Planetary motion in a viscous medium

[Leo Liu]

Homework Statement

none

Relevant Equations

F=ma

The answer to (c) is $-2\pi AGMm$.
Answer to (d)

For sub-question d, I used a different approach and I don't know why the solution to (d) is an appropriate approximation.
What I did was that I use Newton's laws to obtain two differential equation in polar coordinate, as shown:
$$\text{Assume that the planet moves counterclockwise}$$
$$-\frac{GMm}{r^2}\boxed{\hat r}-Amv^2\boxed{\hat\theta} =m(\ddot r-r\dot\theta ^2)\boxed{\hat r}+m(r\ddot\theta+2\dot r \dot\theta)\boxed{\hat\theta}$$
$$\begin{cases}
-A(r\dot\theta)^2=r\ddot\theta+2\dot r\dot\theta\\
-GM/r^2=\ddot r-r\dot\theta^2\\
\end{cases}$$
I would like to know if my solution is correct, and why the official solution works. Thanks a lot.

 

[LCSphysicist]

Homework Statement:: none
Relevant Equations:: F=ma

View attachment 272709
View attachment 272711
The answer to (c) is $-2\pi AGMm$.
Answer to (d)
View attachment 272712

For sub-question d, I used a different approach and I don't know why the solution to (d) is an appropriate approximation.
What I did was that I use Newton's laws to obtain two differential equation in polar coordinate, as shown:
$$\text{Assume that the planet moves counterclockwise}$$
$$-\frac{GMm}{r^2}\boxed{\hat r}-Amv^2\boxed{\hat\theta} =m(\ddot r-r\dot\theta ^2)\boxed{\hat r}+m(r\ddot\theta+2\dot r \dot\theta)\boxed{\hat\theta}$$
$$\begin{cases}
-A(r\dot\theta)^2=r\ddot\theta+2\dot r\dot\theta\\
-GM/r^2=\ddot r-r\dot\theta^2\\
\end{cases}$$
I would like to know if my solution is correct, and why the official solution works. Thanks a lot.

I wouldn't do it by your way, the velocity vector is, actually is$$\dot{r}\widehat{r} + r\dot{\theta}\widehat{\theta}$$, what you show is the velocity in an circular orbital, if it was a circular orbital, the radius wouldn't varies anyway :S

And, what you didn't understand in the approximation? From c we have "the radius varies to low in a period", so it is a understandable approximation.
Remember the differentials theory

$$\Delta R = \frac{dr}{dt}\Delta t$$

 

[Delta2]

I think your approach though it is aiming for the exact solution, has a big disadvantage:
We can't find the exact solution cause the system of ODEs you write has no closed form solution.

So we have to do approximations as the main solution suggests. All the equations for $\Delta r$ and $\Delta t$ are approximations in the given solution for d).

 

[Leo Liu]

what you show is the velocity in an circular orbital, if it was a circular orbital, the radius wouldn't varies anyway :S

Oh yes you are definitely correct. I found out this problem after I could edit my post. Should the first equation be $-A[(r\dot\theta)^2+\dot r ^2]=r\ddot\theta+2\dot r\dot\theta$?

Even this is not accurate because the direction of velocity is not the same as hat theta. I think we have to assume the angle between the tangent and the velocity is phi and make use of the following identity: $\tan\phi=\dot r/r\dot\phi$

And, what you didn't understand in the approximation? From c we have "the radius varies to low in a period", so it is a understandable approximation.
Remember the differentials theory

What I don't understand is why the first equation of the official solution is valid.

Thanks.

 

[LCSphysicist]

Oh yes you are definitely correct. I found out this problem after I could edit my post. Should the first equation be $-A[(r\dot\theta)^2+\dot r ^2]=r\ddot\theta+2\dot r\dot\theta$?

Even this is not accurate because the direction of velocity is not the same as theta.

What I don't understand is why the first equation of the official solution is valid.

Thanks.

The first equation is what they call, in short words, the virial theorem. I recommend you to seek it at wikipedia.
" For gravitational attraction,... the average kinetic energy equals half of the average negative potential energy"
Knowing that E = Epot + Ekin, i think you can understand now how the equation arose ;)
Anyway, while it is approximately true in ellipse orbital, it is exactly true in circular orbits, as follows:
E = mv²/2 - GmM/r
-GmM/r² = -mv²/r (centripetal)
E = GmMr/2r² - GmM/r
E = -GmM/2r

 

[Leo Liu]

The first equation is what they call, in short words, the virial theorem. I recommend you to seek it at wikipedia.

This is exactly what I have been looking for. What puzzled me was the reason a small change in energy corresponded to the force times an infinitesimal movement :P