The dimensions of universal gravitational con | vedclass

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Question

(b) $F = \frac{{G{m_1}{m_2}}}{{{d^2}}} \Rightarrow G = \frac{{F{d^2}}}{{{m_1}{m_2}}}$
$[G] = \frac{{[ML{T^{ - 2}}][{L^2}]}}{{[{M^2}]}} = [{M^{ - 1}}{

Vedclass · Accepted Answer

${M^{ - 1}}{L^3}{T^{ - 2}}$

Vedclass · Answer

(b) $F = \frac{{G{m_1}{m_2}}}{{{d^2}}} \Rightarrow G = \frac{{F{d^2}}}{{{m_1}{m_2}}}$
$[G] = \frac{{[ML{T^{ - 2}}][{L^2}]}}{{[{M^2}]}} = [{M^{ - 1}}{L^3}{T^{ - 2}}]$

The dimensions of universal gravitational constant are

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