Question

The dimensions for the gravitational constant $G$ are ____.

• A. $M^{-1}L^2T^{-2}$
• B. $M^0L^0T^0$
• C. $MT^{-2}$
• D. $ML^2T^{-2}$
• E. $M^{-1}L^3T^{-2}$

Hint:

If you forget the dimensions of Force, remember $F = ma$, so Force is Mass ($M$) $\times$ Acceleration ($LT^{-2}$).

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The dimensional formula for any universal constant can be derived by isolating it in a fundamental physics equation where the dimensions of all other quantities are known.
For the universal gravitational constant $G$, we use Newton's law of universal gravitation.
Step 2: Key Formula or Approach:
Newton's law of gravitation states that the gravitational force $F$ is given by $F = G \frac{m_1 m_2}{r^2}$.
Rearranging this formula to solve for $G$ gives $G = \frac{F \cdot r^2}{m_1 \cdot m_2}$.
Step 3: Detailed Explanation:
We substitute the known dimensional formulas into the rearranged equation:
The dimension of force $F$ is $[MLT^{-2}]$.
The dimension of distance $r$ is $[L]$, so $r^2$ is $[L^2]$.
The dimension of mass $m_1$ and $m_2$ is $[M]$.
Plugging these in, we get:
\[ [G] = \frac{[MLT^{-2}] \cdot [L^2]}{[M] \cdot [M]} \]
Simplifying the terms:
\[ [G] = \frac{[ML^3T^{-2}]}{[M^2]} \]
\[ [G] = [M^{1-2}L^3T^{-2}] \]
\[ [G] = [M^{-1}L^3T^{-2}] \]
Step 4: Final Answer:
The dimensions for the gravitational constant $G$ are $M^{-1}L^{3}T^{-2}$.