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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.