What is the Minimum Distance of Masses After a Split in a Satellite System?

AI Thread Summary
The discussion revolves around a physics problem involving a satellite that splits into two equal masses while maintaining tangential velocity, resulting in increased kinetic energy. The key equation derived for the minimum distance of the masses from the star is R/(1+√(k-1)). Participants express confusion over the application of effective potential and conservation of angular momentum in their calculations. Several attempts to resolve the equations lead to incorrect answers, prompting requests for clarification and verification of the work done. Ultimately, the conversation highlights the complexities of energy conservation in satellite dynamics post-split.
Eitan Levy
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Homework Statement


A satellite with mass of m is circling a star. The radius of the circle is R.
At some moment the mass splits to 2 equal masses (the tangential velocity of the masses doesn't change). As a result of the split the kinetic energy in the system is multiplied by k (k>1). What will be the minimal distance of the masses from the star?
Answer: R/(1+√(k-1)).
zR0Kyms.png


Homework Equations


Ueff=L2/2mr2+V(r)

The Attempt at a Solution


I have honestly never seen a problem similar to this, and have never used effective potential to solve a problem (not saying it's necessary here).

I tried to solve the problem by saying that the energy before the split is -GMm/2R, because the potential energy is -GMm/R and the kinetic energy is GMm/2R. Then I tried to say that the new kinetic energy is kGMm/2R, and that the potential energy stays the same.

Then I wrote the equation -kGMm/2R-GMm/R=-GMm/rmin which gave the wrong answer.

Any help please?
 

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Eitan Levy said:
Then I wrote the equation -kGMm/2R-GMm/R=-GMm/rmin which gave the wrong answer.
This assumes that all of the energy at the point of minimal distance is in the form of potential energy. This cannot be the case as it would ignore conservation of angular momentum.
 
Orodruin said:
This assumes that all of the energy at the point of minimal distance is in the form of potential energy. This cannot be the case as it would ignore conservation of angular momentum.
Thanks.

I tried to fix the equation but it still gives the wrong answer.

The velocity before the explosion is V=√(GM/R). So at the minimal distance we will need to have a tangential velocity of V*R/rmin.

So I got the equation: -GMm/rmin+1/2*m*(V*R/rmin)2=-GMm/R+kGMm/2R.

What is the problem?
 
Eitan Levy said:
-GMm/rmin+1/2*m*(V*R/rmin)2=-GMm/R+kGMm/2R.
Did you even try solving that equation?
 
Orodruin said:
Did you even try solving that equation?
I did, perhaps I made a mistake?

Is it supposed to be correct? Because the solution I got is not close to the correct one.
 
Eitan Levy said:
Because the solution I got is not close to the correct one.
How are we supposed to check your work if you do not provide it?
 
Orodruin said:
How are we supposed to check your work if you do not provide it?
I apologize.

Vxd4yuL.png


I tried using some sites online to verify the solution so I am pretty sure something is wrong with the equation.
 

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