# Find the rotational kinetic energy of the earth about the sun due to its orbit about the sun. The...

## Question:

Find the rotational kinetic energy of the earth about the sun due to its orbit about the sun. The mass of the earth m is {eq}6 \times 10^{24} {/eq} kg, the orbital radius r is {eq}1.5 \times 10^{11} {/eq} m and the rotational period T is 1 year. Treat the earth as a point mass in this problem.

## Rotational Kinetic Energy:

#### Moment of inertia of a point mass:

Moment of inertia of a body is the property of the body ,which resists its rotation about any axis.

Moment of inertia of a point mass about a given axis of rotation is measured as the product of its mass (m) and square of its perpendicular distance (r) from the axis of rotation. It can be given as:

{eq}I = m\, r^{2} {/eq}

Rotational kinetic energy of a body about a given axis of rotation is the energy possessed by a body rotating about the axis of rotation. If {eq}I {/eq} be moment of inertia of a point object of mass {eq}m {/eq} rotating with angular velocity {eq}\omega {/eq} about the axis of rotation at a distance {eq}r {/eq} from axis of rotation, then rotational kinetic energy of the point object is given by:

{eq}KE_R = \frac{1}{2}.I\, \omega^{2}\\ \therefore KE_R = \frac{1}{2}(mr^{2})\omega^{2} {/eq}

## Answer and Explanation:

Mass of earth, {eq}M = 6\times 10^{24}\, \mathrm{kg} {/eq}.

Radius of orbital, {eq}R = 1.5\times 10^{11}\, \mathrm{m} {/eq}

Time period, {eq}T =...

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