$(i)$ How is the force of attraction dependent on the masses of objects and the distance between them?
$(ii)$ The mass of the Earth is $6 \times 10^{24} \, kg$ and that of a natural satellite is $6.5 \times 10^{20} \, kg$. If the distance between them is $3.35 \times 10^{6} \, km$, calculate the force exerted by the Earth on it. $(G = 6.7 \times 10^{-11} \, Nm^{2} kg^{-2})$

(N/A) $(i)$ The force of attraction between two objects is directly proportional to the product of their masses, i.e., $F \propto m_{1} m_{2}$.
The force of attraction between two objects is inversely proportional to the square of the distance between them, i.e., $F \propto \frac{1}{r^{2}}$.
$(ii)$ Given:
Mass of the Earth, $M = 6 \times 10^{24} \, kg$
Mass of the satellite, $m = 6.5 \times 10^{20} \, kg$
Distance between them, $r = 3.35 \times 10^{6} \, km = 3.35 \times 10^{9} \, m$
Gravitational constant, $G = 6.7 \times 10^{-11} \, Nm^{2} kg^{-2}$
Using the formula $F = G \frac{Mm}{r^{2}}$:
$F = \frac{6.7 \times 10^{-11} \times 6 \times 10^{24} \times 6.5 \times 10^{20}}{(3.35 \times 10^{9})^{2}}$
$F = \frac{261.3 \times 10^{33}}{11.2225 \times 10^{18}}$
$F \approx 23.28 \times 10^{15} \, N$ or $2.33 \times 10^{16} \, N$.