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Concept of reduced mass

  1. Mar 29, 2015 #1
    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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  3. Mar 29, 2015 #2

    Orodruin

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    Please provide references for this statement:
    Did you try actually writing down the energy of the system in the CoM system?
     
  4. Mar 29, 2015 #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}##.
     
  5. Mar 29, 2015 #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.
     
  6. Mar 29, 2015 #5

    Orodruin

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    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?
     
  7. Mar 29, 2015 #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"
     
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