Does any object weigh lighter at the Earths equator than at the poles owing to rotation or do they weigh exactly the same.

 

How fast would the Earth have to spin to negate gravitational effects at the equator?

Off subject...but I like the quote Franklin

Yes it is different but marginal

De mass is obvious the same but it weights a little less.

 

you can find the formula here:

http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html#cf

If you can't figure it out I will post it later.

Copied and pasted from a website:

 

Let's take the radius of Earth to be 6.37 million meters. Then the circumference is 2*pi*6.37e6 or 40 million meters. We cover that distance in 24 hours as Earth turns so the tangential speed (Vt) of a person on Earth's surface is 4e7/(24*3600) meters per second. This comes out to be 463 meters per second.

 

Centrifugal force Fc is given by Fc=m*Vt^2/R where m is the mass of the object at Earth's surface and R is the radius of Earth. In our case lets take the mass of a person to be 100kg so Fc=100*463^2/6.37e6. That is Fc=3.365 Newtons.

 

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F=ma so his force is (100kg) (9.8 m/s2)

The force holding the person to the earth is 980N

 

Just substitute values:

980N = 100(x^2) / 6.37e6

 

7901m/s or roughly 17 times the speed at which its spinning now.

 

I did that really quick, so someone correct me if i'm wrong :l

Yes, I got the same but it can be done more simple:

 

'Get more' symbols isn't working for me so it will have to be this way:

 

m.g = m.v.v / r and v= w.r

 

g = w.w.r.r/r

(omega=w=2.pi.f)

 

f = sqrt(g/r) / (2.pi) = 0.00019728 times/sec

 

T = 1/f = 5068.8 sec = 1.41hours or 24h/1.4h =17 times faster.

 

like you see it is mass independant.

 

edit: add on

0.3442 * COS(latitude) * COS(latitude) = percentage that you weight less due to earths rotation.

That number is w.w.r/g*100 or 1/(17*17)*100