Compare the rotational and translational kinetic energy of the Earth. The Earth

Question

Compare the rotational and translational kinetic energy of the Earth. The Earth is approximately a solid sphere, has a mass of 5.98 × 1024 kg, has a radius of 6.38 × 106 m, and completes one revolution about its central axis each day. The Earth-Sun distance is 1.50 × 1011 m Compare the rotational and translational kinetic energy of the Earth. The Earth is approximately a solid sphere, has a mass of 5.98 × 1024 kg, has a radius of 6.38 × 106 m, and completes one revolution about its central axis each day. The Earth-Sun distance is 1.50 × 1011 m Compare the rotational and translational kinetic energy of the Earth. The Earth is approximately a solid sphere, has a mass of 5.98 × 1024 kg, has a radius of 6.38 × 106 m, and completes one revolution about its central axis each day. The Earth-Sun distance is 1.50 × 1011 m

Explanation / Answer

orbital velocity of earth around sun Vo^2 = GM/R

Vo^2 = 6.67 e -11* 1.99 e 30/(1.5 e 11)

Vo = 29.747 km/s

so KE of earth around sun = 0.5 mv^2

KE = 0.5 * 5.97 e 24 * 29747* 29747

KE trans = 2.64 *10^ 33 Joules

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Rotational KE = 0.5 I W^2

Moment of Inertia of solid sphere = 2MR^2/5

I = 2* 5.98 e 24 * 6.38e6 * 6.38e 6/5

I = 9.73 e 37 Kgm^2

Angular velcoity W = 2piR/t

W = (2*3.14* 6.38e6)/(24* 60*60)

W = 463.73 rad/s

So Rotational KE = 0.5 IW^2

ROt KE = 0.5* 9.73 e 37 * 463.75* 463.75

rot KE = 1.043 e 43

so

Comparison is that Trans KE./rot KE = 2.64 e 33/1.043 e 43

= 2.53 e -10

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