The unit of `(G )/(g )`is

Options

A. kg/m
B. `kg//m^2`
C. `m^2 //kg`
D. `m//kg`

Explanation

To find the unit of \(\frac{G}{g}\), we need to understand the definitions of \(G\) and \(g\):

1. **Gravitational Constant (G)**: The universal gravitational constant \(G\) has a value of approximately \(6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2\). Its unit is therefore \(\text{N m}^2/\text{kg}^2\).

2. **Gravitational Acceleration (g)**: The acceleration due to gravity \(g\) at the surface of the Earth is approximately \(9.81 \, \text{m/s}^2\). Its unit is therefore \(\text{m/s}^2\).

Now, we can calculate the unit of \(\frac{G}{g}\):

\[ \frac{G}{g} = \frac{\text{N m}^2/\text{kg}^2}{\text{m/s}^2} \]

Next, we need to express the Newton (N) in terms of its base units:

\[ 1 \, \text{N} = 1 \, \text{kg m/s}^2 \]

Substituting this into the equation:

\[ \frac{G}{g} = \frac{\left( \text{kg m/s}^2 \right) \cdot \text{m}^2}{\text{kg}^2 \cdot \text{m/s}^2} \]

This simplifies to:

\[ \frac{G}{g} = \frac{\text{kg m}^2/s^2}{\text{kg}^2 \cdot \text{m/s}^2} \]

Now, we can cancel out the common units:

- The \(\text{kg}\) in the numerator and one \(\text{kg}\) in the denominator cancel out.
- The \(\text{m}\) in the numerator and one \(\text{m}\) in the denominator cancel out.
- The \(\text{s}^2\) in the numerator and denominator cancel out.

After canceling, we are left with:

\[ \frac{G}{g} = \frac{\text{m}}{\text{kg}} = \text{m}^2/\text{kg} \]

Thus, the unit of \(\frac{G}{g}\) is:

\[ \frac{G}{g} = \text{m}^2/\text{kg} \]

### Final Answer:
The unit of \(\frac{G}{g}\) is \(\text{m}^2/\text{kg}\).