Calculate the magnitude of the angular momentum of me Earth considered as a part

Question

Calculate the magnitude of the angular momentum of me Earth considered as a particle orbiting the sun. The mass of the Earth is 5.97 Times 10^24 kg. Treat it as moving in a circular orbit of radius 1.50 Times 10^11 m at a speed of 2.98 times 10^4 m/s. kgm^2/s Calculate the magnitude of the angular momentum of the Earth due to its rotation around an axis through its north and south poles. Treat the Earth as a uniform sphere of radius 6.38 Times 10^6 m that makes one revolution in 24 hours. Kgm^2/s Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. The clock hand has a length of 15.0 cm and a mass of 5.00 g. Take the second hand to be a slender rod rotating with constant angular velocity about one end. kg-m^2/s

Explanation / Answer

1.(a) The orbital angular momentum of the earth around the sun is

mvr sin(theta) where theta is the angle between the velocity vector and radius vector; this angle is 90 deg so

mvr sin90 = mvr

we are given m and r, we find v from

v = circumference of orbit/period of 1 year

circumference = 2 pi r
period = 1 year =
1 year x 365 days/yr x24 hours/day x 3600s/hr=
3.15x10^7 s

v=2 pi x 1.5x10^11m/3.15x10^7 s = 2.99x10^4m/s

therefore orbital angular momentum =

5.97x10^24 kg x 2.98 x10^4m/s x 1.5x10^11m =
2.68x10^40 kgm^2/s

1(b): Angular momentum of the earth
= I * , where
I = moment of inertia of the earth as a sphere about its axis of rotation = (2/5)MR^2 and
= angular velocity = 2/(24*3600) radian/sec

=> angular momentum
= [(2/5)MR^2] * [2/(24*3600)]
Plugging mass of the earth, M = 5.9742 × 10^24 kg and
radius of the earth, R = 6.38 x 10^5 m,
angular momentum
= [(2/5)*5.9742 x 10^24*6.38^2 x 10^6] * [2/(24*3600)]
= 1.1084 x 10^27 kg-m^2/s.

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