The Earth rotates on its axis with period 86400. s and orbits the Sun in approxi

Question

The Earth rotates on its axis with period 86400. s and orbits the Sun in approximately uniform circular motion of radius 1.495 x 10^8 km and period 3.156 x 10^7 s. Hence, its total kinetic energy consists of the sum of contributions K_spin associated with its rotation and K_orbit associated with its orbital motion. Calculate the ratio -K_spin/K_orbit these two energies. Treat the Earth as a uniform-density sphere (with moment of inertia I = 2/5 MR^2 of radius R_xo = 6371. km and mass M_xo= 5.976 x 10^24 km.

Explanation / Answer

here,

tme peroid of earth on axis, t = 86400 s

radius of circular path, R = 1.495*10^11 m
peroid for path, tp = 3.156*10^7 s

mass of earth, m= 5.976*10^24 kg
radius of earth, r = 6371000 m

From Circular motion, translational velocity of earth while rotating around sun,

v = 2*pi*R/tp
v = (2*pi* 1.495*10^11)/(3.156*10^7)
v = 29763.505 m/s

Angular velocity, w = v/r = 29763.505/6371000
w = 0.004671 rad/s

Orbital Kinetic Energy, Ko = 0.5 * m * v^2
Rotational Kinetic energy, Ks = 0.5*I*w^2 = 0.5*(2/5)*m*r^2*w^2

Ks/Lo = 0.5*(2/5)*m*r^2*w^2 / (0.5 * m * v^2)
Ks/ko = (2/5)*r^2*w^2 / (m * v^2)
Ks/ko = (2/5)*6371000^2*0.004671^2 / (5.976*10^24 * (29763.505)^2)

Ks/ko = 6.691*10^-26

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