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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).

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- Thread starter Shackleford
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- #1

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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).

- #2

Mark44

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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.

- #3

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_{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.

Is my r

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

- #4

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Here's as far as I got.

- #5

Mark44

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Yes, your rIs 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.

- #6

Mark44

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Differentiate both sides with respect to t to get

theta

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

- #7

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_{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.

I don't see where to use theta

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

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- #8

Mark44

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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

- #9

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Look at the page you scanned. It's in the theta-component of acceleration.

Yes. I plugged that in.

If I insert thetatt = (-2K/r3)*rt, I get (-2K/r)*d(r)/dt

- #10

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= 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

- #11

Mark44

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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

[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.

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

- #12

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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.

Plugging in d^2(theta)/dt^2 and d(theta)/dt, I get r * (-2K/r^3) * (dr/dt) + 2 * (dr/dt) * (K/r^2) = 0.

- #13

Mark44

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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.

- #14

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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.

I did not understand the objection. It was a sum of a term and its negative, which is zero. The last one I did is also the sum of a term and its negative, so I'm not quite following you.

I understand the hint in (a), because the derivative of r^2 * (d(theta)/dt), which is a constant, is zero. So, we want to show that the theta-component is zero, and I guess I was missing the whole problem right off the bat by not using the second law.

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Mark44

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Oh. But isn't that the d/dt of the second law as written?

I just used the product rule and chain.

- #17

Mark44

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OK, now I get it. We both came at this from slightly different directions.

- #18

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OK, now I get it. We both came at this from slightly different directions.

Yeah. Sorry about that. Your method was more direct and in tune with the directions.

Thanks for the help so far. I'll post my (b) work in a little bit.

- #19

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dr/dt = dr/d(theta) * d(theta)/dt

d^2(r)/dt^2 = dr/d(theta) * d^2(theta)/dt^2 + d(theta)/dt * [dr/d(theta) * d(theta)/dt]

= dr/d(theta) * d^2(theta)/dt^2 + (d(theta)/dt)^2 * dr/d(theta)

= dr/d(theta) * (-2K/r^3 * dr/dt) + (K/r^2)^2 * dr/d(theta)

- #20

Mark44

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r

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

- #21

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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.

Yeah. I understand the notation. I just haven't taken the time to learn the latex. Maybe I should do that. lol.

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