In Newton's law of universal gravitation, the gravitational force F between two point mass… — JAMB UTME

In Newton's law of universal gravitation, the gravitational force $F$ between two point masses $m_1$ and $m_2$ separated by a distance $r$ is expressed as $F = G \frac{m_1 m_2}{r^2}$ , where $G$ is the universal gravitational constant. What is the dimensional formula of $G$ in terms of mass ( $\text{M}$ ), length ( $\text{L}$ ), and time ( $\text{T}$ )?

$\text{M}^{-1}\text{L}^3\text{T}^{-2}$ Answer
B
$\text{M}^{1}\text{L}^3\text{T}^{-2}$
C
$\text{M}^{-1}\text{L}^2\text{T}^{-2}$
D
$\text{M}^{-1}\text{L}^3\text{T}^{-1}$

Answer

\text{M}^{-1}\text{L}^3\text{T}^{-2}

The dimensional formula of the universal gravitational constant

G

is derived by expressing

G = \frac{F r^2}{m_1 m_2}

. Substituting the basic dimensions of force (

\text{M}\text{L}\text{T}^{-2}

), radius squared (

\text{L}^2

), and mass squared (

\text{M}^2

) gives

\frac{(\text{M}\text{L}\text{T}^{-2})(\text{L}^2)}{\text{M}^2} = \text{M}^{-1}\text{L}^3\text{T}^{-2}

Step-by-Step Solution

Make

G

the subject of the formula

G = \frac{F \cdot r^2}{m_1 \cdot m_2}

To analyze the dimensions of

G

, isolate it on one side of the equation.

Substitute the fundamental dimensions for force, distance, and mass

[F] = \text{M}\text{L}\text{T}^{-2}, [r^2] = \text{L}^2, [m_1 m_2] = \text{M}^2

Force is mass times acceleration, giving dimensions

\text{M}\text{L}\text{T}^{-2}

, while distance squared gives

\text{L}^2

and the product of two masses gives

\text{M}^2

Simplify the dimensional expression

[G] = \frac{(\text{M}\text{L}\text{T}^{-2}) \cdot \text{L}^2}{\text{M}^2} = \text{M}^{1-2} \text{L}^{1+2} \text{T}^{-2} = \text{M}^{-1}\text{L}^3\text{T}^{-2}

Apply laws of indices for fundamental dimensions

\text{M}

\text{L}

, and

\text{T}

Key Concept

Dimensional Analysis of Physical Constants

Estimated Time:1m 15s

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