Question

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 ‘${\omega }_o$’ with respect to its centre and ‘$\omega$’ with respect to the centre of the sun, the total kinetic energy of the
earth is :

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

Answer 1

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

According to the diagram, let S be the sun around which the earth is revolving in a circular orbit, with radius, D.

Mass of earth = M (given)

Radius of earth = M (given)

Angular velocity of earth with respect to its centre = ${\omega }_o$

Angular velocity of earth with respect to sun = $\omega$

Total kinetic energy = ${K.E}_{Translational}+{K.E}_{Rotational}+{K.E}_{Revolutional}$

$K.E=\frac{1}{2}{MV}^2+\frac{1}{2}{I\omega }^2+\frac{1}{2}{{I\omega }_0}^2$

(I - Moment of inertia of earth)

v = linear velocity of earth with respect to sun = $\omega D$

$I=\frac{2}{5}M{R}^2$

$K.E=\frac{1}{2}M{\omega }^2{D}^2+\frac{1}{2}\frac{2}{5}M{R}^2{\omega }^2$

(taking $\frac{M{R}^2{{\omega }_0}^2}{5}$ Common)

K.E=$\frac{M{R}^2{{\omega }_0}^2}{5}\left({\left(\frac{d\omega }{R{\omega }_0}\right)}^2\right(\frac{5}{2})+1+{\left(\frac{\omega }{{\omega }_0}\right)}^2)$

or K.E=$\frac{M{R}^2{{\omega }_0}^2}{5}\left(\right(1+{\left(\frac{\omega }{{\omega }_0}\right)}^2+\frac{5}{2}{\left(\frac{D\omega }{R{\omega }_0}\right)}^2)$