The ratio of acceleration due to gravity at a height 3R above earth's surface to the acceleration due to gravity on the surface o f the earth is (R = radius of earth)

Question

The ratio of acceleration due to gravity at a height 3R above earth's surface to the acceleration due to gravity on the surface o f the earth is (R = radius of earth) (A) (1/9) (B) (1/4) (C) (1/16) (D) (1/3)

Tardigrade Admin · Accepted Answer

The value of acceleration due to gravity changes with height (ie, altitude). Ifg' is the acceleration due to gravity at a point, at height h above the surface of earth, then

g^{'} = \frac{GM}{( R + h ) ^{2}}

but,

g = \frac{GM}{R ^{2}}

∴

\frac{g ^{'}}{g} = \frac{GM}{( R + h )} = \frac{GM}{( R + h ) ^{2}}

Here

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

=

\frac{GM}{( 4 R ) ^{2}} = \frac{GM}{16 R ^{2}} = \frac{g _{e}}{16}

Q. The ratio of acceleration due to gravity at a height 3R above earth's surface to the acceleration due to gravity on the surface o f the earth is (R = radius of earth)

$\frac{1}{9}$

$\frac{1}{4}$

$\frac{1}{16}$

$\frac{1}{3}$

Q. The ratio of acceleration due to gravity at a height 3R above earth's surface to the acceleration due to gravity on the surface o f the earth is (R = radius of earth)

91​

41​

161​

31​

$\frac{1}{9}$

$\frac{1}{4}$

$\frac{1}{16}$

$\frac{1}{3}$