Neil Armstrong, the first man to walk on the moon, described seeing the Earth as follows: "All of a sudden, you could see the whole sphere - Leaving Cert Physics - Question 6 - 2019

A vector quantity has both magnitude and direction, while a scalar quantity has only magnitude.

For example:

• Vector Quantity: Velocity, which indicates both the speed of an object and the direction of its movement.
• Scalar Quantity: Distance, which only tells how far an object has moved without any information about the direction.

An astronaut standing on the moon has a constant speed if they are moving at a steady rate in a circular path. Since the direction of motion is continuously changing as they orbit, their velocity changes despite having a constant speed. This is due to the definition of velocity being a vector quantity, which includes both speed and direction.

Weight can be calculated using the formula:

$W = m \cdot g$

where:

• $m = 90 \, kg$ (mass)
• $g = 9.8 \, m \, s^{-2}$ (acceleration due to gravity)

Thus,

$W = 90 \, kg \cdot 9.8 \, m \, s^{-2} = 882 \, N$

Therefore, Armstrong's weight on Earth is 882 N.

Armstrong's mass remains the same regardless of location. Hence, his mass on the moon is still 90 kg.

The weight of an object depends on the gravitational force acting on it. The gravitational force on the moon is about 1/6th that of Earth. Thus, if Armstrong's weight on Earth is 882 N, his weight on the moon would be:

$W_{moon} = 0.17 \cdot W_{Earth} = 0.17 \cdot 882 \, N = 150.54 \, N$

This reduction in weight is due to the lower gravitational acceleration on the moon, which is approximately 1.63 m/s² as opposed to 9.8 m/s² on Earth.

Pressure is defined as the force exerted per unit area. Mathematically, it can be expressed as:

$P = \frac{F}{A}$

where:

• $P$ = pressure,
• $F$ = force,
• $A$ = area over which the force is distributed.

Using the definition of pressure:

$P = \frac{F}{A}$

From earlier, Armstrong's weight on the moon is approximately 150.54 N. The area of his shoe is 0.03 m². Therefore, substituting the values gives:

$P = \frac{150.54 \, N}{0.03 \, m^2} = 5018 \, Pa$

Thus, the pressure exerted by Armstrong on the surface of the moon is approximately 5018 Pa.