Question

Assume the earth's orbit around the sun is circular and the distance between their centres is D. Mass of the earth is M and its radius is R. If earth has an angular velocity ${\omega }_0$ with respect to its centre and $\omega$ with respect to the centre of the sun, the total kinetic energy of the earth is :

• A. $\frac{M{R}^2{\omega }_0^2}{5}[1+{\left(\frac{\omega }{{\omega }_0}\right)}^2+\frac{5}{2}{\left(\frac{D\omega }{R{\omega }_0}\right)}^2]$
• B. $\frac{M{R}^2{\omega }_0^2}{5}[1+\frac{5}{2}{\left(\frac{D\omega }{R{\omega }_0}\right)}^2]$
• C. $\frac{2}{5}M{R}^2{\omega }_0[1+\frac{5}{2}{\left(\frac{D\omega }{R{\omega }_0}\right)}^2]$
• D. $\frac{2}{5}M{R}^2{\omega }_0^2[1+{\left(\frac{\omega }{{\omega }_0}\right)}^2+\frac{5}{2}{\left(\frac{D\omega }{R{\omega }_0}\right)}^2]$

Answer 1

The correct option is B $\frac{M{R}^2{\omega }_0^2}{5}[1+\frac{5}{2}{\left(\frac{D\omega }{R{\omega }_0}\right)}^2]$

The moment of inertia of a sphere $I=\frac{2}{5}M{R}^2$
And the K.E of a rotating sphere $K=\frac{1}{2}I{\omega }^2$
So, from the question, the total K.E of the earth is the sum of the K.E of the earth w.r.t. its center and w.r.t. the center of the sun.
So. K.E of the earth w.r.t its center $K_1=\frac{1}{2}I{{\omega }_0}^2$
$K_1=\frac{1}{2}\times \frac{2}{5}M{R}^2{{\omega }_0}^2$
K.E. of the earth w.r.t. the center of the sun is
$K_2=\frac{1}{2}M{D}^2{\omega }^2$
So the total K.E of the earth
$K_{total}=K_1+K_2=\frac{1}{2}\times \frac{2}{5}M{R}^2{{\omega }_0}^2+\frac{1}{2}M{D}^2{{\omega }_0}^2$
$K_{total}=\frac{M{R}^2{{\omega }_0}^2}{5}[1+\frac{5}{2}{\left(\frac{D\omega }{R{\omega }_0}\right)}^2]$