Assume the earth's orbit around the sun as circular and the distance between their centres as D. Mass of the earth is M and its radius is R. If earth has an angular velocity with respect to its centre and with respect to the centre of the sun, the total kinetic energy of earth is

Options

A. `(MR^(2))/(5)omega_(0)^(2)[1+((omega)/omega_(0))^(2)+(5)/(2)((Domega)/(Romega_(0)))^(2)]`
B. `(MR^(2))/(5)omega_(0)^(2)[1+(5)/(2)((Domega)/(Romega_(0)))^(2)]`
C. `(2)/(3)MR^(2)omega_(0)^(2)[1+(5)/(2)((Domega)/(Romega_(0)))^(2)]`
D. `(2)/(5)MR^(2)omega_(0)^(2)[1+((omega)/omega_(0))^(2)+(5)/(2)((Domega)/(Romega_(0)))^(2)]`

Answer

(b)

Explanation

Kinetic energy of earth with respect to its centre is,
`KE_(1)=(1)/(2)Iomega_(0)^(2)`
`= (1)/(2)xx(2)/(5)MR^(2)xxomega_(0)^(2)`
Kinetic energy of earth with respect to the sun is,
`KE_(1)=(1)/(2)Iomega_(0)^(2)`
`(1)/(2)xxMR^(2)((D^(2))/(R^(2)))omega^(2)`
Total kinetic energy `K.E_(Total)`
`KE_(1)+KE_(2)`
`= (1)/(2)xx(2)/(5)MR^(2)xxomega_(0)^(2)+(1)/(2)xxMR^(2)((D^(2))/(R^(2)))omega^(2)`
` (MR^(2)omega_(0)^(2))/(5)[1+(5)/(2)((Domega)/(Romega_(0))^(2))]`