How can I simplify vector analysis and understand the notation better?

[Shackleford]

I'm stuck on both problems. Here's my work thus far. I have no idea he got rtt in (b).

 

[Mark44]

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.

 

[Shackleford]

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.

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.

 

[Shackleford]

Here's as far as I got.

 

[Mark44]

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.

Yes, your r_t looks fine. I'll look over the other part and get back to you.

 

[Mark44]

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/r², where K is the constant referred to.

Differentiate both sides with respect to t to get
theta_tt = (-2K/r³)*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.

 

[Shackleford]

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/r², where K is the constant referred to.

Differentiate both sides with respect to t to get
theta_tt = (-2K/r³)*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_tt = (-2K/r³)*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/r³)*r_t, I get (-2K/r)*d(r)/dt

 

[Mark44]

I don't see where to use theta_tt = (-2K/r³)*r_t.

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

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/r³)*r_t, I get (-2K/r)*d(r)/dt

 

[Shackleford]

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

 

[Shackleford]

Actually, 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

 

[Mark44]

Actually, I just showed that the theta-component

= r^2 * d^2(theta)/dt^2 + d(theta)/dt * 2r * 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 is
$$r \frac{d^2 \theta}{dt^2} + 2\frac{dr}{dt} \frac{d\theta}{dt}$$

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

 

[Shackleford]

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
$$r \frac{d^2 \theta}{dt^2} + 2\frac{dr}{dt} \frac{d\theta}{dt}$$

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.

 

[Mark44]

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.

 

[Shackleford]

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.

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.

 

[Mark44]

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.

 

[Shackleford]

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.

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

I just used the product rule and chain.

 

[Mark44]

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

 

[Shackleford]

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.

 

[Shackleford]

I'm basically doing the same thing here.

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)

 

[Mark44]

You seem to be on the wrong track here. The goal for part b is to show that
r_tt - r($\theta_t$)² ~ 1/r²

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.

 

[Shackleford]

You seem to be on the wrong track here. The goal for part b is to show that
r_tt - r($\theta_t$)² ~ 1/r²

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.