Question

Derive a relation for acceleration due to gravity. How their values vary with?
(i) Mass of the planet
(ii) Size of the planet

Answer 1

Step 1: Finding the relation of Acceleration using Universal law of gravitation

If we drop a body of mass $\text{'}m\text{'}$ from a distance say from the centre of the earth having mass $M,$ then the force exerted by the earth on body is given by

$F=G\times \frac{M\times m}{R^2}$ (by the universal law of gravitation) ……. (I)

We know that this force exerted by the earth will produce an acceleration in the stones because of which stone will fall downwards.

Force $=$Mass$\times$Acceleration

$F=m\times a$

or Acceleration (of stone), $a=\frac{F}{m}$ ……. (ii)

From equation (i)

$\frac{F}{m}=\frac{G\times M}{R^2}$

$a=\frac{G\times M}{R^2}$ {putting $\frac{F}{m}=a$}

$a=\frac{G\times M}{R^2}$ …… (iii)

Step 2: Replacing $\text{'}a\text{'}$ by $\text{'}g\text{'}$

We know that acceleration produced by the earth is called as acceleration due to gravity and is denoted by $\text{'}g\text{'}.$Hence replacing $\text{'}a\text{'}$ by $\text{'}g\text{'}$ in eq (iii).

$g=\frac{G\times M}{R^2}$

where $G=$gravitational constants

$M=$mass of the earth

$R=$radius of the earth.

Step 3: Acceleration due to gravity

(i) $g=\frac{G\times M}{R^2}$

means $g$ directly proportional to the Mass of the planet. Hence, $\text{'}g\text{'}$ will increase as mass of a planet increases.

(ii) As acceleration due to gravity is given by

$g=\frac{G\times M}{R^2}$

means $g$ is inversely proportional is the square of the radius of the planet. Hence, the value $\text{'}g\text{'}$ decreases as the size of the planet increases.