Question

Earth has a mass of 5.97 x 1024 kg and a radius of 6.38 x 106 m. Assume it is a uniform solid sphere. The distance of Earth from the sun is 1.50 x 1011 m. (Assume Earth completes a single rotation in 24.0 hours and orbits the Sun once every 365 Earth days.) (a) Calculate the angular momentum of Earth in its orbit around the Sun kg m2/s (b) Calculate the angular momentum of Earth on its axis kg m/s Additional Materials eBook

Answer 1

Angular momentum of earth in its orbit around the sun will be given by:

Angular momentum = L = I*w

I = moment of inertia = Me*R^2

Me = mass of earth = 5.98*10^24 kg

R = distance between sun and earth = 1.50*10^11 m

w = angular velocity = 2*pi/T

w = 2*pi/1 yr

1 yr = 3.154*10^7 sec

w = 2*pi/(3.154*10^7)

So,

L = I*w

L = 5.98*10^24*(1.50*10^11)^2*2*pi/(3.154*10^7)

L = 2.68*10^40 kg-m^2/sec

Part B.

Angular momentum of earth on its axis will be

about the axis of earth, assuming earth is a sphere

I = 2*Me*Re^2/5

Re = radius of earth = 6.38*10^6 m

Now w = 2*pi/1 day

1 day = 86400 sec

w = 2*pi/86400 sec

So,

L = [2*5.98*10^24*(6.38*10^6)^2/5]*(2*pi/86400)

L = 7.08*10^33 kg-m^2/sec

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