View Full Version : Integration help please
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.
What you've written is completely incomprehensible.
What are "theta" and R?
What is u and t?
ComputerGeek
Oct3-04, 06:12 PM
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.
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. =\
A good guess; I think it is some sort of separable differential equation he started up with, but I'm not sure..
Odyssey: Could you state the original problem?
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! :)
Is this how it should be?
t-t_{0}=\int_{\theta_{0}}^{\theta(t)}\frac{m}{L}R^{2 }(u)du
Is this how it should be?
t-t_{0}=\int_{\theta_{0}}^{\theta(t)}\frac{m}{L}R^{2 }(u)du
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.
Allright!
So, am I correct in assuming that "R" is the unknown radius in some orbit expressed as a function of the angle?
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)
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)
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)]
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.
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!
Note:
DO NOT KEEP t AS THE INDEPENDENT VARIABLE!!
Switch to using the angle.
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?
If H is some constant related to your A, yes; then the potential goes as the inverse square of the radius.
Is this how it should be?
t-t_{0}=\int_{\theta_{0}}^{\theta(t)}\frac{m}{L}R^{2 }(u)du
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?
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:
Careful!
There's a factor of "2" somewhere which you just forgot..
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.
It's about getting from your F to your V
(In addition, be careful about the sign as well..)
t-t_{0}=\int_{R_{0}}^{R(t)}\frac{du}{\sqrt{2mEL^{-2}u^4-u^2-2mL^{-2}u^4V(u)}}
where, u = R of theta. So should I plug the V = -(1/2)Amr^-2 into V(u)? :confused:
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.....?
What's your argument?
Hummm, this is where I am stuck. I don't know how to proceed with the integral 3 posts above. :confused:
Integration of that integral would allow me to solve for R in terms of theta, which will in turn tell me the shape of the orbit of any test particle about the origin, in terms of its energy and angular momentum.
But the thing is, I don't know how to use the term V(u) in the integral, where u = R(\Theta)
I meant why you think the sign should be positive.
Is it true that V = -\int F? =\
Wait. If that's true then it should be negative. AHHHH!! ><
t-t_{0}=\int_{R_{0}}^{R(t)}\frac{du}{\sqrt{2mEL^{-2}u^4-u^2-2mL^{-2}u^4V(u)}}
where, u = R of theta. So should I plug the V = -(1/2)Amr^-2 into V(u)? :confused:
Now I have the integral. plug the V = -(1/2)Amr^-2 into V(u)? How should I proceed from here? :confused:
Thank you again for the help!
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