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Difficulty: MediumFundamental and Derived Units

According to Newton's law of universal gravitation, the gravitational force FF between two masses m1m_1 and m2m_2 separated by a distance rr is given by F=Gm1m2r2F = \frac{G m_1 m_2}{r^2}. Which of the following correctly expresses the derived SI unit of the universal gravitational constant, GG, in terms of fundamental (base) units?

  1. kg−1⋅m3⋅s−2\text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}Answer
  2. B
    kg⋅m3⋅s−2\text{kg}\cdot\text{m}^3\cdot\text{s}^{-2}
  3. C
    kg−1⋅m2⋅s−2\text{kg}^{-1}\cdot\text{m}^2\cdot\text{s}^{-2}
  4. D
    kg⋅m⋅s−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}

Answer

The SI unit of the universal gravitational constant GG expressed in fundamental base units is kg−1⋅m3⋅s−2\text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}.
The expression kg−1⋅m3⋅s−2\text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2} is correct because rearranging F=Gm1m2r2F = \frac{G m_1 m_2}{r^2} yields G=Fr2m1m2G = \frac{F r^2}{m_1 m_2}. Replacing FF with its base equivalent kg⋅m⋅s−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}, rr with m\text{m}, and m1,m2m_1, m_2 with kg\text{kg} gives (kg⋅m⋅s−2)⋅m2kg2=kg−1⋅m3⋅s−2\frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot\text{m}^2}{\text{kg}^2} = \text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}.

Step-by-Step Solution

1
Rearrange the gravitational formula to solve for GG
G=F⋅r2m1⋅m2G = \frac{F \cdot r^2}{m_1 \cdot m_2}
To express the unit of GG, we need it in terms of quantities with known units.
2
Substitute the SI unit for force in fundamental base units
\text{Unit of } F = \text{N} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}
Force is mass times acceleration (F=maF = ma), so its base unit is kg⋅m⋅s−2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}.
3
Substitute all base units into the formula for GG
\text{Unit of } G = \frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \cdot \text{m}^2}{\text{kg} \cdot \text{kg}} = \frac{\text{kg}\cdot\text{m}^3\cdot\text{s}^{-2}}{\text{kg}^2}
Combining length terms (m⋅m2=m3\text{m} \cdot \text{m}^2 = \text{m}^3) and mass terms in the denominator.
4
Simplify the mass exponent using laws of indices
\text{Unit of } G = \text{kg}^{1 - 2}\cdot\text{m}^3\cdot\text{s}^{-2} = \text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}
Dividing by kg2\text{kg}^2 subtracts 2 from the mass exponent.

Key Concept

Derivation of Derived Units in Base SI Units
Estimated Time:1m 15s
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Topics

Fundamental and Derived Units

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