The gravitational pull between two bodies is 20N. Find the gravitational pull when their distance of separation is doubled.

The gravitational force ($F$) between two bodies is inversely proportional to the square of the distance ($r$) between their centers.

$$F \propto \frac{1}{r^2}$$

This can be written as:

$$F = G \frac{m_1 m_2}{r^2}$$

where $G$ is the gravitational constant, and $m_1$ and $m_2$ are the masses of the two bodies.

Let the initial gravitational pull be $F_1 = 20 \text{ N}$ at an initial distance $r_1$.
So,

$$F_1 = G \frac{m_1 m_2}{r_1^2} = 20 \text{ N}$$

The new distance of separation is $r_2 = 2r_1$.
Let the new gravitational pull be $F_2$.

$$F_2 = G \frac{m_1 m_2}{r_2^2}$$

Substitute $r_2 = 2r_1$ into the equation:

$$F_2 = G \frac{m_1 m_2}{(2r_1)^2}$$ $$F_2 = G \frac{m_1 m_2}{4r_1^2}$$ $$F_2 = \frac{1}{4} \left( G \frac{m_1 m_2}{r_1^2} \right)$$

From Step 2, we know that $G \frac{m_1 m_2}{r_1^2} = F_1$.
So,

$$F_2 = \frac{1}{4} F_1$$ $$F_2 = \frac{1}{4} (20 \text{ N})$$ $$F_2 = 5 \text{ N}$$

The final answer is $\text{D) 5N}$.