Angular Momentum of the earth at the equator

  • Thread starter vesperaka
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  • #1
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Homework Statement



For a 98 kg person standing at the equator, what is the magnitude of the angular momentum about Earth's center due to Earth's rotation? Assume the Earth has a radius of 6.4 x 106 m.


Homework Equations



Angular Momentum = rmv * sin(theta)



The Attempt at a Solution



r= 6.4*10^6 m
m = 98kg
Vangular (of earth) = 7.27*10^5 rad/s

I multiplied these 3 values together and got 45597.44 kgm^2/s as my answer but it's wrong. Do I have to incorporate anything about the sin of the angle into my equation (I figured it would just be sin(90) = 1 though...)? I don't know what else to do :\
 

Answers and Replies

  • #2
alphysicist
Homework Helper
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Hi vesperaka,

Homework Statement



For a 98 kg person standing at the equator, what is the magnitude of the angular momentum about Earth's center due to Earth's rotation? Assume the Earth has a radius of 6.4 x 106 m.


Homework Equations



Angular Momentum = rmv * sin(theta)



The Attempt at a Solution



r= 6.4*10^6 m
m = 98kg
Vangular (of earth) = 7.27*10^5 rad/s

This number does not look right to me. You have units of radians per second, and so the angular velocity of the earth is 7.27*10^(-5) rad/s.

But this is not the v that is in your equation (it does not have the same units, for example). So you need to either find the speed v from this angular velocity, or use an equation for the angular momentum that has the angular velocity in it.
 

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