Gravitation - The Gravitational Constant

Two spheres of masses $4000 \text{ kg}$ and $1000 \text{ kg}$ are placed at a distance. If the distance between their centers is $2 \text{ m}$, calculate the gravitational force of attraction between them. Also, find the difference in their masses using vertical subtraction.

Given: $m_1 = 4000 \text{ kg}$ $m_2 = 1000 \text{ kg}$ $r = 2 \text{ m}$ $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$

Calculation of Force: $F = G \frac{m_1 m_2}{r^2}$ $F = 6.67 \times 10^{-11} \times \frac{4000 \times 1000}{2^2}$ $F = 6.67 \times 10^{-11} \times \frac{4 \times 10^6}{4}$ $F = 6.67 \times 10^{-11} \times 10^6$ $F = 6.67 \times 10^{-5} \text{ N}$

Calculation of mass difference: $\begin{array}{r} 4000 \\ - 1000 \\ \hline 3000 \end{array}$

The gravitational force is calculated using the formula $F = \frac{G m_1 m_2}{r^2}$. The masses are multiplied and divided by the square of the distance, then multiplied by the constant $G$. The mass difference is calculated using simple vertical subtraction as shown.

Determine the dimensions of the Universal Gravitational Constant $G$ from the gravitational force formula.

From the formula $F = G \frac{m_1 m_2}{r^2}$, we can isolate $G$: $G = \frac{F r^2}{m_1 m_2}$ Dimensions of Force $[F] = [M L T^{-2}]$ Dimensions of distance squared $[r^2] = [L^2]$ Dimensions of product of masses $[m_1 m_2] = [M^2]$ Substituting these: $[G] = \frac{[M L T^{-2}] [L^2]}{[M^2]}$ $[G] = [M^{-1} L^3 T^{-2}]$

By rearranging the gravitational force formula and substituting the standard dimensions for Force, Length, and Mass, we derive the units and dimensions for the constant $G$.