Question

Why the value of acceleration due to gravity is $9.8m/s$ on earth.

Answer 1

Step1: Analysing the gravitational force

1. Consider two bodies of mass: one heavy $\left(M\right)$ and one lighter $\left(m\right)$. There is a distance $\left(R\right)$ separating these two bodies.
2. The gravitational force acting on both the bodies will be given by: $F_G=\frac{GMm}{R^2}$ , where ($G$) is the universal gravitational constant, given by: $G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$.
3. Due to the gravitational force's attracting nature, it will operate on the bodies in such a way that the separation between them will grow smaller.
4. We will compare the gravitational force to the general force equation, which says that force is determined by the body's mass multiplied by its acceleration. That is: $F=ma$

Step1: Acceleration of the lighter body $\left(m\right)$

The acceleration of the lighter body $\left(m\right)$ towards the heavier body of mass $\left(M\right)$ is given by:
$F_G=F\Rightarrow \frac{GMm}{R^2}=m{a}_m\Rightarrow a_m=\frac{GM}{R^2}$
Step2: Acceleration of the heavier mass

The acceleration of the heavier mass $\left(M\right)$ will be: $F_G=F\Rightarrow \frac{GMm}{R^2}=M{a}_M\Rightarrow a_M=\frac{GM}{R^2}$
Step3: Comparison between lighter body and heavier body

1. If the lighter object has a mass significantly lower than the heavier object, then the acceleration of the bigger mass object with respect to the lesser mass object can be roughly calculated to be zero.
2. To understand it better, let’s consider, the heavier object to be Earth, having a mass of $M_e=5.972\times {10}^{24}kg$ and the lighter mass be us having a mass of $m=70kg$. The radius of the earth is $R_e=6400km=6.4\times {10}^{6}m$
3. The acceleration of the lighter mass $\left(m\right)$ would be: $a_m=\frac{GM}{R^2}\Rightarrow a_m=\frac{(6.67\times {10}^{-11}N{m}^{2}k{g}^{-2})(5.972\times {10}^{24}kg)}{{(6.4\times {10}^{6}m)}^2}\Rightarrow a_m\approx 9.8m{s}^{-2}$

Step4: Acceleration of earth

Similarly, the corresponding acceleration of earth is $\approx {10}^{23}$ times lesser than the value of $a_m$, hence approximately equal to zero.

Hence, the acceleration due to the gravity of the earth is $g=a_m=9.8m{s}^{-2}$.