# Answer to Question #26257 in Mechanics | Relativity for zameer

Question #26257

Express the rotational kinetic energy of the earth in terms of its period of rotation. The moment

of inertia of the earth about its spin axis is 8.04 × 1037 kg m2. Calculate its rotational kinetic

energy

of inertia of the earth about its spin axis is 8.04 × 1037 kg m2. Calculate its rotational kinetic

energy

Expert's answer

The rotational kinetic energy equals:

Ec = I*w^2/2

I -moment of inertia

w - angular speed

In another side:

w = 2*pi/T

pi = 3.14

T - period

Finally:

Ec = I*(2*pi/T)^2/2 = 2*I(pi/T)^2

For the Earth -& T = 24 hours = 24*60 minutes = 24*60*60 seconds

Ec = 2* 8.04 × 10^37 kg m2 * (3.14/(24*3600))^2 =& 2.13*10^29 J

Answer: Ec = 2.13*10^29 J

Ec = I*w^2/2

I -moment of inertia

w - angular speed

In another side:

w = 2*pi/T

pi = 3.14

T - period

Finally:

Ec = I*(2*pi/T)^2/2 = 2*I(pi/T)^2

For the Earth -& T = 24 hours = 24*60 minutes = 24*60*60 seconds

Ec = 2* 8.04 × 10^37 kg m2 * (3.14/(24*3600))^2 =& 2.13*10^29 J

Answer: Ec = 2.13*10^29 J

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