Answer: 7.364 x 10^-4 N
Explanation:
[tex]$Given:$\begin{aligned}m_{1} &=7.5 \times 10^{5} \mathrm{~kg} \\m_{2} &=9.2 \times 10^{7} \mathrm{~kg} \\r &=2.5 \times 10^{3} \mathrm{~m}\end{aligned}$Here, $m_{1}, m_{2}$, and $r$ are the first mass, the second mass, and the distance[/tex]
[tex]$The gravitational force exerted between the masses is determined as follows.$F=G \frac{m_{1} m_{2}}{r^{2}}$Here, $G$ is the gravitational constant.Substitute the known values.$\begin{aligned}F &=\left(6.67 \times 0^{-11} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{kg}^{2}\right) \frac{\left(7.5 \times 10^{5} \mathrm{~kg}\right)\left(2.5 \times 10^{7} \mathrm{~kg}\right)}{\left(2.5 \times 10^{3} \mathrm{~m}\right)^{2}} \\&\bold{=7.364 \times 10^{-4} \bold{\mathrm{~N}}}\end{aligned}$[/tex]
