The average distance separating Earth and the Moon is 384000 km. What is the net gravitational force exerted by Earth and the Moon on a 3.00 x 10⁴ kg spaceship located halfway between them? (Mass of earth-5.98 x 10²⁴ kg and mass of moon=7.36 x 10²² kg)

Since the spaceship is located at halfway between the Earth and Moon, the distance of the spaceship from both the Earth and the Moon is,

$r=\frac{R}{2}=\frac{384000 \mathrm{~km}}{2}=192000 \mathrm{~km}$

The net gravitational force exerted by the Earth and Moon on the spaceship can be calculated as:

$F_{n e t}=\frac{G m_{s}\left(M_{E}-M_{M}\right)}{r^{2}}$

Here, $G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{~m}^{2} / \mathrm{kg}^{2}$ is the gravitational constant.

Substitute the values,

$\begin{array}{l} F_{n e t}=\frac{\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{~m}^{2} / \mathrm{kg}^{2}\right)\left(3 \times 10^{4} \mathrm{~kg}\right)\left(5.98 \times 10^{24} \mathrm{~kg}-7.36 \times 10^{22} \mathrm{~kg}\right)}{\left[192000 \times 10^{3} \right]^{2}} \\ F_{n e t}=320.6 \mathrm{~N} \end{array}$