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Magnitude of the Angular Velocity of the Earth in a Circular Orbit Around the Sun

  • Thread starter MissEuropa
  • Start date
1. The problem statement, all variables and given/known data
A)Calculate the magnitude of the angular momentum of the earth in a circular orbit around the sun.

B) Is it reasonable to model it as a particle?
Yes, considering the size of the Earth in comparison of its orbit around the sun it is reasonable to model it is a particle.

2. Relevant equations
L = I*ω (when considering the Earth as a rigid body)
L=r*p
I=mE*rE


3. The attempt at a solution
Assuming the Earth has a completely circular orbit:
The Earth's orbit is 2∏ radians, it completes this orbit in one year.
1 year = 3.145*107 s
2∏/3.156*107 = 1.99*10-7rad/s
Average linear speed = ω*r (radius of orbit which is 1AU or 1.49*1011
(1.99*107)*(1.49*1011)=2.977*1018m/s

So, using the formula L=r x p
(1.49*1011)*(2.977*1018)*sin90°=4.45*1029m^2/s

But, this was not the answer masteringphysics was looking for.

So, using the formula L=Iω
I calculated I (mass of the Earth*radius) = (3.81*1031)*(1.99*10-7) = 7.58*1024 kg*m2/s

And that's also not the answer masteringphysics was looking for.
I'm not sure if I made a mistake somewhere in my calculations or if I'm not using the right values for mass, radius of Earth etc. Because Earth's orbit is non circular I'm not sure at what point in Earth's orbit they wanted us to consider as the radius. I choose one AU because it's an averaged value.

Thoughts?
 

tms

644
17
Average linear speed = ω*r (radius of orbit which is 1AU or 1.49*1011
(1.99*107)*(1.49*1011)=2.977*1018m/s
Always check your answers for sanity. What is the speed of light? This number is obviously wrong. In fact, it is exactly 14 orders of magnitude wrong. That should give you a hint.
 
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D H

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Assuming the Earth has a completely circular orbit:
The Earth's orbit is 2∏ radians, it completes this orbit in one year.
1 year = 3.145*107 s
2∏/3.156*107 = 1.99*10-7rad/s
Average linear speed = ω*r (radius of orbit which is 1AU or 1.49*1011
(1.99*107)*(1.49*1011)=2.977*1018m/s

So, using the formula L=r x p
(1.49*1011)*(2.977*1018)*sin90°=4.45*1029m^2/s
Momentum has units of mass*velocity or mass*length/time. Angular momentum has units of mass*length2/time. Where's the mass in your answer?
 
Momentum has units of mass*velocity or mass*length/time. Angular momentum has units of mass*length2/time. Where's the mass in your answer?
Just caught that. I'm reworking now and I'll update soon. :-) thanks for the help
 
Congrats!http://www.infoocean.info/avatar1.jpg [Broken]
Thankies!
So, I started from the beginning again, as not to confuse myself. Calculated the inertia of the Earth (which is why I was missing a mass component in my final answer) and got it right.

Symbolic math is the way to go for the most part, then plugging in numbers afterward.
I ended up with 2.67*1040 kg*m2/s

Thanks for the help again.
 
Last edited by a moderator:

tms

644
17
Symbolic math is the way to go for the most part, then plugging in numbers afterward.
Absolutely. Most students never seem to learn that.
 

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