What will be the acceleration due to gravity at a depth d, where g is acceleration due to gravity on the surface of earth?

Question

What will be the acceleration due to gravity at a depth d, where g is acceleration due to gravity on the surface of earth? (A) (g/[1+(d)R]2) (B) g[1-(2 h/R)] (C) (g/[1-(d)R]2) (D) g[1-(d/R)]

Tardigrade Admin · Accepted Answer

g =(4/3) π ρ GR at earth surface g=(4/3) π ρ G(R-d) at depth d of earth g=g[1-(d/R)]

Q. What will be the acceleration due to gravity at a depth d, where g is acceleration due to gravity on the surface of earth?

$\frac{g}{[ 1 + \frac{d}{R} ] ^{2}}$

$g [1 - \frac{2 h}{R}]$

$\frac{g}{[ 1 - \frac{d}{R} ] ^{2}}$

$g [1 - \frac{d}{R}]$

$g = \frac{4}{3} π ρGR$ at earth surface
$g = \frac{4}{3} π ρG (R - d)$ at depth $d$ of earth
$g = g [1 - \frac{d}{R}]$

Q. What will be the acceleration due to gravity at a depth d, where g is acceleration due to gravity on the surface of earth?

[1+Rd​]2g​

g[1−R2h​]

[1−Rd​]2g​

g[1−Rd​]

g=34​πρGR at earth surface g=34​πρG(R−d) at depth d of earth g=g[1−Rd​]

$\frac{g}{[ 1 + \frac{d}{R} ] ^{2}}$

$g [1 - \frac{2 h}{R}]$

$\frac{g}{[ 1 - \frac{d}{R} ] ^{2}}$

$g [1 - \frac{d}{R}]$

$g = \frac{4}{3} π ρGR$ at earth surface
$g = \frac{4}{3} π ρG (R - d)$ at depth $d$ of earth
$g = g [1 - \frac{d}{R}]$