Question

Calculate the value of 'g' using the given values.
( Mass of earth = 6×10^24kg. Radius of earth = 6.4 × 10^6 m^2., G= 6.7 × 10^-11 Nm^2 kg^-2.)

Answer 1

AnswEr :

Force on any Object exerted by the Earth kept on the Surface is equal to :

$\blacksquare\quad\huge\boxed{\sf{F = GM\dfrac{m}{R ^{2} }}}$

$\bf{where}\begin{cases}\text{G = Universal Gravitational Constant}\\\text{M = Mass of the Earth}\\ \text{R = Radius of the Earth}\\\text{m = Mass of the Object on the Surface} \end{cases}$

• But we know that, F = ma i.e. (Mass × Acceleration due to Gravity)

$\rule{300}{1}$

• Therefore, the Value of g will be :

$\implies \large\sf{F = GM\dfrac{m}{R ^{2}}} \\ \\\implies \large\sf{\cancel{m}a = GM\dfrac{\cancel{m}}{R ^{2}}} \\ \\\implies \large\sf{a = GM\dfrac{1}{R ^{2}}} \\ \\ \quad \scriptsize\maltese\quad \text{putting the given values} \\ \\\implies \large\sf{a = \dfrac{6.7 \times {10}^{ -11} \times 6 \times {10}^{24}}{({6.4 \times {10}^{6})}^{2}}} \\ \\\implies \large\sf{a = \dfrac{40.2 \times {10}^{(24 -11)}}{40.96 \times {10}^{12}}}\\\\\implies \large\sf{a = \dfrac{40.2 \times {10}^{(13-12)}}{40.96}} \\ \\\implies \large\sf{a = \cancel\dfrac{40.2 \times 10}{40.96}} \\ \\\implies \large \boxed{ \green{\sf{a = 9.8 \:m{s}^{ - 2} }}}$

⠀

∴ Therefore, Value of g (Acceleration due to Gravity) will be 9.8 m/s².

Answer 2

$\huge{\underline{Answer}}$

Acceleration due to gravity at the surface of a planet is given by the formula .

g = $\frac{GM}{{R}^{2}}$

• G = universal gravitational constant.
• M = Mass of planet
• R = radius of planet

→ For earth

$\frac{6.7 \times {10}^{ - 11} \times 6 \times {10}^{24} }{ {(6378 \times 1 {0}^{3}) }^{2} }$

→ By solving this we get

9.78 m/s²

Similarly acceleration due to gravity (g) of moon is

1.67 m/s²