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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 IC-W11D1-1 Table Problem Angular Momentum for the Earth Solution The earth, of mass me = 5.97 !1024 kg and (mean) radius Re = 6.38 !106 m , moves in a nearly circular orbit of radius rs,e = 1.50 !1011 m around the sun with a period Torbit = 365.25 days , and spins about its axis in a period Tspin = 23 hr 56 min , the axis inclined to the normal to the plane of its orbit around the sun by 23.5° (see figure below). What is the ratio of the magnitudes of the orbital angular momentum to the spin angular momentum? Solution: If we approximate the earth as a uniform sphere, then the moment of inertia of the earth about its center of mass is 2 I cm = me Re2 . (1) 5 If we choose the point S to be at the center of the sun, and assume the orbit is circular, then the orbital angular momentum is ! ! ! Lorbital = rS , cm ! p total = rs, e r̂ ! me vcm "ˆ = rs, e me vcm k̂ . S (2) The velocity of the center of mass of the earth about the sun is related to the orbital angular velocity by vcm = rs,e ! orbit , (3) where the orbital angular velocity is ! orbit = 2" 2" = Torbit (365.25 d)(8.640 # 104 s $ d %1 ) %7 (4) %1 = 1.991 # 10 rad $ s . The orbital angular momentum is then ! Lorbital = me rs,2e ! orbit k̂ S = (5.97 " 1024 kg)(1.50 " 1011 m)2 (1.991 " 10#7 rad $ s #1 ) k̂ (5) = (2.68 " 1040 kg $ m 2 $ s #1 ) k̂. The spin angular momentum is given by ! ! 2 Lspin = I cm ! spin = me Re2 ! spin n̂ , cm 5 (6) where n̂ is a unit normal pointing along the axis of rotation of the earth and "spin = 2! 2! = = 7.293 $10#5 rad % s #1 . Tspin 8.616 $104 s (7) The spin angular momentum is then ! 2 Lspin = (5.97 ! 1024 kg)(6.38 ! 106 m)2 (7.293 ! 10"5 rad # s "1 ) n̂ cm 5 = (7.10 ! 1033 kg # m 2 # s "1 ) n̂. (8) The ratio of the magnitudes of the orbital angular momentum to the spin angular momentum is greater than a million, 2 me rs,e2 ! orbit Lorbital 5 rs,e Tspin S = = = 3.77 "106 , spin 2 2 Lcm (2 / 5)me Re !spin 2 Re Torbit (9) as this ratio is proportional to the square of the ratio of the distance to the sun to the radius of the earth. The total angular momentum is then ! 2 Ltotal = me rs,e2 ! orbit kˆ + me Re2!spin nˆ . S 5 (10) Two different values have been used for one “day;” in converting the orbit period from days to seconds, the value for the solar day, Tsolar = 86, 400s was used. In converting the earth’s spin angular frequency, the sidereal day, Tsidereal = Tspin = 86,160s was used. The two periods, the solar day from noon to noon and the sidereal day from the difference between the times that a fixed star is at the same place in the sky, do differ in the third significant figure.