Fifty years ago the Apollo 13 mission to the Moon captured the public's imagination when technical issues led to the aborting of the planned Moon landing - Leaving Cert Physics - Question a - 2020

Given:

• Mass of Earth, $M = 6.0 \times 10^{24} \text{ kg}$
• Radius of the Moon’s orbit, $R = 3.85 \times 10^{8} \text{ m}$
• Gravitational constant, $G = 6.674 \times 10^{-11} \text{ m}^3\text{ kg}^{-1}\text{ s}^{-2}$

Using the derived formula:

$T = 2\pi\sqrt{\frac{R^3}{GM}}$

Substituting the values:

$T = 2\pi\sqrt{\frac{(3.85 \times 10^8)^3}{(6.674 \times 10^{-11})(6.0 \times 10^{24})}}$

Calculating:

$T = 2\pi\sqrt{\frac{5.7309 \times 10^{25}}{4.0044 \times 10^{14}}}$

$T \approx 2.37 \times 10^6 \text{ seconds} \approx 27.5 \text{ days}$

First, calculate the distance from the center of the Moon:

• Radius of the Moon: $R_{moon} = 1740 \text{ km} = 1.74 \times 10^6 \text{ m}$
• Height above the surface: $h = 250 \text{ km} = 2.5 \times 10^5 \text{ m}$
• Total distance from the center: $R = R_{moon} + h = 1.74 \times 10^6 + 2.5 \times 10^5 = 1.99 \times 10^6 \text{ m}$

Using Newton's law of gravitation:

$F = \frac{GmM_{moon}}{R^2}$ Where $M_{moon} = 7.3 \times 10^{22} \text{kg}$:

$F = \frac{(6.674 \times 10^{-11})(80)(7.3 \times 10^{22})}{(1.99 \times 10^6)^2}$

Calculating:

$F \approx 78.73 \text{ N}$

Astronauts experience weightlessness when they orbit the Moon due to free-fall conditions. Both the astronaut and the spaceship are in free fall towards the Moon, falling along the same path as the Moon's curvature. Although gravity is still acting on them, they are moving forward fast enough such that the curve of their path matches the curvature of the Moon.

This creates a scenario where the astronauts do not exert a normal force on the spaceship, leading to the sensation of weightlessness. It is important to note that weightlessness does not mean the absence of gravitational force; rather, it indicates that there is no net force acting on the astronauts in the context of their motion.