The variation of acceleration due to gravity $g$ with distance $d$ from centre of the earth is best represented by (R=Earth's radius) :

Question

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Answer 1

With depth

$g_{1} = g (1 - \frac{( d )}{R})$

As depth $d$ goes on increasing $g_{1}$ goes on decreasing, it remains maximum at the surface of Earth .The above equation is in the form of straight line.

With height

$g_{2} = g (1 - \frac{2 h}{R})$

$= g - \frac{2 g h}{R}$

$g_{2} \propto \frac{1}{R}$ (Hyperbola)

Acceleration due to gravity goes on decreasing as the $h$ above Earth surface increases.
Hence the correct answer is option A.

Answer 2

With depth

g_{1} = g (1 - \frac{( d )}{R})

As depth

d

goes on increasing

g_{1}

goes on decreasing, it remains maximum at the surface of Earth .The above equation is in the form of straight line.

With height

g_{2} = g (1 - \frac{2 h}{R})

= g - \frac{2 g h}{R}

g_{2} \propto \frac{1}{R}

(Hyperbola)

Acceleration due to gravity goes on decreasing as the

h

above Earth surface increases.

Hence the correct answer is option A.