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:
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\( F \) (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: \[ [F] = [M][LT^{-2}] = [MLT^{-2}]. \]
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\( m_1 \) and \( m_2 \) are both masses, so their dimensions are also \([M]\).
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\( r \) (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: \[ \boxed{[M^{-1}L^3T^{-2}]}. \]
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