Q6.11 Angular momentum (S)
Calculate:
- the angular momentum of the earth relative to the centre of the sun (assuming that the earth’s orbit is circular, and treating the earth as a point mass);
- the angular momentum of an 80 kg person standing on the earth’s equator, relative to the centre of the earth;
- as in (2) but sited in the Appleton Tower, Edinburgh (latitude: 56∘).
[Mass of earth = 5.98×1024 kg; Radius of earth = 6.37×106 m; Distance of earth from the sun = 1.5×1011 m]
Hint
Reveal
Solution
Reveal
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Solution
- The earth, mass m=6×1024 kg, is in circular orbit round
the sun in an orbit of radius is r=1.5×1011 m.
The speed of the earth along the circumference of the circle is v=rω, with ω=2π rad/year=2×10-7 rad s-1;andv=3×104 m s-1.Sinceis always perpendicular to
,
The person of mass 80 kg is in circular orbit about the centre of the earth in an orbit of radius r=6.4×106 m.
The orbital angular velocity is
ω=2π rad/day=7.3×10-5 rad s-1The orbital speed is
v=rω=467 ms-1Again,is always perpendicular to
, so
L=mrv=2.4×1011 kg m2 s-1The direction ofis perpendicular to both
and
, along the earth’s rotational axis.
The person is in circular orbit, not about the centre of the earth C, but about A (see diagram).
The radius of this orbit is r⊥=rcos56∘=3.58×106 mThe orbital angular velocity about A isω=2π rad/day =7.3×10-5 rad s-1and the speed isv=r⊥ω=261 m s-1The angular momentum about C isNowand
are always perpendicular, so
L=mvr=1.34×1011 kg m2 s-1The direction of
is perpendicular to both
and
. In this case the direction is not constant, but changes as the earth rotates (see diagram).