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The unit quantity of $\frac{G}{g}$ is?

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kg m^{- 2}

m^{2} k g^{- 1}

kg m^{- 1}

kg m^{2}

answer is B.

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Detailed Solution

The unit quantity of

\frac{G}{g}

m^{2} k g^{- 1}

.
The force of attraction between any two bodies is given by Newton’s law of gravity

F = \frac{G m_{1} m_{2}}{r^{2}}

m_{1}

and

m_{2}

are masses

G

is universal gravitational constant is

6.75 \times 10^{- 11} N m^{2} / k g^{2}

r

is separation between bodies
The variation of acceleration due to gravity at surface of earth is

g = \frac{GM}{R^{2}}

= 9.8 \frac{m}{s^{2}}

The units of quantity

\frac{G}{g}

\Rightarrow \frac{N m^{2} k g^{- 2}}{m s^{- 2}}

\Rightarrow Nm s^{2} k g^{- 2}

\Rightarrow kgm s^{- 2} \times m s^{2} k g^{- 2}

\Rightarrow m^{2} k g^{- 1}

Hence the units of quantity

\frac{G}{g}

m^{2} k g^{- 1}

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