The value of acceleration due to gravity 'g', at earth's surface is 10 m / s 2 . Its value at the centre of the earth which is assumed to be sphere of radius ' R' and uniform mass density, is

Question

The value of acceleration due to gravity 'g', at earth's surface is 10 m / s 2 . Its value at the centre of the earth which is assumed to be sphere of radius ' R' and uniform mass density, is (A) 2.5 R m / s 2 (B) 5 R m / s 2 (C) 10 R m / s 2 (D) 0

Tardigrade Admin · Accepted Answer

Acceleration due to gravity at a depth h from the earthâs surface g'=g(1-(h/R))=g(1-(R/R))=g(1-1)=0.

Q. The value of acceleration due to gravity $^{'} g^{'}$ , at earth's surface is $10 m / s^{2} .$ Its value at the centre of the earth which is assumed to be sphere of radius $^{'} R^{'}$ and uniform mass density, is

$2.5 R m / s^{2}$

$5 R m / s^{2}$

$10 R m / s^{2}$

0

Acceleration due to gravity at a depth h from the earth’s surface
$g^{'} = g (1 - \frac{h}{R}) = g (1 - \frac{R}{R}) = g (1 - 1) = 0.$

Q. The value of acceleration due to gravity ′g′, at earth's surface is 10m/s2. Its value at the centre of the earth which is assumed to be sphere of radius ′R′ and uniform mass density, is

2.5Rm/s2

5Rm/s2

10Rm/s2

0

Acceleration due to gravity at a depth h from the earth’s surface g′=g(1−Rh​)=g(1−RR​)=g(1−1)=0.

Q. The value of acceleration due to gravity $^{'} g^{'}$ , at earth's surface is $10 m / s^{2} .$ Its value at the centre of the earth which is assumed to be sphere of radius $^{'} R^{'}$ and uniform mass density, is

$2.5 R m / s^{2}$

$5 R m / s^{2}$

$10 R m / s^{2}$

Acceleration due to gravity at a depth h from the earth’s surface
$g^{'} = g (1 - \frac{h}{R}) = g (1 - \frac{R}{R}) = g (1 - 1) = 0.$