Question

A plant of mass m_(1) revolves round the sun of mass m_(2). The distance between the sun the planet is r. Considering the motion of the sun find the total energy of the system assuming the orbits to be circular.

Answer 1

Given:-

1. Mass of planet is m_1
2. Mass of sun is m_2
3. Distance between sun and planet is r.

To find:-

1. Total energy of the system assuming orbits are circular.

Answer:-

1. Velocity of planet is given by $V = \sqrt{\frac{G.m_{2} }{r} }$
2. We know that Total energy = Potential Energy+ Kinetic Energy
3. Potential Energy= $\frac{-G. m1. m2}{r}$  
4. Kinetic Energy= $\frac{1}{2} m_{1} . v^{2}$
5. Substituting V from 1 in Kinetic Energy we get,
6. Kinetic Energy= $\frac{Gm_{1}m_{2} }{2r} }$
7. Total Energy= Kinetic Energy+ Potential Energy
8. Substituting from 3 and 6,
9. Total Energy= $\frac{-G. m_1.m_2}{2r}$.

Answer 2

Hence the value of total energy of the system is E = − Gm1m2 / r

Explanation:

Both the planet and the sun revolve around their center of mass with same angular velocity (say ω)

r = r1 + r2    --------(1)

m1r1 ω^2 = m2r2ω^2  = Gm1m2 / r^2   -------(2)

Solving Equations. (1) and (2), we get

r1 − r( m2 / m1+m2)

ω^2  =  G(m1+m2) / r^3

Now, total energy of the system is

E = P.E + K.E

Or E = −Gm1m2 / r + 1 / 2m1r1^2ω^2 + 1 / 2m2r2^2ω^2

Substituting the value of r1,r2 and ω2, we get

E = − Gm1m2 / r

Hence the value of total energy of the system is E = − Gm1m2 / r