Question

The SI unit of $G$ is$N{m^2}k{g^{ - 2}}$ . Which of the following can also be used as the SI unit of G?
A. ${m^3}k{g^{ - 1}}{s^{ - 2}}$
B. ${m^2}k{g^{ - 2}}{s^{ - 1}}$
C. $mk{g^{ - 3}}{s^{ - 1}}$
D. ${m^2}k{g^{ - 3}}{s^{ - 2}}$

Answer 1

Gravitational force is the force of attraction. The gravitational constant G depends upon this force of attraction. Also, G is directly proportional to the mass of bodies and the gravitational force and inversely proportional to the distance between the two bodies.

Formula used:
The gravitational constant can also be written as:$G = \dfrac{{g{r^2}}}{M}$

Complete step by step answer:
attraction between two bodies of masses m and M is given by
$F = \dfrac{{GMm}}{{{r^2}}}$ $\to (1)$
Where, G is the gravitational constant and r is the radius between two given masses. This law is the same as Coulomb's Law of charges. Near the earth’s surface, the force in vertical direction due to gravity is proportional to the mass of the object and is nearly independent for heights at least as small as the radius of the earth.
Rearranging this equation in terms of G. Therefore, G will be as follows
$G = \dfrac{{F{r^2}}}{{Mm}}$
Thus, the unit of G is given as: $N{m^2}k{g^{ - 2}}$ . ( also mentioned in the problem)
The acceleration due to gravity is given by:$g = \dfrac{{GM}}{{{r^2}}}$ $\to (2)$
Therefore, equation (1) can be written as
$F = mg$
Consider equation (2) written in terms of G
$G = \dfrac{{g{r^2}}}{M}$
Therefore the units will be
$G = \dfrac{{\dfrac{m}{{{s^2}}} \times {m^2}}}{{kg}} = {m^3}{s^{ - 2}}k{g^{ - 1}}$

So, the correct answer is “Option A”.

Note:
Force can also be written as mass times the gravitational acceleration by Newton’s laws, where $g$ is the acceleration due to gravity given by $g = \dfrac{{GM}}{{{r^2}}}$. Here, $M$ is considered to be the mass of the Earth. The value of $g$ given by the value $9.8m/{s^2}$ is a constant only when $M$ is the mass of Earth.