Question

The value of ''g'' at a height one earth's radius above the surface of the earth is
A)1/4g
B)1/3g
C)1/2g
D)2g

Answer 1

Answer:

your answer is a) 1/4 g. hope it help

Answer 2

Acceleration due to gravity: The acceleration achieved by any object due to the gravitational force of attraction by the earth is called acceleration due to gravity by the earth.

As each planet has a different mass and radius, so the acceleration due to gravity will be different for a different planet.

Acceleration due to the gravity of the earth having mass (M) and radius (R) on earth's surface is given by:

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

Here G is the universal gravitational constant.

Acceleration due to gravity at any depth (h) of the earth's surface whose distance from the center of the earth is r is given by:

Acceleration due to gravity at depth $(g')=\frac{gr}{R}$

Acceleration due to gravity at height (h') whose distance from the center of the earth is r is given by:

Acceleration due to gravity at height $(g')=\frac{gR^2}{r^2}$

When G is a universal gravitational constant, $r=(R-h)$ and $r'=(R+h)$

Calculation:

New gravity above the surface of earth g':

$g'=g(1+\frac{h}{R_e} )^{-2}$

$g'=\frac{g}{(1+\frac{h}{R_e} )^2}$

When $h=R_e$

$g'=\frac{g}{(1+\frac{R_2}{R_e} )^2}$

$g'=\frac{g}{2^2}$

$g'=\frac{g}{4}$

Alternate Method

Acceleration due to gravity $=g=\frac{GM}{R^2}$

R= Distance from the center of the earth.

At a height = R above the earth's surface $g=\frac{GM}{(R+R)^2}$

$g'=\frac{GM}{4(R^2)}=\frac{g}{4}$

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