The gravitational force $$F$$ that an object of mass $$m$$ placed at a distance $$r$$ (from the center of Earth) experiences can be calculated using the universal law of gravitation, given by:

$$F = \frac{G M m}{r^2}$$

Where:

• $$G$$ is the gravitational constant ($$6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2$$)
• $$M$$ is the mass of the Earth (approximately $$5.972 \times 10^{24} \text{kg}$$)
• $$m$$ is the mass of the object
• $$r$$ is the distance from the center of the Earth to the object

However, when the object is at a distance $$2R$$ from the surface of the Earth, the total distance from the center of the Earth $$r'$$ becomes $$R + 2R = 3R$$. This is because the radius of the Earth $$R$$ is the distance from the Earth's center to its surface, so if the object is $$2R$$ above the surface, the total distance from the center is $$3R$$.

At the surface of the Earth, the gravitational force ($$F_{earth}$$) that acts on an object is given by its weight, which can be calculated using the formula $$F_{earth} = m \cdot g$$, where $$g$$ is the acceleration due to gravity on the surface of the Earth. Given that $$g = 10 \text{m/s}^2$$ and the mass of the body $$m = 90 \text{kg}$$, we get:

$$F_{earth} = 90 \text{kg} \cdot 10 \text{m/s}^2 = 900 \text{N}$$

To find the gravitational pull at a distance $$2R$$ from the Earth's surface, we use the fact that gravitational force varies inversely as the square of the distance from the center of the Earth. Since the distance triples ($$3R$$ from $$R$$), the gravitational force becomes $$\frac{1}{3^2} = \frac{1}{9}$$ of the force at the surface.

Therefore, the gravitational pull $$F_{2R}$$ on the body when placed at $$2R$$ from the Earth's surface is:

$$F_{2R} = \frac{F_{earth}}{9} = \frac{900 \text{N}}{9} = 100 \text{N}$$

Thus, the gravitational pull on the body when it is placed at a distance of $$2R$$ from the Earth's surface is 100 N. So, the correct option is:

Option C: 100 N