Answer to Question #213578 in Mechanics | Relativity for Sam 63x

Question

Newton's law of universal gravitation gives the force between two objects of mass 1 and mass 2 separated by distance how as F=G(m1 m2)/R2 use dimensional analysis to find the unit of gravitational

Answer 1

Explanations & Calculations

• R is a length, hence its dimension should be $\small [R] = L$

• Force is defined as $\small F = ma$, the mass has the dimension $\small [m]=M$, and acceleration is defined as the rate of change of velocity. Since the change in velocity results in just another value of velocity, it has the same dimension as the velocity: $\small [\Delta v]=[v]$. And the definition for velocity: rate of change in displacement, $v = \frac{\Delta s}{\Delta t}$ gives dimension of velocity to be $\small [V]=\large \frac{[\Delta s]}{[\Delta t]}=\frac{[s]}{[t]}=\frac{L}{T}=\small LT^{-1}$. Dimension of the diference in those parameters follow the same nature discribed for the change in velocity.
• Then we receive the dimension of acceleration to be $\small [a]=\large \frac{[\Delta v]}{[\Delta t]}=\frac{LT^{-1}}{T}=\small LT^{-2}$
• Then the dimension of force will be $\small [F]=M.LT^{-2}$
• $\small [m_1\times m_2]= M^2$

• Finally, a true equation dimension of both sides separated by an equal mark should have the same dimension. Therefore,

$\qquad\qquad \begin{aligned} \small [F]&=\small [G.\frac{m_1m_2}{R^2}]\\ &=\small [G].[\frac{m_1m_2}{R^2}]\\ \small MLT^{-2}&=\small {G}.(\frac{M^2}{L^2})\\ \small [G]&=\small M^{-1}L^3T^{-2} \end{aligned}$