If acceleration due to gravity at distance d[

Question

If acceleration due to gravity at distance d[ < R] from the centre of earth is β, then its value at distance d above the surface of earth will be [where R is radius of earth] (A) (β R2/(R+d)3) (B) (β R/2 d) (C) (β d/(R+d)2) (D) (β R3/d(R+d)2)

Tardigrade Admin · Accepted Answer

Here, gd=(G M/R3) d=β, d < R ... (1) gd prime=(G M/(R+d)2), d>R .... (2) Using (1), G M=(β R3/d) gd prime=(β R3/d(R+d)2)

Q. If acceleration due to gravity at distance $d [< R]$ from the centre of earth is $β$ , then its value at distance $d$ above the surface of earth will be [where $R$ is radius of earth]

$\frac{β R ^{2}}{( R + d ) ^{3}}$

$\frac{βR}{2 d}$

$\frac{β d}{( R + d ) ^{2}}$

$\frac{β R ^{3}}{d ( R + d ) ^{2}}$

Here, $g_{d} = \frac{GM}{R ^{3}} d = β, d < R$ ... (1)
$g_{d}^{'} = \frac{GM}{( R + d ) ^{2}}, d > R$ .... (2)
Using (1), $GM = \frac{β R ^{3}}{d}$
$g_{d}^{'} = \frac{β R ^{3}}{d ( R + d ) ^{2}}$

Q. If acceleration due to gravity at distance d[<R] from the centre of earth is β, then its value at distance d above the surface of earth will be [where R is radius of earth]

(R+d)3βR2​

2dβR​

(R+d)2βd​

d(R+d)2βR3​