1. Oct 3, 2004

### Odyssey

How do I intergrate this?

$$t - to = \int_\Theta ^\Theta(t) \frac{m}{L} R(u)^2 du$$

u is $$\Theta(t)$$, $$\Theta$$ is the angle at time zero, and t is time. Can I solve for R in terms of t?

Thanks.

Last edited: Oct 3, 2004
2. Oct 3, 2004

### arildno

What you've written is completely incomprehensible.
What are "theta" and R?
What is u and t?

3. Oct 3, 2004

### ComputerGeek

I think he is doing solids of revolution. that is the only thing that I can think of as to why there would be an R in there...but from theta to theta makes no sense to me.

4. Oct 3, 2004

### Odyssey

Sorry for the confusion of the integral. I am new to Latex...and I don't know how to do the "integral from theta to theta(t)" (theta of t). I didn't mean theta to theta. =\

5. Oct 3, 2004

### arildno

A good guess; I think it is some sort of separable differential equation he started up with, but I'm not sure..

6. Oct 3, 2004

### arildno

Odyssey: Could you state the original problem?

7. Oct 3, 2004

### Odyssey

Thanks for the help. I appreciate it! =D

OK, my problem is actually much more complicated than the integral.

Suppose a point mass m experiences a net force F = -Amr^-3.
How should I go about in describing its orbits for E>0, E <0, and E = 0 for the non-zero angular momentum cases?

My prof gave me an example on the gravitational force law, which is an inverse square law. I am asked to find out how the particle would move (say Earth) if it is under the influence of an inverse CUBIC law. The integral I (tried to) give above is the angular momentum of the particle. And yes, it is a differential equation! :)

Last edited: Oct 3, 2004
8. Oct 3, 2004

### arildno

Is this how it should be?
$$t-t_{0}=\int_{\theta_{0}}^{\theta(t)}\frac{m}{L}R^{2}(u)du$$

9. Oct 3, 2004

### Odyssey

Precisely! Yes, that's how it should be! :rofl:

Hummmm, I did some research on the Net, and I *think* it wouldn't be very pleasant if we live in a 1/r^3 world. The Earth would spiral into the Sun!?!?!? I am asked to find the equation of the orbit of such a world. Can you give me some pointers so I can start this problem please?

Thx for the help again.

Last edited: Oct 3, 2004
10. Oct 3, 2004

### arildno

Allright!
So, am I correct in assuming that "R" is the unknown radius in some orbit expressed as a function of the angle?

11. Oct 3, 2004

### Odyssey

Yes, precisely. R is a function of the angle, which is not limited to just between 0 and 2pi.

Let me get some more info on the problem. That might help! :) (will be posted here in minutes)

Last edited: Oct 3, 2004
12. Oct 3, 2004

### arildno

Well, then you can't integrate any further.
If you don't know what you integrate, you can't know what the integral would be.
(That is, what you've got is a dead-end)

13. Oct 3, 2004

### Odyssey

Total energy is conserved for a conservative force.

E = (1/2) m [r'(t)]^2 + (1/2) m [r(t)]^2 [theta'(t)]^2 + V (r(t))

Angular momentum is conserved for a central force.

L = m [r(t)]^2 [theta'(t)]^2

we can view r may be a function of theta, assuming that L is non-zero.

then E = (1/2) m[R'(theta(t))]^2 [theta'(t)]^2 + (1/2) m [R(theta(t))]^2 [theta'(t)]^2 + V(R(theta(t)))
and
L = m [R(theta(t))]^2 [theta'(t)]

14. Oct 3, 2004

### Odyssey

Hummm, under a 1/r^3 force, and using angular momentum/total energy conservation (writting the equations in polar coordinate form too), I separate the variables theta and t, and try to integrate to solve for R in terms of t (in trying to get an equation of the orbit). Am i heading into the right direction? That's how I got the integral at the very top in the first place.

15. Oct 3, 2004

### arildno

True enough; try to eliminate your time dependence, (using angular momentum equation),
$$\frac{d\theta}{dt}=\frac{L}{mR^{2}}$$

You should then by able to find a differential equation for R expressed in the new independent variable $$\theta$$
(Basically, this is the same technique used to find the orbits for an inverse-square field)
Good luck!

16. Oct 3, 2004

### arildno

Note:
DO NOT KEEP t AS THE INDEPENDENT VARIABLE!!
Switch to using the angle.

17. Oct 3, 2004

### Odyssey

OK, let me try to work this out........thanks for the help arildno!! =D

I need the potential V(R) too. Just to verify the work. Is V(R) = Hmr^-2?

Last edited: Oct 3, 2004
18. Oct 3, 2004

### arildno

If H is some constant related to your A, yes; then the potential goes as the inverse square of the radius.

19. Oct 3, 2004

### Odyssey

Actually I just found out the integral should be

$$t-t_{0}=\int_{\theta_{0}}^{\theta(t)}\frac{m}{L}R(u)^{2}du$$

So do I still treat R as a constant? No?

20. Oct 3, 2004

### Odyssey

OOps! Sorry, I used two variables for the same constant! A = H! Silly me. :tongue: