You're right; I meant to have a 1/2 in front of the l^2/mr^2 term. For some reason I'm having trouble with Latex.

 

[alphysicist]

Thanks!
You're right; I meant to have a 1/2 in front of the l^2/mr^2 term. For some reason I'm having trouble with Latex.

That's right. And here, angular momentum is not conserved, so you should eliminate L, and get an expression for the total energy that has r (but not L or v).

 

[bodensee9]

Okay, since I will have
\frac{mv^2}{r} = \frac{GmM}{r^2},
so then my v^{2} = \frac{GM}{r}.
Then plugging in this means that my total energy at any given r is
\frac{-GmM}{2r}.
Then I would have:
\frac{-GmM}{2r_{i}} = \frac{-GmM}{2r_{f}} - 2r_{i}f\pi?

Where r_{i} is the initial radius and r_{f} is the new radius after one revolution has passed.
Then I will have

GmM(\frac{r_{f}-r_{i}}{2r_{i}r_{f}}) = 2r_{i}f\pi.

So then can I make the approximation that since r_final and r_initial won't differ very much from revolution to revolution, so effectively in the denominator I have 2r_{i}^2? And then I can simply the expression to get

(r_{f}-r_{i}) = \frac{4r_{i}^3f\pi}{GmM}

Thanks! Sorry this is taking so long.

 

[Redbelly98]

You're very nearly right.

A couple of questions:

Do you expect that:

• the initial energy is less than or greater than the final energy?
• the final radius is smaller or larger than the initial radius?

Think about the answers to those questions, and whether these equations are consistant with or contradict them:

\frac{-GmM}{2r_{i}} = \frac{-GmM}{2r_{f}} - 2r_{i}f\pi?

(r_{f}-r_{i}) = \frac{4r_{i}^3f\pi}{GmM}

 

[bodensee9]

Um, I thought that the potential energy will be less than the initial energy (because the done by friction will actually go into the increase in kinetic energy?) and I thought the final radius will be smaller than the initial radius...? Thanks.

 

[bodensee9]

oh right, I should have -GmM/r_initial - 2(pi)r_initial*f = -GmM/r_final. Thanks!