Question

Differentiate between Universal gravitational constant ‘G’ and acceleration due to gravity ‘g’.

Answer 1

Hint: A constant is independent of all variable quantities. ‘G’ is a constant appearing in Newton’s law of gravitation. Newton's law of gravitation is given as: $F=G\cdot\dfrac{ m_1\cdot m_2}{r^2}$. Whereas ‘g’ is the acceleration of a body freely falling under the influence of the force of gravity.
Use Newton’s second law of motion to obtain acceleration due to gravity ‘g’.

Formula used: $F=G\cdot\dfrac{ m_1\cdot m_2}{r^2}$

Complete solution Step-by-Step:
Before we start answering the question, it is important to understand what is asked. The question is asking to provide differences between Universal gravitational constant ‘G’, which appears in Newton's law of gravitation and acceleration due to gravity ‘g’, which is derived using Newton’s second law stating: Force is equal to mass multiplied by acceleration.

$F=ma\quad\quad{\cdots\cdots\cdots\cdots (\text{I})}$
where ‘m’ is the mass of the body and ‘a’ is its acceleration, when force applied is ‘F’.

The question is asked because both Universal gravitation constant ‘G’ and acceleration due to gravity ‘g’ are represented by the same alphabet, so it may be confusing to understand the difference. To develop a better understanding, we substitute $m_1=m, m_2=M$ in Newton’s Law of gravitation to get:

$F=G\cdot\dfrac{mM}{r^2}\quad\quad{\cdots\cdots\cdots\cdots (\text{II})}$

where, ‘F’ is the force on the body, ‘m’ is the mass of the first body under consideration, ‘M’ is the mass of the second body and ‘r’ is the distance between the two bodies.

From equation I and II, we conclude:

$ma=G\cdot\dfrac{mM}{r^2}$

Let us substitute $a=g$ for acceleration due to gravity:

$mg=G$
$\dfrac{mM}{r^2}g=\dfrac{GM}{r^2}$

With this knowledge, we are in position to summarise the difference between universal gravitational constant ‘G’ and acceleration due to gravity ‘g’.

S. No	Universal gravitational constant(G)	Acceleration due to gravity($g=\dfrac{GM}{r^2}$)
1.)	This is a constant quantity.	This quantity may vary from place to place
2.)	Value of ‘G’ is independent of all factors in the universe and does not change	Value of acceleration ‘g’ depends on the mass of the body (M) as well as the distance between the bodies applying gravitational force on each other (r).
3.)	Universal gravitation constant is a scalar quantity.	Acceleration due to gravity is a vector quantity.
4.)	Value of universal gravitation constant is: $G=6.7\times 10^{-11}\mathrm{\;Nm^2/kg^2}$	Acceleration due to gravity near the equator of earth is calculated by substituting mass of earth $M=6\times10^{24}\mathrm{\;kg}$, radius of earth $r=6.4\times10^{6}\mathrm{\;m}$ and universal gravitation constant $G=6.7\times10^{-11}\mathrm{\;Nm^2/kg^2}$ in the formula: $g=\dfrac{GM}{r^2}$.The value obtained is:$g=\mathrm{9.8\;m/s^2}$

Note:
1. The student may be confused as to what the question is asking by the capital ‘G’ and small ‘g’. A prior understanding of the chapter ‘Universal Law of Gravitation’ is needed to attempt this question.
2. While dealing with this type of question, we always assume all the events taking place in an inertial frame as defined by Newton’s first law.