Integration help please

  • Thread starter Odyssey
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  • #1
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How do I intergrate this?

[tex] t - to = \int_\Theta ^\Theta(t) \frac{m}{L} R(u)^2 du [/tex]

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

Thanks.
 
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  • #2
arildno
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What you've written is completely incomprehensible.
What are "theta" and R?
What is u and t?
 
  • #3
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arildno said:
What you've written is completely incomprehensible.
What are "theta" and R?
What is u and t?

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
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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
arildno
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A good guess; I think it is some sort of separable differential equation he started up with, but I'm not sure..
 
  • #6
arildno
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Odyssey: Could you state the original problem?
 
  • #7
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arildno said:
Odyssey: Could you state the original problem?

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? :confused:

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! :)
 
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  • #8
arildno
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Is this how it should be?
[tex]t-t_{0}=\int_{\theta_{0}}^{\theta(t)}\frac{m}{L}R^{2}(u)du[/tex]
 
  • #9
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arildno said:
Is this how it should be?
[tex]t-t_{0}=\int_{\theta_{0}}^{\theta(t)}\frac{m}{L}R^{2}(u)du[/tex]

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.
 
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  • #10
arildno
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Allright!
So, am I correct in assuming that "R" is the unknown radius in some orbit expressed as a function of the angle?
 
  • #11
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arildno said:
Allright!
So, am I correct in assuming that "R" is the unknown radius in some orbit expressed as a function of the angle?

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)
 
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  • #12
arildno
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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
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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
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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
arildno
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True enough; try to eliminate your time dependence, (using angular momentum equation),
[tex]\frac{d\theta}{dt}=\frac{L}{mR^{2}}[/tex]

You should then by able to find a differential equation for R expressed in the new independent variable [tex]\theta[/tex]
(Basically, this is the same technique used to find the orbits for an inverse-square field)
Good luck!
 
  • #16
arildno
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Note:
DO NOT KEEP t AS THE INDEPENDENT VARIABLE!!
Switch to using the angle.
 
  • #17
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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?
 
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  • #18
arildno
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If H is some constant related to your A, yes; then the potential goes as the inverse square of the radius.
 
  • #19
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arildno said:
Is this how it should be?
[tex]t-t_{0}=\int_{\theta_{0}}^{\theta(t)}\frac{m}{L}R^{2}(u)du[/tex]

Actually I just found out the integral should be

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

So do I still treat R as a constant? No?
 
  • #20
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arildno said:
If H is some constant related to your A, yes; then the potential goes as the inverse square of the radius.

OOps! Sorry, I used two variables for the same constant! A = H! Silly me. :tongue:
 
  • #21
arildno
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Careful!
There's a factor of "2" somewhere which you just forgot..
 
  • #22
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arildno said:
Careful!
There's a factor of "2" somewhere which you just forgot..

In the potential? :confused: Let me try to get the energy equation to integrate.
 
  • #23
arildno
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It's about getting from your F to your V
(In addition, be careful about the sign as well..)
 
  • #24
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[tex]t-t_{0}=\int_{R_{0}}^{R(t)}\frac{du}{\sqrt{2mEL^{-2}u^4-u^2-2mL^{-2}u^4V(u)}}[/tex]

where, u = R of theta. So should I plug the V = -(1/2)Amr^-2 into V(u)? :confused:
 
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  • #25
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arildno said:
It's about getting from your F to your V
(In addition, be careful about the sign as well..)

Ah yes, (1/2)Amr^-2. The sign...it should be positive.....?
 

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