Can You Solve for R in Terms of t Using This Integral?

In summary, the conversation revolved around a mathematical problem involving an integral that describes the angular momentum of a particle under the influence of a net force. The original problem involved finding the equation of motion for a particle under an inverse cubic force. The conversation included discussions on the potential energy and conservation of energy and angular momentum. The conclusion was that the integral needed to be solved in order to determine the shape of the particle's orbit.
  • #1
Odyssey
87
0
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
What you've written is completely incomprehensible.
What are "theta" and R?
What is u and t?
 
  • #3
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
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
A good guess; I think it is some sort of separable differential equation he started up with, but I'm not sure..
 
  • #6
Odyssey: Could you state the original problem?
 
  • #7
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
Is this how it should be?
[tex]t-t_{0}=\int_{\theta_{0}}^{\theta(t)}\frac{m}{L}R^{2}(u)du[/tex]
 
  • #9
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! :smile:


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
Allright!
So, am I correct in assuming that "R" is the unknown radius in some orbit expressed as a function of the angle?
 
  • #11
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
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
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
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
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
Note:
DO NOT KEEP t AS THE INDEPENDENT VARIABLE!
Switch to using the angle.
 
  • #17
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
If H is some constant related to your A, yes; then the potential goes as the inverse square of the radius.
 
  • #19
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
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. :-p
 
  • #21
Careful!
There's a factor of "2" somewhere which you just forgot..
 
  • #22
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
It's about getting from your F to your V
(In addition, be careful about the sign as well..)
 
  • #24
[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
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...?
 
  • #26
What's your argument?
 
  • #27
arildno said:
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 [tex]V(u)[/tex] in the integral, where [tex] u = R(\Theta)[/tex]
 
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  • #28
I meant why you think the sign should be positive.
 
  • #29
Is it true that [tex] V = -\int F[/tex]? =\
 
  • #30
Wait. If that's true then it should be negative. AHHHH! ><
 
  • #31
Odyssey said:
[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:

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!
 

FAQ: Can You Solve for R in Terms of t Using This Integral?

What is integration?

Integration is a mathematical process used to find the area under a curve or the accumulation of a quantity over a given interval. It is the inverse operation of differentiation.

Why do we need to solve for R in integration?

In some integration problems, the variable R represents the upper limit of integration. Solving for R allows us to find the definite integral, which gives us the exact value of the accumulated quantity over the given interval.

What are the steps involved in solving for R in integration?

The steps involved in solving for R in integration include: 1) setting up the integration problem, 2) applying the appropriate integration rule, 3) solving for the indefinite integral, 4) evaluating the indefinite integral at the upper and lower limits of integration, and 5) subtracting the lower limit value from the upper limit value to find the definite integral.

What are some common integration rules used to solve for R?

Some common integration rules used to solve for R include the power rule, the substitution rule, and the integration by parts rule. These rules allow us to integrate different types of functions, such as polynomials, trigonometric functions, and exponential functions.

How can I check my solution for R in integration?

You can check your solution for R by plugging the value of R back into the original integration problem and evaluating it. If the result matches the given value of the definite integral, then your solution for R is correct.

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