Question

Arrange the following in the increasing order of the value of g (1) on the earth's surface (2) at the centre of
earth (3) at a height twice the radius of the earth surface of the earth (4) at a height half the radio
the earth from the surface of earth
1)3,2,4,1
2)1,2,3,4
3) 2,3,4, 1
4)4,3,2,1​

Answer 1

Answer:

3rd option is correct /

Answer 2

The increasing order of the value of g will be 2<3<4<1 (option 3).

• The value of g on the earth's surface is $9.8m/s^{2}$.

• The value of g at the centre of the earth is zero.

        As we know,

        $g'=g(1-dR)$

        if $d=R$( at the centre of the earth)

        then,

         $g'=g(1-1) \\g'= 0$

• If Earth’s radius is twice its current size, its surface gravity will be = $\frac{1}{4}g$i.e. 0.25 times its current surface gravity. Since g is $9.8m/s^{2}$, the new value of g at a height twice the radius of the earth will be $2.4525 m/s^{2}$.

• The value of g at a height equal to half the radius of earth from the surface of the earth is $40m/s^{2}$.

        Since, the acceleration on the surface of the earth due to gravity is    given as, $g=\frac{GM}{R^{2} }$

        Therefore, the acceleration at a height due to gravity is given as,

         $g'=\frac{GM}{h^{2} }\\ = \frac{GM}{(\frac{R}{2})^{2} }\\=4g\\=4(10)\\=40m/s^{2}$