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Give the dimensional formula for Gravitational constant G.

Answer
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Hint: Gravitational constant is the proportionality constant in Newton’s law of Gravitation. It relates to the force of attraction of two bodies having masses and separated at a distance. Hence the dimensional formula is depending on the dimensions of force, mass, and distance.

Complete step by step answer:
We know that Newton’s law of gravitation states that the force of attraction or repulsion between two bodies is proportional to the product of two masses and inversely proportional to the square of the distance between them. This can be expressed as,
$F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$
Where, $F$ is the force of attraction, ${m_1}$ and ${m_2}$ are the masses of two bodies, $r$ is the distance between the bodies, and $G$ is the proportionality constant called the Gravitational constant.
Thus the Gravitational constant, $G = \dfrac{{F{r^2}}}{{{m_1}{m_2}}}$
Let’s find the dimensional formula for force. According to Newton’s second law of motion, the expression for force is given as,
$F = ma$
Where $m$ is the mass and $a$ is the acceleration.
The dimensional formula for acceleration is given as $\left[ {L{T^{ - 2}}} \right]$ where $\left[ L \right]$ is for length and $\left[ T \right]$ is for time.
Therefore the dimensional formula for force is $\left[ {ML{T^{ - 2}}} \right]$
Where, $\left[ M \right]$ is for mass.
For the distance, we take the dimension for length. Here the distance is squared. Hence the dimension is $\left[ {{L^2}} \right]$.
Substituting all the dimensions in the formula for Gravitational constant, we get
$G=\dfrac{{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}{{\left[ M \right]\left[ M \right]}}$
$\Rightarrow G = \dfrac{{\left[ {{L^3}{T^{ - 2}}} \right]}}{{\left[ M \right]}}$
$\Rightarrow G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$

$\therefore$ The dimensional formula for Gravitation constant is $\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$.

Note:
Alternative method:
The SI unit for Gravitational constant is ${\text{N}}{{\text{m}}^{\text{2}}}{\text{/k}}{{\text{g}}^{\text{2}}}$. Newton is the SI unit of force.
Substituting the dimensions for the SI units we can find the dimensional formula directly.
Therefore,
$G= \dfrac{{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}{{\left[ M \right]\left[ M \right]}} $
$\Rightarrow G = \dfrac{{\left[ {{L^3}{T^{ - 2}}} \right]}}{{\left[ M \right]}} $
$\Rightarrow G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right] $
Thus the dimensional formula for Gravitation constant is $\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$.