Question

discuss variation of g with depth
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Answer 1

Answer:

the acceleration due to gravity on the surface of earth is given by:

g=GM/R^2.......(1)

let 'Q' be the density of the material of the earth

now, mass= volume× density

M= 4/3πR^3 ×p

substituting in equation (1) we get

g=G/R^2×4/3πR^3×p

g=4/3πGRp.......(2 )

gd= 4/3Gπ(R-d)p........(3)

dividing equation 3by2

we get

gd/g=R-d/R= (1-d/R)

gd=g(1-d/R)

Answer 2

Explanation:

The acceleration due to gravity on the surface of the earth is given by.

$g \: = \frac{gm}{r {}^{2} } (1)$

$let \: 'Q' \: be \: the \: density \: of \: the \: material \: \\ of \: the \: earth.</p><p> \\ </p><p> Now, mass = volume × density$

$M= \frac{4}{3} \: πR </p><p></p><p> ×ρ$

$Substituting \: in \: equation \: ( 1 ) \: we \: get$

$g \: = \frac{G}{r {}^{2} } \times \frac{4}{3} {}^{3} πR {}^{3} \times p = \frac{4}{3} πGRρ$

$g= \frac{4}{3} \: </p><p> </p><p> πGRρ \: ( 2)$

$gd \: = \frac{4}{3} πG(R - d)ρ \: (3)$

$Dividing \: equation \: ( 3 ) \: by \: ( 2 ) we \: get$

$gd \: = \frac{r - d}{r} = (1 - \frac{d}{r} )$

$\\ gd = g \: (1 - \frac{g}{r} )$