Earth is approximately a solid sphere. It has a mass of 5.98 x 1024 kg, has a radius of 6.38 x 106 m, and completes one rotation about its central axis each day. The orbit of Earth is approximately a circle with a radius of 1.5 x 1012 m, and Earth completes one orbit around the Sun each year. Calculate the ratio of the rotational and translational kinetic energies of Earth in its orbit.

The rario of the rotational and translational kinetic energies of Earth in its orbit is 1.81*10^-11

Explanation:

Rotational Kinetic Energy (RKE) = 1/2Iw^2

I (moment of inertia) = mr^2 = m (6.38*10^6) ^2 = 40.7044*10^12m

w (angular velocity) = v/r

w^2 = v^2/r^2 = v^2 / (1.5*10^12) ^2 = v^2/2.25*10^24

RKE = 1/2 (40.7044*10^12m * v^2/2.25*10^24) = 1/2 (1.81*10^-11mv^2)

Translational Kinetic Energy (TKE) = 1/2mv^2

RKE/TKE = 1/2 (1.81*10^-11mv^2) * 2/mv^2 = 1.81*10^-11

Write down the values given in the question

Mass of earth, m = 5.98 * 10^24 Kg

radius, r = 6.38 * 10^6 m

distance from sun, d = 1.5 * 10^11 m

a)

Rotational kinetic energy of earth on axis = 0.5 * I * w^2

Rotational kinetic energy of earth on axis = 0.5 * (2/3 * m * r^2) * (2pi / T) ^2

Rotational kinetic energy of earth on axis = 0.5 * (2/3 * 5.98 * 10^24 * (6.38 * 10^6) ^2) * (2pi / (24 * 3600)) ^2

Rotational kinetic energy of earth on axis = 4.29 * 10^29 J

the Rotational kinetic energy of earth on axis is 4.29 * 10^29 J

b)

For the earth going around the sun

Kinetic energy of earth = 0.5 * I * w^2

Kinetic energy of earth = 0.5 * m * d^2 * (2pi / T) ^2

Kinetic energy of earth = 0.5 * 5.98 * 10^24 * (1.5 * 10^11) ^2 * (2pi / (365 * 24 * 3600)) ^2

Kinetic energy of earth = 2.671 * 10^33 J

the Kinetic energy of earth is 2.671 * 10^33 J