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

In summary, the problem involves a satellite with mass m orbiting a star at radius R. When the mass splits into two equal masses, the kinetic energy in the system is multiplied by a factor of k. The minimum distance of the masses from the star can be calculated using the equation R/(1+√(k-1)). The incorrect equation -kGMm/2R-GMm/R=-GMm/rmin ignores conservation of angular momentum and the correct equation is -GMm/rmin+1/2*m*(V*R/rmin)^2=-GMm/R+kGMm/2R. Rationalizing the denominator of this equation gives the correct solution of 1/(√(k-1)+1).
  • #1
Eitan Levy
259
11

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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  • #2
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.
 
  • #3
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?
 
  • #4
Eitan Levy said:
-GMm/rmin+1/2*m*(V*R/rmin)2=-GMm/R+kGMm/2R.
Did you even try solving that equation?
 
  • #5
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.
 
  • #6
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?
 
  • #7
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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  • #9

What is the "Effective Potential Problem"?

The effective potential problem is a concept in physics that is used to describe the behavior of a system of particles under the influence of a central force. It involves calculating the effective potential energy of the system, which takes into account both the potential energy of the particles and the kinetic energy of their motion.

Why is the "Effective Potential Problem" important?

The effective potential problem is important because it allows us to understand and predict the behavior of systems under the influence of a central force, such as the gravitational force between planets or the electrostatic force between charged particles. It also helps us to solve problems in classical mechanics and quantum mechanics.

How is the "Effective Potential Problem" solved?

The effective potential problem can be solved by using mathematical techniques such as calculus and differential equations. The first step is to write down the equations of motion for the system, which describe how the particles move under the influence of the central force. Then, by manipulating these equations, we can derive an expression for the effective potential energy and use it to analyze the behavior of the system.

What are some applications of the "Effective Potential Problem"?

The effective potential problem has many applications in physics, including the study of planetary motion, the behavior of atoms and molecules, and the motion of particles in a potential well. It is also used in fields such as astrophysics, quantum mechanics, and fluid dynamics.

What are some common misconceptions about the "Effective Potential Problem"?

One common misconception about the effective potential problem is that it only applies to systems with circular or elliptical orbits. In reality, it can be used to analyze the behavior of any system under the influence of a central force, regardless of the shape of its orbit. Another misconception is that the effective potential energy is the same as the total energy of the system, when in fact it is just a mathematical tool used to simplify the analysis of the system.

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