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

In summary, the energy of the system is still positive even when the two bodies are replaced by a single body of mass.
  • #1
AdityaDev
527
33
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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  • #2
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?
 
  • #3
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}##.
 
  • #4
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.
 
  • #5
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?
 
  • #6
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"
 

1. What is the concept of reduced mass?

The concept of reduced mass is a measure of the effective mass of a system that is undergoing a two-body interaction. It takes into account the masses of both bodies as well as their distance from each other.

2. How is reduced mass calculated?

The formula for reduced mass is μ = (m1m2)/(m1+m2), where m1 and m2 are the masses of the two bodies in the system. This formula is derived from Newton's second law of motion.

3. Why is the concept of reduced mass important?

The concept of reduced mass is important because it helps simplify the equations used to describe the motion of two bodies interacting with each other. By using reduced mass, the equations become more manageable and easier to solve.

4. What is the relationship between reduced mass and relative motion?

Reduced mass is directly related to relative motion, as it is used to calculate the acceleration and forces acting on each body in a two-body system. It also takes into account the distance between the two bodies, which affects their relative motion.

5. Can reduced mass be negative?

No, reduced mass cannot be negative. It is a positive value that represents the effective mass of the system. If one of the masses in the system is significantly larger than the other, the reduced mass will be closer to the smaller mass but it will always be a positive value.

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