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.

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

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student

the closer the planet is to the sun, the larger the gravity