(N/A) The acceleration due to gravity inside the Earth at a distance $r$ from the center is given by $g' = \frac{GM}{R^3} r$. Thus,$g' \propto r$ (directly proportional).
$(b)$ The weight of an object is $W = mg$. Since $g = \frac{GM}{R^2}$,if the radius $R$ becomes $R' = \frac{R}{2}$,the new acceleration due to gravity $g'$ is:
$g' = \frac{GM}{(R/2)^2} = 4 \frac{GM}{R^2} = 4g$.
Therefore,the weight $W' = mg' = 4mg$,which is $4$ times the original weight.
$(c)$ The orbital velocity $v_0$ is given by $v_0 = \sqrt{\frac{GM_e}{r}}$.
For a geostationary satellite,$r \approx 42260 \times 10^3 \ m$.
Substituting $G = 6.67 \times 10^{-11} \ N \ m^2/kg^2$ and $M_e = 5.98 \times 10^{24} \ kg$:
$v_0 = \sqrt{\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{42260 \times 10^3}} \approx 3070 \ m/s \approx 3 \ km/s$.