A vertical mine shaft's bottom should be displayed.dug to depthD, the measured value of ag will be&nbsp;ag=ags(1−D/R),&nbsp;ags&nbsp;being the surface value. Assume that Earth is a uniform sphere of radius R.

Let at the bottom of a vertical mine shaft dug todepth D, the measured value of acceleration be agThe net separation from the centre of planet ( r ) = ( R - D )Let ' ag ' be the gravitational acceleration at thatpoint ag=G×m′(R−D)2But m′=r×(4×p3)×(R−D)3So, ag=G×r×(4×p3×(R−D)Here; ags=4×p×G×R×r3Therefore; ag= ags(1 - D/R)Assumethat Earth is a uniform sphere of radius R

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Albarellak

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2020-12-12

A vertical mine shaft's bottom should be displayed.

dug to depthD, the measured value of ag will be $a g = a_{g s} (1 - D / R)$ , $a_{g s}$ being the surface value. Assume that Earth is a uniform sphere of radius R.

Answer & Explanation

Ezra Herbert

Skilled2020-12-13Added 99 answers

Let at the bottom of a vertical mine shaft dug todepth D, the measured value of acceleration be ag
The net separation from the centre of planet ( r ) = ( R - D )
Let ' ag ' be the gravitational acceleration at thatpoint

a g = \frac{G \times m^{'}}{(R - D) 2}

But

m^{'} = r \times (\frac{4 \times p}{3}) \times (R - D) 3

So,

a g = G \times r \times (\frac{4 \times p}{3} \times (R - D)

Here;

a g s = \frac{4 \times p \times G \times R \times r}{3}

Therefore;
ag= ags(1 - D/R)
Assumethat Earth is a uniform sphere of radius R

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