This topic is a struggle because my Professor made me rewrite my paper because she doesn't want forces included. Please help because I am struggling to understand this. I have googled and this is so foreign to me.
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2Related: https://physics.stackexchange.com/q/8074/2451 and links therein. – Qmechanic Oct 01 '17 at 13:07
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What's the topic of the paper? – garyp Oct 01 '17 at 14:09
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1Also related: https://physics.stackexchange.com/questions/141856 – David Hammen Oct 01 '17 at 15:00
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This is not intended to be a comprehensive answer, but is meant to give you a basis on which to build.
An object at the equator, even if stationary relative to the Earth, is accelerating towards the centre of the Earth with an acceleration of $R \omega ^2$ in which $R$ is the Earth's equatorial radius and $\omega$ is the angular velocity of its daily rotation about its axis. [The pull of the Earth's gravity on the object is much greater than the force needed to give it this acceleration.] When you let go of the object from a small height above the equator, the acceleration that you measure relative to the earth, $g_{app}$, will be smaller than the total acceleration, $g_{grav}$ due to the gravitational pull of the Earth, because this pull also has to supply the acceleration $R \omega ^2$. Thus… $$g_{app}=g_{grav}-R \omega ^2$$ At the poles, there is no $R \omega ^2$ because an object at either pole isn't moving in a circle, so $g_{app}$ is greater than at then equator.
However, the discrepancy between $g_{app}$ at poles and equator is found to be greater than $R \omega ^2$. Roughly twice as much! This is because there is another factor involved… The radius of the Earth is greater at the equator than the poles, so the rest of the Earth is further from the object, and, according to the inverse square law, $g_{grav}$ itself is smaller at the equator than the poles!
Philip Wood
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I am going to assume that you were talking about gravitational acceleration.
The short answer is :
Earth isn’t completely round.
Let's look at equation of gravitational acceleration, $$g=\frac{GM}{r^2}$$
Where $r$ is radius. Since Earth isn’t completely round hence radius changes depending on location. When talking about mountain the radius increases let radius of earth at that point is $a$ and height of the mountain is $h$. So the equation is $$g=\frac{GM}{(a+h)^2}$$
If you do numerical Solution tyen you will get 9.81,9.79,9.78 and some other values.
Billy Istiak
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