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

AI Thread Summary
The discussion centers on the concept of reduced mass in a two-body system, specifically regarding the total energy calculation when one body revolves around another. It highlights that when using reduced mass, the total energy appears positive, which contradicts the expectation of a negative total energy from gravitational interactions. The participants clarify that the correct approach involves considering the kinetic energy of the system in the center of mass frame, leading to the conclusion that the total energy should indeed be negative. References to textbooks, such as DC Pandey's "Understanding Physics for JEE," are mentioned to support the calculations and concepts discussed. Ultimately, the conversation emphasizes the importance of correctly applying the principles of gravitational forces and energy in circular motion.
AdityaDev
Messages
527
Reaction score
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?
 
Physics news on Phys.org
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"
 
I think it's easist first to watch a short vidio clip I find these videos very relaxing to watch .. I got to thinking is this being done in the most efficient way? The sand has to be suspended in the water to move it to the outlet ... The faster the water , the more turbulance and the sand stays suspended, so it seems to me the rule of thumb is the hose be aimed towards the outlet at all times .. Many times the workers hit the sand directly which will greatly reduce the water...
Back
Top