1. Gravitational force at distance $R+h$ from Earth's center

The astronaut is at height $h$ above the Earth's surface, so his distance from the center of the Earth is

$r=R+h.$

By Newton's law of gravitation, the gravitational pull on the astronaut is

$F_G=\frac{GMm}{(R+h)^2}.$

2. Check whether it is zero

Although the astronaut feels weightless, that does not mean gravitational force is zero. Weightlessness in orbit occurs because the astronaut and satellite are in free fall together.

So option A is false.

3. Compare with $\dfrac{GMm}{R^2}$

Since $h>0$, we have

$(R+h)^2>R^2.$

Therefore,

$\frac{GMm}{(R+h)^2}<\frac{GMm}{R^2}.$

Also, gravitational force is clearly positive, so

$0<\frac{GMm}{(R+h)^2}<\frac{GMm}{R^2}.$

Hence option B is true.

4. Check option C

Option C says

$\frac{GMm}{(R+h)^2}<F_G<\frac{GMm}{R^2}.$

But we already found

$F_G=\frac{GMm}{(R+h)^2}.$

So the left inequality is not strict; it is actually equal. Therefore option C is false.

5. Check option D

From the formula directly,

$F_G=\frac{GMm}{(R+h)^2}.$

So option D is true.

Final Answer

Correct options are B and D.