If the unit of length be doubled then the numerical value of the universal gravitation constant G will become (with respect to present value)

Options

A. Double
B. Half
C. `8` times
D. `1//8` times

Explanation

To solve the problem, we need to understand how the universal gravitation constant \( G \) is defined and how the units affect its numerical value. The universal gravitation constant \( G \) is given by the formula:

\[ G = \frac{F \cdot r^2}{m_1 \cdot m_2} \]

where:
- \( F \) is the gravitational force,
- \( r \) is the distance between the centers of two masses,
- \( m_1 \) and \( m_2 \) are the masses of the two objects.

### Step 1: Identify the units of \( G \)

The SI unit of \( G \) is:

\[ G = \frac{N \cdot m^2}{kg^2} \]

Where:
- \( N \) (Newton) can be expressed as \( kg \cdot m/s^2 \).

Thus, the units of \( G \) can be rewritten as:

\[ G = \frac{kg \cdot m/s^2 \cdot m^2}{kg^2} = \frac{m^3}{kg \cdot s^2} \]

### Step 2: Analyze the effect of doubling the unit of length

If we double the unit of length (let's denote the new unit of length as \( L' = 2L \)), we need to see how this affects the numerical value of \( G \).

1. The distance \( r \) in the formula for \( G \) will also double, so \( r' = 2r \).
2. The mass units remain unchanged, so \( m_1 \) and \( m_2 \) remain the same.
3. The force \( F \) is proportional to the product of the masses and inversely proportional to the square of the distance. Thus, when the distance doubles, the gravitational force becomes:

\[ F' = \frac{G \cdot m_1 \cdot m_2}{(2r)^2} = \frac{G \cdot m_1 \cdot m_2}{4r^2} = \frac{1}{4} F \]

### Step 3: Substitute into the equation for \( G \)

Now, substituting back into the equation for \( G \):

\[ G' = \frac{F' \cdot (r')^2}{m_1 \cdot m_2} = \frac{\left(\frac{1}{4} F\right) \cdot (2r)^2}{m_1 \cdot m_2} \]

Calculating this gives:

\[ G' = \frac{\left(\frac{1}{4} F\right) \cdot 4r^2}{m_1 \cdot m_2} = \frac{F \cdot r^2}{m_1 \cdot m_2} = G \]

### Conclusion

Thus, when the unit of length is doubled, the numerical value of the universal gravitation constant \( G \) remains unchanged. Therefore, the answer is:

\[ \text{The numerical value of } G \text{ will remain the same.} \]