Question

The SI unit of the universal gravitational constant $G$:
(A) $Nmk{g^{ - 2}}$
(B) $N{m^2}k{g^{ - 2}}$
(C) $N{m^2}k{g^{ - 1}}$
(D) $Nmk{g^{ - 1}}$

Answer 1

Hint: The SI unit of the universal gravitational constant $G$ can be determined by using the formula of Newton’s of the gravitational force which lies between the two objects having the mass ${m_1}$ and ${m_2}$, which are separated by the distance $r$, then the SI unit of gravitational constant is determined.

Useful formula:
By Newton’s law of gravitation,
$F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
Where, $F$ is the gravitational force, $G$ is the gravitational constant, ${m_1}$ is the mass of the first body, ${m_2}$ is the mass of the second body and $r$ is the distance between the two body.

Complete step by step solution:
By Newton’s law of gravitation,
$F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}\,................\left( 1 \right)$
From the above equation, the gravitational constant is written as,
$G = \dfrac{{F \times {r^2}}}{{{m_1}{m_2}}}\,...................\left( 2 \right)$
The SI unit of force is $N$.
The SI unit of distance is $m$.
The SI unit of mass is $kg$.
By substituting the SI unit of force, distance and mass in the equation (2), then the equation (2) is written as,
$G = \dfrac{{N{m^2}}}{{kg \times kg}}$
On multiplying, then the above equation is written as,
$G = \dfrac{{N{m^2}}}{{k{g^2}}}$
The above equation is also written as,
$G = N{m^2}k{g^{ - 2}}\,..............\left( 3 \right)$
Thus, the above equation shows the SI unit of the gravitational constant.

Hence, the option (B) is the correct answer.

Note: The SI unit of the gravitational constant is also given by another term. In equation (3), the SI unit of the force is Newton and it is also written as the product of mass and acceleration. So, the SI unit of the force will become a kilogram meter per square of the second $\left( {kgm{s^{ - 2}}} \right)$. Now, this unit is substituted in the equation (3), the SI unit of gravitational constant is ${m^3}{s^{ - 2}}k{g^{ - 1}}$.