Why Does the Reduced Mass Concept Yield a Positive Total Energy?

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Discussion Overview

The discussion revolves around the concept of reduced mass in the context of gravitational interactions between two bodies, specifically addressing why the total energy calculated using reduced mass appears positive, while traditional formulations yield negative total energy. The scope includes theoretical considerations and mathematical reasoning related to gravitational systems and energy calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that when considering a planet revolving around a star, the system can be simplified to a single body of reduced mass ##\mu##, leading to a total energy expression of ##\frac{1}{2}\mu v^2##, which is positive.
  • Another participant questions the validity of this total energy expression and suggests writing down the energy of the system in the center of mass (CoM) frame.
  • A different participant asserts that the energy of a body in circular motion should indeed be ##\frac{1}{2}\mu v^2##, but presents conflicting results when calculating total energy using different methods, leading to confusion about the signs of the energy terms.
  • One participant acknowledges a misunderstanding regarding the application of gravitational force in the context of a single body, indicating a need for clarification.
  • Another participant emphasizes the need to replace the two bodies with a single body of reduced mass moving in a central gravitational potential and requests references for the claims made.
  • A participant cites a textbook, DC Pandey's "Understanding Physics for JEE Main and Advanced," and shares equations related to the kinetic energy of the system and the relationship between the masses and their respective distances.

Areas of Agreement / Disagreement

Participants express differing views on the application of the reduced mass concept and its implications for total energy calculations. There is no consensus on the correct interpretation or resolution of the positive total energy issue.

Contextual Notes

Participants reference various equations and methods for calculating energy, but there are indications of missing assumptions and potential errors in the application of gravitational principles. The discussion reflects uncertainty regarding the correct approach to energy calculations in this context.

AdityaDev
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According to the concept: when a planet revolves around a star, and when both the bodies move in circular orbits due to the interaction between each other, both the bodies can be replaced by a single body of mass ##\mu## revolving in a circular orbit of radius equal to the distance between both the bodies.
so the total energy of the system becomes ##\frac{1}{2}\mu v^2##
But it is positive. The total energy has to be be negetive
##-\frac{GMm}{r}+E_1+E_2## where e1, e2 are kinetic energies of the bodies comes out to be negetive after solving.
Why do I get positive sign using concept of reduced mass?
 
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Please provide references for this statement:
AdityaDev said:
so the total energy of the system becomes ##\frac{1}{2}\mu v^2##

Did you try actually writing down the energy of the system in the CoM system?
 
Only the statement is given.
The energy of the body in circular motion should be ½μν2 right? Since both the bodies can be replaced by a single body of reduced mass ##\mu##
I got the total energy using ##-\frac{GMm}{r}+\frac{1}{2}I_1\omega^2+\frac{1}{2}I_2\omega^2=-\frac{GMm}{2r}## which is also given in my textbook.
On using ½μν2, i get ##\frac{GMm}{2r}##.
 
I understand now. I tried using the relation $$mv^2/r=GMm/R^2$$ which is wrong since there is no force of gravitation when you just have a single body.
 
No no no, you replace the two bodies with a single body of mass ##\mu## moving in a central gravitational potential. You still need to provide the reference, what book are you using?
 
DC Pandey, Understanding Physics for JEE main and advanced.
I did the 1/2μν2 part myself. I just posted what i understood so that you could correct me. This is taken from the textbook (exactly the same including errors):

$$m_1r_1=m_2r_2$$
$$m_1r_1\omega^2=m_2r_2\omega^2=\frac{Gm_1m_2}{r^2}$$
$$L=(I_1+I_2)\omega^2=\mu r^2\omega^2$$
Kinetic Energy of the system, ##K=\frac{1}{2}\mu r^2\omega^2##
"Thus, the two bodies can be replaced by a single body whose mass is equal to the reduced mass. The single body revolve in a circular orbit whose radius is equal to the distance between the bodies and force of circular motion is equal to force of interaction between the two bodies for actual separation"
 

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