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Consider the Law of Gravitation where the force of gravity F is:

F = G(m1m2/r^{2})

Which of the following represents the gravitational constant G in the fundamental dimensions of mass (M), length (L) and time (T)?

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Correct Answer: **A.** *M*^{-1}L^{3}T^{-2}

PHY 2.4

This is a classic example of dimensional analysis (DA). Expect to use DA to one degree or the other during the real MCAT. First, you should know that the unit of force is a newton which is also a kg m/s^{2} (even if you did not memorize that, it is easy to work out since F = ma so the units of F must be kg multiplied by acceleration which is m/s^{2}). Of course the 2 m’s in the Law of Gravitation represent masses (M) and the r represents a distance or length (L). So we get:

F = G(m1m2/r^{2})

Now transferring to the fundamental quantities except for G:

MLT^{-2} = G(M)^{2}L^{-2}

Divide both sides by (M)^{2}L^{-2} and isolate for G:

G = M^{-1}L^{3}T^{-2}

… which is the same as m^{3}kg^{-1}s^{-2}.

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