Question

Derive an expression for acceleration due to gravity
in terms of mass of the earth and the radius of the
earth.​

Answer 1

Answer:

The acceleration produced in the motion of a body falling freely under the force of gravity is called the acceleration due to gravity. It is denoted by 'g'. ** Consider the earth to be a sphere of mass M and radius R. Now, suppose a body of mass 'm' situated at distance 'r' from the centre of the earth.

Explanation:

Answer 2

Explanation:

Let a body P of mass m be situated at a depth h below the earth's surface.

The gravitational force of attraction on a body inside a spherical shell is always zero. Therefore, the body P experiences gravitational attraction only due to the inner solid sphere. The mass of this sphere is

M′ =volume×density=4/3π(Re−h)^3ρ

where ρ is mean density of the earth. Therefore, according to Newton's law of Gravitation, the force of attraction on the body P is

$\frac{GM′m}{ (Re−h)2} =G \frac{\frac{4}{3} \pi(Re−h)ρm}{ {(Re−h)}^{2} }</p><p>= \frac{4}{3} πG(Re−h)ρm</p><p></p><p>$

This force must be equal to the weight of the body mg ′ , where g′is the acceleration due to gravity at a depth h below the surface of the earth. Thus,

$mg = \frac{4}{3} \pi \: g (Re−h)ρm</p><p></p><p>$

Similarly, if a body be at the surface of the earth (h=0) where acceleration due to gravity is g, then

$mg = \frac{4}{3} \pi \: g \: (Re−h)ρm</p><p></p><p>$

Dividing eq.(i) by (ii) we have

⇒

$\frac{g1}{g} = \frac{(Re</p><p></p><p> - h)}{Re} = g1 = g(1 \times \frac{h}{ \:Re} )$

anushka Here