The acceleration of a body due to attraction of the earth (radius R ) at a distance 2R from the surface of the earth is ( g= acceleration due to gravity at the surface of the earth)

Question

The acceleration of a body due to attraction of the earth (radius R ) at a distance 2R from the surface of the earth is ( g= acceleration due to gravity at the surface of the earth) (A) (g/9) (B) (g/3) (C) (g/4) (D) g

Tardigrade Admin · Accepted Answer

gâ²=(g R2/(R + h)2) gâ²=(g R2/(R + 2 R)2) gâ²=(g R2/(3 R)2) gâ²=(g/9)

Q. The acceleration of a body due to attraction of the earth (radius $R$ ) at a distance $2 R$ from the surface of the earth is ( $g =$ acceleration due to gravity at the surface of the earth)

$\frac{g}{9}$

$\frac{g}{3}$

$\frac{g}{4}$

$g$

$g' = \frac{g R ^{2}}{( R + h ) ^{2}}$
$g' = \frac{g R ^{2}}{( R + 2 R ) ^{2}}$
$g' = \frac{g R ^{2}}{( 3 R ) ^{2}}$
$g' = \frac{g}{9}$

Q. The acceleration of a body due to attraction of the earth (radius R ) at a distance 2R from the surface of the earth is ( g= acceleration due to gravity at the surface of the earth)

9g​

3g​

4g​

g

g′=(R+h)2gR2​ g′=(R+2R)2gR2​ g′=(3R)2gR2​ g′=9g​

Q. The acceleration of a body due to attraction of the earth (radius $R$ ) at a distance $2 R$ from the surface of the earth is ( $g =$ acceleration due to gravity at the surface of the earth)

$\frac{g}{9}$

$\frac{g}{3}$

$\frac{g}{4}$

$g$

$g' = \frac{g R ^{2}}{( R + h ) ^{2}}$
$g' = \frac{g R ^{2}}{( R + 2 R ) ^{2}}$
$g' = \frac{g R ^{2}}{( 3 R ) ^{2}}$
$g' = \frac{g}{9}$