An astronaut of mass $$m$$ is working on a satellite orbiting the earth at a distance $$h$$ from the earth's surface. The radius of the earth is $$R$$, while its mass is $$M$$. The gravitational pull $$F_G$$ on the astronaut is

We start by recalling Newton’s universal law of gravitation, which states that the attractive force between two point masses is

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

where $$G$$ is the universal gravitational constant, $$m_1$$ and $$m_2$$ are the two masses involved, and $$r$$ is the separation between their centres.

In the present situation the two masses are the earth of mass $$M$$ and the astronaut of mass $$m$$. The distance between their centres is not the earth’s radius $$R$$ alone, because the satellite is orbiting at a height $$h$$ above the earth’s surface. Hence the centre-to-centre distance becomes

$$r = R + h.$$

Substituting $$m_1 = M$$, $$m_2 = m$$, and $$r = R+h$$ into the gravitational formula, we obtain

$$F_G = \dfrac{G\,M\,m}{(R+h)^{2}}.$$

This expression gives the exact magnitude of the gravitational pull on the astronaut. It is clearly a positive, non-zero quantity as long as $$h$$ is finite. The sense of weightlessness experienced in orbit arises because both the satellite and the astronaut are in free fall with the same acceleration, not because the gravitational force vanishes.

Among the given alternatives, Option C reproduces precisely the derived formula, while the remaining choices either set the force to zero or express an inequality instead of the exact value.

Hence, the correct answer is Option C.