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Answer Questions · Q7

Q.What are the dimensions of the universal gravitational constant?

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From F=Gm1m2r2F=\dfrac{Gm_1m_2}{r^2}, rearrange for G: G=Fr2m1m2G=\dfrac{Fr^2}{m_1m_2}. Force has dimensions [F]=[MLT−2][F]=[MLT^{-2}] (from F=maF=ma), r2r^2 has dimensions [L2][L^2], and m1m2m_1m_2 has dimensions [M2][M^2]. Substituting: [G]=[MLT−2][L2][M2]=[ML3T−2][M2]=[M−1L3T−2][G]=\dfrac{[MLT^{-2}][L^2]}{[M^2]}=\dfrac{[ML^3T^{-2}]}{[M^2]}=[M^{-1}L^3T^{-2}]. [!ANSWER] [G]=[L3M−1T−2][G]=[L^3M^{-1}T^{-2}], i.e. GG has units N m^2/kg^2 in SI.

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