We know that the acceleration due to gravity at the surface of the Earth is 9.8m

Question

We know that the acceleration due to gravity at the surface of the Earth is 9.8m/s2. We also know that the acceleration due to gravity falls off as 1/r2 above the surface of the Earth, with r being measured from the center of the planet.

There's one other domain, what's the value of gravity below the surface of the Earth? Sir Isaac Newton showed that the M term in gravity for the Earth can be set equal to the amount of mass contained inside a sphere of radius r centered upon the middle of the Earth in cases where we are investigating rEarth. Please find the value of g when we are 1/2 of the way from the center to surface of the Earth.

Explanation / Answer

Let g' be the g inside the Earth at half its radius.
Let g be the g on the surface of the Earth.
Where g is the acceleration due to gravity.

Formula: g'=g(1-h/Re)
where h=half the radius of the Earth.
and Re=radius of the Earth.

Now we can write h as 1/2 Re.
So the Re gets Cancelled.
And therefore g'=1/2g
That is, the value of g is half of that of the value on the surface of the Earth.

Mathematically proved: g at surface of the earth=9.8m/s^2
Therefore, at the centre of earth, g'=4.9m/s^2.

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