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Relation between the two Radii Rf and Ro 
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#1
Feb113, 05:10 AM

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1. The problem statement, all variables and given/known data
A planet orbits a massive star in a highly elliptical orbit, i.e, the total orbital energy is close to Zero. The initial distance of closest approach is 'Ro'. Energy is dissipated through tidal motions until the orbit is circularized with a final radius of 'Rf'. Assume the orbital angular momentum is conserved during the circularization process, Then, what is the relation between Rf and Ro ? 2. Relevant equations 1.) GMm/2Ro 2.) GMm/Rf 3. The attempt at a solution I tried subtracting the gravitational energy from the Energy of a circular orbit, to account for the dissipation, and then equated it to zero(since it was originally almost 0). I ended up with 'Rf=2Ro' But I am not sure about my method. Felt like I have left a lot of loose ends. Also the Angular momentum conservation hint was never used. I believe I have reached the wrong answer(or at best, the right answer the wrong way.) Need help in solving this problem. If anybody can help me set up the necessary equation(not jut the final answer), I would appreciate it. Thank you in advance :) 


#2
Feb113, 12:33 PM

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Energy is lost, so energy conservation is useless here. Angular momentum, on the other hand, is conserved, so I would use this. What is the initial angular momentum? Which circular orbit has the same angular momentum?



#3
Feb113, 08:38 PM

P: 18

Applying it here I get, mv_{1}Ro*Sinθ=mv_{2}Rf Implying, Ro/Rf=v_{2}/v_{1}sinθ Is that right ? I'll give a heads up, I am a bit of a douche, so you'll have to be a bit patient with me. But I will certainly try my best to come up with the right equations :) And Thank you for replying.. 


#4
Feb213, 01:28 PM

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Relation between the two Radii Rf and Ro
Right.
Some parts of the equation can be calculated with the known initial conditions, and you can get another equation for the circular motion. 


#5
Feb213, 08:33 PM

P: 18

Ok, so I make θ=90° and get Ro/Rf=v_{2}/v_{1}.
Now if I can get v_{1} or v_{2} in terms of each other, then I can get my answer. So I look for another relationship, I try the force equation, and end up with, mv_{1}^{2}/Ro=GMm/Ro^{2} which even when simplified doesn't take me much further. Then I tried the total energy, GMm/2Ro≈0 Again this led to nowhere ! This is where I get stuck :) Am I supposed to write in terms of Kinetic energy so as to include a velocity term ? Even so, everything goes down to 0 because of the RHS X) 


#6
Feb213, 10:33 PM

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#7
Feb313, 05:57 AM

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#8
Feb313, 08:36 PM

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K.E=(1/2)P.E
K.E.= (1/2)Mv_{1}^{2} P.E.=GMm/Ro So (since total energy is ≈0), (1/2)Mv_{1}^{2}=GMm/2Ro Even if I do the same with v_{2} and Rf, at best I reach Rf=Ro. But I am unable to understand how to set up equations differently for an Elliptical or bit, and specifically for a circular orbit. All the equations I seem to know are for circular orbits. So I am going wrong somewhere with my calculation of v1 and Ro, isn't it ? X) 


#9
Feb413, 02:09 AM

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The relation between v1 and R0 is correct.
Going back to Ro/Rf=v2/v1 again, you have Ro as fixed parameter, and you can express v1 and v2 in terms of Ro and Rf, respectively. The constant GMm should cancel if you do this, and you get a relation between Ro and Rf only. 


#10
Feb413, 02:26 AM

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[itex]\displaystyle \text{P.E.}=G\frac{mM}{R_f}\ .[/itex]The kinetic energy for the same orbit is [itex]\displaystyle \text{K.E.}=\frac{1}{2}m{v_f}^2=G\frac{mM}{2R_f}\ .[/itex]So the total mechanical energy for a circular orbit of radius, R_{f} is [itex]What is the angular momentum for this circular orbit? 


#11
Feb413, 09:24 AM

P: 18

My solution seems to have gone into a circular orbit as well. I started by calculating the angular momentum of a circular orbit, and now I am back to finding the angular momentum of a circular orbit ! X)
Anyway, how do I write v1 and v2 in terms of Ro and Rf ? I can so far only write v1 in terms of Ro and v2 in terms of Rf, still stalling any possible headway I can make. And even after manipulating algebraically(cancelling the GMm), I can reach Ro=Rf ! Is that the right relation ? :0 I am not sure it is.. 


#12
Feb413, 10:47 AM

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R0=Rf is wrong, you forgot a factor of 2 somewhere (might become a factor of sqrt(2) in the process).



#13
Feb413, 08:17 PM

P: 18

Ok, let me start over again.
So far I have, 1.) R_{0}/R_{f}=v_{2}/v_{1} 2.) P.E.=GMm/R_{f} 3.) K.E=(1/2)mv_{22}=GMm/2R_{f} 4.) Total Energy=GMm/2R_{f} 5.) Angular Momentum(in general)=MvRSinθ (θ=90°for a circle) 6.) v_{1}^{2}=Gm/R_{0} 7.) Centripetal force=Mv_{2}^{2}/R_{f} 8.) Gravitational Force(in general)=GMm/R^{2} So now which equations do I equate to which ones in what order and get my answer ? Obviously energy with energy; force with force and momentum with momentum. Equation (1.) comes from Conservation of angular momentum. I am unable to use Equations (2.), (3.) and (4.) in a productive way on paper like you guys have done it in your head ! 


#14
Feb513, 02:15 PM

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With 0 and f as indices everywhere:
(1) ##\frac{R_0}{R_f}=\frac{v_f}{v_0}## (from conserved angular momentum) (2) ##\frac{1}{2}v_0^2=\frac{GM}{R_0}## (from negligible initial energy, v is the velocity at the radius R_{0}) (3) ##\frac{1}{2}v_f^2=\frac{GM}{2R_f}## (circular final orbit) Dividing equation 2 by equation 3 gives ##\frac{v_0^2}{v_f^2}=\frac{2R_f}{R_0}## and this is equivalent to $$\frac{v_0}{v_f}=\sqrt{\frac{2R_f}{R_0}}$$ Using this in (1) will give you the correct answer. 


#15
Feb513, 08:32 PM

P: 18

Now, at the risk of sounding stupid, can you explain to me how you got Equation 2 ? That was my culprit. I wrote equation 2(for R_{0} and v_{1}) the same as its written in equation 3(for R_{f} and v_{2}). I don't want to make the same mistakes the next time I form such equations(I prefer making new ones :D).
Got the final answer though, R_{f}=2R_{0}. Thanks a lot mfb and SammyS. Appreciate it :) 


#16
Feb513, 10:57 PM

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The planet comes from very far where it's total orbital energy is close to Zero. Both the K.E. and P.E. being nearly zero. At closet approach its P.E. is [itex]\displaystyle \ \ G\frac{mM}{R_0}\,,\ [/itex] so it loses how much P.E. ? That's how much K.E. it gains. 


#17
Feb613, 04:44 AM

P: 18

Ok. Thats kind of how I had started and gotten R_{f}=2R_{0} even before I had asked this question here. But my concepts were all messed up and so were my equations.
Thanks a lot for your patience :) 


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