Question

Calculate the value of $g$ on the surface of earth.

Answer 1

Step 1: Findind the relation of Acceleration and the 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 the body is given by

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

Step 2: Finding Force

We know that this force exerted by the earth will produce an acceleration in the stone because of which the 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)

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)

$g=\frac{G\times M}{R^2}$ ……. (iv)

where $G=$gravitational constants,

$M=$mass of the earth

$R=$ radius of the earth

Step 3: Putting the values of $G,M,R$ in equation $\left(iv\right)$

We have, Gravitation constant,$G=6.7\times {10}^{-11}N{m}^2/k{g}^2$

Mass of the earth, $M=6\times {10}^{24}kg$

Radius of the earth, $R=6.4\times {10}^{6}m$

$$\begin{gathered} g=\frac{6.7\times {10}^{-11}\times 6\times {10}^{24}}{{(6.4\times {10}^6)}^2} \\ g=9.8m/s^2 \end{gathered}$$

Thus, the value of acceleration due to gravity is $9.8m/s^2.$