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Question 2
I think it just means, "write \vec{F} = m\vec{r}''". But it doesn't mean to draw a graph, where did you get that idea?
The Earth's atmosphere [...] circular orbit.
I don't get what these physicists were thinking. After the storm, the new orbit is an ellipse, and r=3a is its aphelie, so r'=0 there naturally! They don't have to exert no force to put its radial velocity to 0 "instanteneously"; it already is. The only way to restore the new orbit is by instantaneously restoring its angular velocity
when the satellite is at r=3a.
By solving the equation [...] occur).
If we break down the equation of motion written in the last question into radial and angular part, we get the two ugly coupled following ode:
-\frac{\lambda}{r^2}-kr' = r''-r(\phi ')^2
-kr\phi '=r\phi '-2r'\phi '
Setting k =0, they become instead
-\frac{\lambda}{r^2} = r''-r(\phi ')^2
0=r\phi '-2r'\phi '
Now, we are told that the new angular momentum is half the old one: L_f = L_i/2. Angular momentum in polar coordinates is simply L=mr^2\phi ' \Leftrightarrow (r\phi ')^2=L^2/(m^2r^3). So we can substitute (r\phi ')^2 in our first equation for (L_i/2)^2/(m^2r^3) and it becomes
r'' = -\frac{\lambda}{r^2} + \frac{(L_i/2)^2}{m^2r^3}
which is an equation for r only, hurray. Now, we assume that r can be written as a function of phi, and we make the good old substitution r=1/u. Now the equation is just that of a forced harmonic oscillator:
\frac{d^2u}{d\phi^2}+u=-\frac{m\lambda}{(L_i/2)^2}
It's an inhomogeneous ode of second order with constant coefficient, so the general solution is just the solution of the associated homogeneous equation + a particular solution of the inomogeneous one. The homogeneous equation associated is just that of an harmonic oscillator of unitary angular frequency:
\frac{d^2u}{d\phi^2}+u=0
We know the general solution to be Acos(\phi + \phi_0).
Meanwhile, an obvious particular solution to \frac{d^2u}{d\phi^2}+u=-\frac{m\lambda}{(L_i/2)^2} is just u=-\frac{m\lambda}{(L_i/2)^2}, making the general solution
u(\phi) = -\frac{m\lambda}{(L_i/2)^2}+Acos(\phi + \phi_0)
Choose a coordinate system such that when r=3a, \phi = 0. Then you have the "initial" conditions
u(0)=\frac{1}{3a}
\frac{du}{d\phi}(0)=0
The second condition is because when the orbit was a circle of radius r=3a, \frac{du}{d\phi} was always 0 (otherwise the orbit would not be a circle).
From these, you can find A and \phi_0=0.
Hopefully, you've seen this derivation before and you know that in general, the orbit can be either an ellipse, a circle, an hyperbolea or a parabola depending on the energy of the particle (ellipse corresponding to lowest energy, circle higher even, hyperbolea higher even and parabola highest). We know that initially, the orbit was a circle (which is a kind of ellipse). Then rose the atmosphere which induced friction on the satellite, exerting negative work on it and thus diminishing its total mechanical energy. So the new orbit will necessarily be an ellipse, with the inert mass located at one of the foci!
Hopefully, you also know that the equation of an ellipse of half major axis b (along the x axis), semi minor axis \sqrt{a^2-(a\epsilon)^2} and focal distance 2b\epsilon in polar coordinates is
\frac{1}{r} = \frac{1}{b(1-\epsilon^2)}+\frac{\epsilon}{b(1-\epsilon^2)}cos(\phi)
(with 0 \leq \epsilon \leq 1 [the circle corresponds to the limiting case \epsilon = 0])
Compare the general equation of the ellipse with your general solution for u and solve for \epsilon and b. Now you have all you need to draw the ellipse. Finally, the last question is to determine wheter in this new orbit, the satellite will be further influenced by the atmosphere. This is equivalent to asking "is the perhelie of the orbit lesser than 'a'?", right? The perihelie is defined as the point of the orbit closest to the inert mass. So it is a minimum of the function r(\phi) and thus a max of u(\phi). Find it. Is it greater than 'a'? If so, great, the satelite won't fall on our heads! (Not until the next freak solar storm at least.
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but I really do appreciate it :)
