Question

If F=Gmn/r`2 find the dimensions of G

Answer 1

To find the dimensions of the gravitational constant \( G \) in the equation

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

we start by recognizing the dimensions of the other variables involved:

1. \( F \) (force) has dimensions given by Newton's second law, \( F = ma \):
- Mass \( m \) has dimensions \([M]\) (kilograms),
- Acceleration \( a \) has dimensions of length per time squared \([LT^{-2}]\).

Therefore, the dimensions of force \( F \) are:
\[
[F] = [M][LT^{-2}] = [MLT^{-2}].
\]

2. \( m_1 \) and \( m_2 \) are both masses, so their dimensions are also \([M]\).

3. \( r \) (distance) has dimensions of length:
\[
[r] = [L].
\]

Now substituting into the equation \( F = \frac{G m_1 m_2}{r^2} \):

Rearranging it for \( G \):
\[
G = \frac{F r^2}{m_1 m_2}.
\]

Substituting the dimensions we have:
- \( [F] = [MLT^{-2}] \),
- \( [r^2] = [L^2] \),
- \( [m_1] = [M] \) and \( [m_2] = [M] \) implies \( [m_1 m_2] = [M^2] \).

Now substituting these into the expression for \( G \):
\[
[G] = \frac{[MLT^{-2}] [L^2]}{[M^2]}.
\]

This simplifies to:
\[
[G] = \frac{[ML^3T^{-2}]}{[M^2]} = [M^{-1}L^3T^{-2}].
\]

So, the dimensions of \( G \) are:
\[
\boxed{[M^{-1}L^3T^{-2}]}.
\]