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 it’s radius is $R$ . If earth has an angular velocity ${\omega _ \circ }$ with respect to its centre and $\omega$ with respect to the centre of the sun, the total kinetic energy of the earth is:
A. $\dfrac{{M{R^2}{\omega _ \circ }^2}}{5}[1 + {(\dfrac{\omega }{{{\omega _ \circ }}})^2} + \dfrac{5}{2}{(\dfrac{{D\omega }}{{R{\omega _ \circ }}})^2}]$
B. $\dfrac{{M{R^2}{\omega _ \circ }^2}}{5}[1 + \dfrac{5}{2}{(\dfrac{{D\omega }}{{R{\omega _ \circ }}})^2}]$
C. $\dfrac{2}{5}M{R^2}{\omega _ \circ }[1 + \dfrac{5}{2}{(\dfrac{{D\omega }}{{R{\omega _ \circ }}})^2}]$
D. $\dfrac{2}{5}M{R^2}{\omega _ \circ }^2[1 + {(\dfrac{\omega }{{{\omega _ \circ }}})^2} + \dfrac{5}{2}{(\dfrac{{D\omega }}{{R{\omega _ \circ }}})^2}]$

Answer 1

To solve this question, first we will rewrite the given facts of the question. and then we will write the formula of Kinetic Energy as the Translational, Rotational and Revolutional Kinetic Energy is included.

Complete step by step solution:
Given facts-
Distance between the centres of earth and the sun is $D$ .
Mass of the Earth is $M$ .
Radius of the Earth is $R$ .
Angular Velocity of the earth is ${\omega _ \circ }$ with respect to it’s centre.
And, Angular Velocity of the earth is $\omega$ with respect to the centre of the sun.
Now, we will find the total kinetic energy of the earth.
So, $Total\,Kinetic\,Energy = K.{E_{Translational}} + K.{E_{Rotational}} + K.{E_{\operatorname{Re} volutional}}$
$K.E = \dfrac{1}{2}M{V^2} + \dfrac{1}{2}I{\omega ^2} + \dfrac{1}{2}I{\omega _ \circ }^2$
here, $I$ is the moment of inertia of the earth.
$V$ is the linear velocity of earth with respect to sun $= \omega D$
$I = \dfrac{2}{5}M{R^2}$
Now,
$K.E = \dfrac{1}{2}M{\omega ^2}{D^2} + \dfrac{1}{2}.\dfrac{2}{5}M{R^2}{\omega ^2} + \dfrac{1}{2}.\dfrac{2}{5}M{R^2}{\omega _ \circ }^2$
(taking $\dfrac{{M{R^2}{\omega _ \circ }^2}}{5}$ as common)
$K.E = \dfrac{{M{R^2}{\omega _ \circ }^2}}{5}({(\dfrac{{d\omega }}{{R{\omega _ \circ }}})^2}(\dfrac{5}{2}) + 1 + {(\dfrac{\omega }{{{\omega _ \circ }}})^2})$
or,
$K.E = \dfrac{{M{R^2}{\omega _ \circ }^2}}{5}[1 + {(\dfrac{\omega }{{{\omega _ \circ }}})^2} + \dfrac{5}{2}{(\dfrac{{D\omega }}{{R{\omega _ \circ }}})^2}]$
Hence, the correct option is A. $\dfrac{{M{R^2}{\omega _ \circ }^2}}{5}[1 + {(\dfrac{\omega }{{{\omega _ \circ }}})^2} + \dfrac{5}{2}{(\dfrac{{D\omega }}{{R{\omega _ \circ }}})^2}]$ .

Note:
The kinetic energy due to the rotation of an object is part of its total kinetic energy. Rotational kinetic energy is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity. A rolling object has both translational and rotational kinetic energy.