- #1
Shackleford
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I'm stuck on both problems. Here's my work thus far. I have no idea he got rtt in (b).
Use the hint. You have r_{t}, so differentiate it again with respect to t to get r_{tt}. You'll need to undo your substitution for u to get back into terms of theta. You'll need to use the product rule and then quotient rule. Also d/dt(dtheta/dt) = d^2(theta)/dt^2.
One other thing: you'll need to use the equation r^2 d(theta)/dt = const ==> d(theta)/dt = const/r^2.
I haven't worked it all the way through, but these are the things I would do.
Yes, your r_{t} looks fine. I'll look over the other part and get back to you.Is my r_{t} correct?
dr/d(theta) = [(a e sin(theta))/(1 + e cos(theta))^2 ] d(theta)/dt
I differentiated with respect to t again and didn't get anything close to the r_{tt} on the sheet. I'll scan it in a few minutes.
On second thought, even though your work was fine, I don't think what you're doing is the way to go about this problem. Let's look at the a part, since you said you were stuck on it, too. From Kepler's 2nd law, we have theta_{t} = K/r^{2}, where K is the constant referred to.
Differentiate both sides with respect to t to get
theta_{tt} = (-2K/r^{3})*r_{t}
Using this and Kepler's 2nd law in the form above, you can show that the theta-component of the acceleration is zero.
When you get to that point, you have shown that a = [r_{tt} - r(theta_{t})^{2}]u_{r} and the b part is pretty straightforward after that.
Look at the page you scanned. It's in the theta-component of acceleration.I don't see where to use theta_{tt} = (-2K/r^{3})*r_{t}.
The only way I can see to get the theta-component form is if I do the following and divide by 1/r.
d(r^2 * d(theta)/dt)/dt = r^2 * d^2(theta)/dt^2 + d(theta)/dt * d(r^2)/dt =
r^2 * d^2(theta)/dt^2 + d(theta)/dt * (2r*d(r)/dt
If I insert theta_{tt} = (-2K/r^{3})*r_{t}, I get (-2K/r)*d(r)/dt
Look at the page you scanned. It's in the theta-component of acceleration.
If I insert thetatt = (-2K/r3)*rt, I get (-2K/r)*d(r)/dt
That's not quite right; you have an extra factor of r all the way through. Also, you didn't need to show it, since it is given right near the beginning of the page you scanned. From that page, the theta-component of acceleration isActually, I just showed that the theta-component
= r^2 * d^2(theta)/dt^2 + d(theta)/dt * 2r * d(r)/dt
Now, plugging in my d^2(theta)/dt^2 and d(theta)/dt from the law gives
= r^2 * (-2K/r^3) + (K/r^2) * 2r * d(r)/dt
= -2K/r * d(r)/dt + 2K/r * d(r)/dt = 0
That's not quite right; you have an extra factor of r all the way through. Also, you didn't need to show it, since it is given right near the beginning of the page you scanned. From that page, the theta-component of acceleration is
[tex]r \frac{d^2 \theta}{dt^2} + 2\frac{dr}{dt} \frac{d\theta}{dt}[/tex]
I outlined what you can do to show that this expression is zero.
OK, that's what you needed to to. Did you understand my objection to your work in post 10?
Anyway, that takes care of part a. Now you're ready to take on part b. Use the same substitution from Kepler's 2nd law - d(theta)/dt = K/r^2 - and work on the radial component of acceleration.
My objection was that the expression you showed was zero wasn't the right expression. Yours had an extra factor of r for some reason.
OK, now I get it. We both came at this from slightly different directions.
You seem to be on the wrong track here. The goal for part b is to show that
r_{tt} - r([itex]\theta_t[/itex])^{2} ~ 1/r^{2}
Don't overthink this - it's much simpler than the first part.
The subscript notation is a lot less cumbersome than the d/dt etc. notation you're using. I don't know if you're not understanding it, but they are using r_{tt} to mean the same thing as d^2(r)/dt^2.