If g is the acceleration due to gravity on the surface of earth, its value at a height equal to double the radius of earth is

Question

If g is the acceleration due to gravity on the surface of earth, its value at a height equal to double the radius of earth is (A)  g  (B)  (g/2)  (C)  (g/3)  (D)  (g/9)

Tardigrade Admin · Accepted Answer

Acceleration due to gravity at a height h from the surface of the earth g'=g (1/(1+(h)R)2) Given h=2 R ∴ g'=g (1/(1+2)2) or g'=(g/9)

Q. If $g$ is the acceleration due to gravity on the surface of earth, its value at a height equal to double the radius of earth is

$g$

$\frac{g}{2}$

$\frac{g}{3}$

$\frac{g}{9}$

Acceleration due to gravity at a height $h$ from the surface of the earth
$g^{'} = g \frac{1}{( 1 + \frac{h}{R} ) ^{2}}$
Given $h = 2 R$
$∴ g^{'} = g \frac{1}{( 1 + 2 ) ^{2}}$
or $g^{'} = \frac{g}{9}$

Q. If g is the acceleration due to gravity on the surface of earth, its value at a height equal to double the radius of earth is

g

2g​

3g​

9g​

Acceleration due to gravity at a height h from the surface of the earth g′=g(1+Rh​)21​ Given h=2R ∴g′=g(1+2)21​ or g′=9g​

Q. If $g$ is the acceleration due to gravity on the surface of earth, its value at a height equal to double the radius of earth is

$g$

$\frac{g}{2}$

$\frac{g}{3}$

$\frac{g}{9}$

Acceleration due to gravity at a height $h$ from the surface of the earth
$g^{'} = g \frac{1}{( 1 + \frac{h}{R} ) ^{2}}$
Given $h = 2 R$
$∴ g^{'} = g \frac{1}{( 1 + 2 ) ^{2}}$
or $g^{'} = \frac{g}{9}$