I.E. Irodov Advanced Gravitation Problem

[zorro]

Homework Statement

An artificial satellite of mass m of a planet of mass M, revolves in a circular orbit whose radius is n times the radius R of the planet. In the process of motion, the satellite experiences a slight resistance due to cosmic dust. Assuming the resistive force on satellite depends on velocity as F=av² where a is constant, calculate how long the satellite will stay in orbit before it falls onto the planet's surface.

The Attempt at a Solution

Click this link
http://latex.codecogs.com/png.latex?av^{2}&space;=&space;m\frac{\partial&space;v}{\partial&space;t}\\\\&space;\int_{0}^{t}adt&space;=&space;\int_{v_{1}}^{v_{2}}&space;\frac{m}{v^{2}}\\\\v_{1}=\sqrt{\frac{GM}{nR}}\\\\v_{2}=&space;\frac{GM}{R}\left&space;(&space;2-\frac{1}{n}&space;\right&space;)\\\\t(incorrect)=&space;\frac{m}{a}\sqrt{\frac{Rn}{GM}}\left&space;(&space;1-\frac{1}{\sqrt{2n-1}}&space;\right&space;)\\\\t(correct)=&space;\frac{m\sqrt{R}\left&space;[&space;\sqrt{n}-1&space;\right&space;]}{a\sqrt{GM}}

I got v₂ by conserving energy in the orbit and on the surface of earth.
On substituting these values in my second equation, the answer is coming wrong. Please explain me my mistake.

 

[hikaru1221]

The wrong answer comes from the assumption that gravity doesn't do work on the satellite, which also means assuming that the orbit of the satellite is circular. But that's not the case. The actual orbit of satellite is spiral, though locally, it looks like circular. That is, the orbit is slightly slanted to the circular orbit at the same point (see the picture). The angle between the orbits is small, thus, the component of gravity along the spiral orbit is small. But friction is also small. When there are 2 things small, we simply cannot eliminate either one of them.

If we apply Newton's 2nd law (as you did) here, I suppose it would be quite mathematically crazy, as the angle is an unknown function of time/position. The best approach might be energy approach. Calculate the work done by friction and see if we can arrive at the expected result

Hint:
1/ Due to that locally the spiral orbit is approximately circular, we can assume that in 1 round, the orbit is nearly circular. Let r denote the approximate radius of the orbit. Since the orbit is circular and friction is small, we can also assume that v doesn't change much in this round. From here, calculate v.
2/ Write dA, the element work done by friction in time dt in term of r.
3/ Write dE, the energy increment of the satellite in term of r and dr.
4/ dA = dE, then integrate.

 

[zorro]

Well, I don't really understand how gravity will do any work in spiral path of satellite. The motion of the satellite at every small interval is perpendicular to the gravitational field. So no work is done by gravity.

I just now figured out that we could not use the energy conservation law as friction does some work (which is offcourse not negligible).
I calculated the velocity v₂ as a function of radial distance r by using the binding energy equation i.e. mv²/r = GMm/r²
On substituting this value in the second equation with r=R at t=t , I got the correct answer!

 

[hikaru1221]

Well, I don't really understand how gravity will do any work in spiral path of satellite. The motion of the satellite at every small interval is perpendicular to the gravitational field. So no work is done by gravity.

That's where you got it wrong. As I pointed out, the component of gravity along the orbit is of course very small, but so is friction. If you neglect gravity, you have to neglect friction. Your 1st equation $$\alpha v^2 = mdv/dt$$ neglected gravity, but not friction.

I just now figured out that we could not use the energy conservation law as friction does some work (which is offcourse not negligible).
I calculated the velocity v2 as a function of radial distance r by using the binding energy equation i.e. mv2/r = GMm/r2
On substituting this value in the second equation with r=R at t=t , I got the correct answer!

Well you'd better check your calculation. The sign must be consistent. I'll show you that your equation is wrong.

We have:

$$v_1 = \sqrt{\frac{GM}{nR}}$$

$$v_2 = \sqrt{\frac{GM}{R}}$$

From Newton's 2nd law: $$m\frac{dv}{dt} = -av^2$$
(notice the minus sign? v is DECREASING, and friction ceases the motion, so there must be a minus sign!)

Again, do the integration, we have: $$t = \frac{m}{a}\sqrt{\frac{R}{GM}}(1-\sqrt{n})$$ which is wrong.

Here the constant a must be positive (a>0). You can verify that by looking at the expected result. In order that t>0, we must have a>0. Since a>0, there must be minus sign in the expression of frictional force. The question only gives you the magnitude of the force.

P.S.: This funny situation with the sign happens due to the fact that the component of gravity along the orbit has some connection with friction. It's actually a very surprising result. If you add gravity to the Newton's 2nd law's equation, the sign will change (it's like -A added 2A gives A ).

 

[zorro]

I don't think v decreases. The kinetic energy of the satellite increases in its spiral path as it approaches the earth. So v must increase.

You can also observe from the equation of v₁ and v₂ that v₂ > v₁.

But I also have a doubt, the total mechanical energy of the satellite decreases due to resistance from cosmic dust, as the satellite approaches earth, its potential energy decreases. Its kinetic energy should also decrease for a net decrease in total mechanical energy, which is found to violate the above result. Is it that the decrease in potential energy is more than the increase in kinetic energy?

 

[hikaru1221]

I don't think v decreases. The kinetic energy of the satellite increases in its spiral path as it approaches the earth. So v must increase.

You can also observe from the equation of v₁ and v₂ that v₂ > v₁.

That's exactly where your equation went wrong. If you only consider friction, speed definitely decreases. But there is gravity, and that's why it actually increases. See what's wrong with your equation?

But I also have a doubt, the total mechanical energy of the satellite decreases due to resistance from cosmic dust, as the satellite approaches earth, its potential energy decreases. Its kinetic energy should also decrease for a net decrease in total mechanical energy, which is found to violate the above result. Is it that the decrease in potential energy is more than the increase in kinetic energy?

Yes. You get this hint: -A = -2A + A? Potential energy decreases a lot (-2A), plus the work done by friction (-A), resulting in an increase in kinetic energy (A). If you don't take into account gravity, what you have is this: -A = A, which is so wrong. If you neglect the signs, we have: |-A| = |A|, which is so "right".

 

[zorro]

I understood it. Thanks hikaru