Question

S.I unit of gravitational constant $G$ is:
(A) $\dfrac{{{\text{N}}\,{{\text{m}}^2}}}{{{\text{kg}}}}$
(B) $\dfrac{{{\text{N}}\,{{\text{m}}^2}}}{{{\text{k}}{{\text{g}}^{\text{2}}}}}$
(C) $\dfrac{{{\text{N}}\,{\text{m}}}}{{{\text{k}}{{\text{g}}^{\text{2}}}}}$
(D) $\dfrac{{{\text{N}}\,{\text{m}}}}{{{\text{kg}}}}$

Answer 1

First of all, we will use the expression which states that the force of attraction is proportional to the product of the masses of the two bodies and inversely proportional to the square of the distance between them. We will rearrange the expression to find $G$ and substitute the respective units to get the desired answer.

Complete step by step solution:
Let us discuss something about this gravitational constant. As we all know that the gravity force is generally insignificant for smaller masses, but it is of high significance in case of the huge masses, like planets, natural satellites, stars, etc. The gravitational force is the binding force of all the elements present in the universe.
Let us proceed to answer the question. We know that the force between the two bodies is directly proportional to the product of their masses. However, the same force is inversely proportional to the square of the distance. We can say that the force between the two bodies increases when the mass of the bodies becomes heavier. Again, the force decreases in magnitude with the increase in the distance between them. To remove the sign of proportionality, a constant called gravitational constant was introduced in the equation.
Let us elaborate the following.Let the two masses be ${m_{\text{1}}}$ and ${m_2}$ . The distance between them is $r$ . The force which acts between the two bodies is $F$ .
The unit of force is Newton $\left( {\text{N}} \right)$ .
The unit of mass is kilograms $\left( {{\text{kg}}} \right)$ .
The unit of distance is metres $\left( {\text{m}} \right)$ .
Now, we will write the expression which gives the gravitational force:
$F = G \times \dfrac{{{m_1} \times {m_2}}}{{{r^2}}}$ …… (1)
After modifying the above equation, we get:
$G = \dfrac{{F \times {r^2}}}{{{m_1} \times {m_2}}}$ …… (2)
We will now put the respective units in the equation (2) and we get:
$G = \dfrac{{N \times {m^2}}}{{kg \times {\text{kg}}}} \\\ \therefore G = \dfrac{{{\text{N}}\,{{\text{m}}^2}}}{{{\text{k}}{{\text{g}}^2}}}$
Hence, the unit of gravitational constant is $\dfrac{{{\text{N}}\,{{\text{m}}^2}}}{{{\text{k}}{{\text{g}}^2}}}$ .

The correct option is B.

Note: While answering this question, one should remember that even $G$ is a constant, still it will have respective units. The accepted value of $G$ is $6.67 \times {10^{ - 11}}\,{\text{N}}\,{{\text{m}}^2}\,{\text{k}}{{\text{g}}^{ - 2}}$ . The fact that the force of gravitational attraction is only appreciable for objects of large mass accounts for its smallness.