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

A, B

Explanation

Both the planet and the sun revolve around their center of mass with same angular velocity (say `omega`)
`r = r_(1) + r_(2)` ..(i)
`m_(1)r_(1) omega^(2) = m_(2) r_(2) omega^(2) = (Gm_(1) m_(2))/(r_(2))` ..(ii)
Solving Eqs. (i) and (ii), we get
`r_(1) - r(m_(2)/(m_(1) + m_(2)))`
`omega^(2) = (G(m_(1) + m_(2)))/(r^(3))`
Now, total energy of the system is
`E = PE + KE` or `E = - (Gm_(1)m_(2))/(r) + (1)/(2) m_(1) r_(1)^(2)omega^(2) + (1)/(2) m_(2)r_(2)^(2)omega^(2)`
Substituting the value of `r_(1),r_(2)` and `omega^(2)`, we get
`E = - (Gm_(1)m_(2))/(r)`