Using dimensional analysis, determine the possible form of the gravitational force between two point masses $M$ and $m$ separated by a distance $r$, given that the force must depend only on $M$, $m$, $r$, and the universal gravitational constant $G$.

Step-by-step solution

1. Let $Fpropto{G}^a{M}^b{m}^c{r}^d$
2. Write dimensions: $[F]=ML{T}^{-2}$, $[G]=L^3{M}^{-1}{T}^{-2}$, $[M]=[m]=M$, $[r]=L$
3. Substitute: $(L^3{M}^{-1}{T}^{-2}{)}^a{M}^{b+c}{L}^d=ML{T}^{-2}$
4. Equate exponents: $M:-a+b+c=1$\ $L:3a+d=1$\ $T:-2a=-2\to a=1$
5. Solve: $-1+b+c=1\to b+c=2$; $3(1)+d=1\to d=-2$
6. With $b=c=1$ to keep symmetry, we get $F=kGMm/r^2$
7. The constant $k$ is dimensionless; choose $k=1$ for Newton’s law

Concept tested: Dimensional analysis

Published 4 October 2026 · How this solution was prepared and checked