Question

What is the dimension of the Gravitational constant?

Answer 1

Derivation:
The formula for gravitational force is given by 2F=G⋅r2m1​⋅m2​​.
Alternatively, we can express G using dimensional analysis:
Starting with 2F=G⋅r2m1​⋅m2​​, we rearrange it to solve for G:
2G=m1​⋅m2​F⋅r2​ . . . . . (1)
Here, G represents the Universal Gravitational Constant.
Now, let's analyze the dimensions of the involved quantities:
Mass has dimensions: [M] Radius has dimensions: [L] Force has dimensions: [M⋅L⋅T−2]
Substituting these dimensions into equation (1), we get:
G=[M]⋅[M][M⋅L⋅T−2]⋅[L]2​
Simplifying further:
G=[M−1⋅L3⋅T−2]
Hence, the dimension of the Universal Gravitational Constant (G) is [M−1⋅L3⋅T−2].
M^-1L³T^-2

Answer 2

The dimension of the Gravitational constant, denoted as \( G \), can be derived from Newton's law of universal gravitation, which states:
\[F = G \frac{m_1 m_2}{r^2}\]
Here, \( F \) is the force between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \).
To find the dimension of \( G \), we need to express \( G \) in terms of the fundamental quantities: mass (M), length (L), and time (T).
Rearranging the formula to solve for \( G \), we get:
\[G = \frac{F r^2}{m_1 m_2}\]
Now, let's break down the dimensions of each term:
- The dimension of force \( F \) is:
 \[ [F] = \text{MLT}^{-2} \]
- The dimension of distance \( r \) is:
 \[ [r] = \text{L} \]
- The dimension of mass \( m_1 \) and \( m_2 \) is:
 \[ [m_1] = \text{M}, \quad [m_2] = \text{M} \]
Substituting these into the equation for \( G \):
\[[G] = \frac{[F] [r]^2}{[m_1] [m_2]}\]
Substitute the dimensions:
\[[G] = \frac{\text{MLT}^{-2} \cdot \text{L}^2}{\text{M} \cdot \text{M}}\]
Simplify the expression:
\[[G] = \frac{\text{MLT}^{-2} \cdot \text{L}^2}{\text{M}^2} = \frac{\text{ML}^3 \text{T}^{-2}}{\text{M}^2} = \text{M}^{-1} \text{L}^3 \text{T}^{-2}\]
So, the dimension of the Gravitational constant \( G \) is:
\[[G] = \text{M}^{-1} \text{L}^3 \text{T}^{-2}\]