Question

Calculate the value of g using the values of $ G $ , $ M $ and $ R $ namely universal gravitational constant $ G = 6.7 \times {10^{ - 11}}N{m^2}k{g^2} $ , mass of the earth $ M = 6 \times {10^{24}}kg $ , and radius of the earth $ R = 6.4 \times {10^6}m $ .

Answer 1

Hint : The question requires calculation of the value of acceleration due to gravity by using the formula which is given by $ {F_{Gravity}} = G\dfrac{{{M_{Earth}}m}}{{{R_{Earth}}^2}} $ . We will be using the concept of Newton’s law of gravitation in this solution.

Formula Used: The following formulas are used to solve this question.
 $ F = ma $ where $ F $ is the force acting on a body, $ m $ is the mass of the body and $ a $ is the acceleration of the body.
 $ {F_{Gravity}} = G\dfrac{{{M_{Earth}}m}}{{{R_{Earth}}^2}} $ where $ {F_{Gravity}} $ is the force of gravity of the earth, $ G $ is the universal gravitational constant, $ {M_{Earth}} $ is the mass of the earth, $ m $ is the mass of the object on which the force of gravity acts and $ {R_{Earth}} $ is the radius of the earth.

Complete step by step answer
The force on a body is the product of the mass and acceleration of a body.
Thus, we can write that, $ F = ma $ where $ F $ is the force acting on a body, $ m $ is the mass of the body and $ a $ is the acceleration of the body.
The acceleration that acts on a body falls solely under the influence of gravity is known as the acceleration due to gravity and is represented by $ g $ . The body is a freely falling body. The motion of a freely falling body is an example of a motion where the acceleration does not change with the function of time. It is thus a uniformly accelerated motion.
According to Newton’s Law of Gravitation,
 $ {{\text{F}}_{{\text{gravity}}}}{\text{ = G}}\dfrac{{{{\text{M}}_{{\text{Earth}}}}{\text{m}}}}{{{{\text{R}}_{{\text{Earth}}}}^{\text{2}}}} $ where $ {{\text{F}}_{{\text{gravity}}}} $ is the force of gravity, $ {\text{G}} $ is the gravitational constant, $ {{\text{M}}_{{\text{Earth}}}} $ is the mass of the Earth, $ {{\text{R}}_{{\text{Earth}}}} $ is the radius of the Earth, $ {\text{m}} $ is the mass of the body.
For a body falling freely under the influence of gravity, force acting on the body is equal to the product of its mass and its acceleration, which is equal to the acceleration due to gravity.
It can be written that acceleration $ a $ is the acceleration due to gravity $ g $ .
 $ \therefore $ Force $ F = mg $ where , $ m $ is the mass and acceleration due to gravity $ g $ .
Now let us equate this force to the force of gravity.
 $ {{\text{F}}_{{\text{gravity}}}} = {\text{F}} $
 $ \Rightarrow $ $ {\text{G}}\dfrac{{{{\text{M}}_{{\text{Earth}}}}{\text{m}}}}{{{{\text{R}}_{{\text{Earth}}}}^{\text{2}}}} = mg $
Since $ m $ on both sides gets cancelled,
 $ \Rightarrow {\text{G}}\dfrac{{{{\text{M}}_{{\text{Earth}}}}}}{{{{\text{R}}_{{\text{Earth}}}}^{\text{2}}}} = g $
Substituting the values of $ G $ , $ {M_{Earth}} $ and $ {R_{Earth}} $ in the equation above, we get,
 $ \Rightarrow g = 6.7 \times {10^{ - 11}}\dfrac{{\left( {6 \times {{10}^{24}}} \right)}}{{{{\left( {6.4 \times {{10}^6}} \right)}^2}}}Nk{g^{ - 1}} $
We know that, $ 1N = 1kgm{s^{ - 2}} $
 $ \therefore \;g = 0.98144 \times {10^{ - 11 + 24 - 12}}m{s^{ - 2}} $
 $ \Rightarrow g = 9.8144m{s^{ - 2}} $
 $ \therefore $ The value of acceleration due to gravity $ g $ is $ 9.8144m{s^{ - 2}} $ .

Note
Let us consider the height from which the freely falling body is dropped be $ h $ . Since $ {R_{Earth}} \gg h $ thus, Newton’s Law of Gravitation remains the same. This is why the increase in separation between the centre of the Earth and the body can be ignored.