Integral for the linear speed of the Earth

matai
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I need to make an integral to fine the speed of the earth. Say the radius is 6378137 meters. How would I account for things closer to the axis that have a radius of 0.0001 meters? I don't think I can just take the speed at the radius. So I found that the Earth rotates at 6.963448857E-4 revs/min then I convert that to radians to get 0.000072921065826007 radians/second, the angular speed. To get linear velocity I multiply ω by the radius, but I need to integrate from 0 to r? So would it be ∫ 0.000072921065826007*r dr a=0, b =6378137? I get 1.483237507 × 10^9. Does this seem right?
 
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The linear speed of Earth relative to what? Note that the rotation has absolutely nothing to do with the linear velocity of the Earth as a whole. The overall momentum of the Earth is zero in the Earth's centre of momentum frame by definition.
 
Orodruin said:
The linear speed of Earth relative to what? Note that the rotation has absolutely nothing to do with the linear velocity of the Earth as a whole. The overall momentum of the Earth is zero in the Earth's centre of momentum frame by definition.
I have to find the kinetic energy. And have that the mass is 5.9928982144×10^24kg from using the shell method and assuming that the Earth has a constant density of 5514 kg/m^3. So I need to linear speed, to put into the equation KE=1/2mv^2
 
What you need is the moment of inertia of the Earth. Note that assuming constant density is a pretty bad approximation (in particular as the core is denser - which significantly lowers the moment of inertia) and that the entire shell at radius ##r## is not moving with the same linear velocity.

matai said:
5.9928982144×10^24kg
This is way too many significant digits. The current precision in the value for the mass of the Earth is on the level of 0.01%.
 
Orodruin said:
What you need is the moment of inertia of the Earth. Note that assuming constant density is a pretty bad approximation (in particular as the core is denser - which significantly lowers the moment of inertia) and that the entire shell at radius ##r## is not moving with the same linear velocity.This is way too many significant digits. The current precision in the value for the mass of the Earth is on the level of 0.01%.
yeah I use inertia, but my calc teacher said she wants it in more of math terms so she didn't accept it when i did it with inertia
 
Well, why don't you show how to find the moment of inertia then? You should use calculus for that.
 
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