Question

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.

Answer 1

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

Step-by-step 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

Answer 2

Step-by-step explanation:

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