Explain Newton's law of universal gravitation and calculate the gravitational force on a 60 kg parachutist at 3630 m above Earth.

Mass of parachutist, $m_1 = 60 \text{ kg}$
Height above Earth's surface, $h = 3630 \text{ m}$
Radius of Earth, $R = 6371 \text{ km} = 6371 \times 10^3 \text{ m}$
Gravitational constant, $G = 6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2$
Mass of Earth, $M_2 = 5.972 \times 10^{24} \text{ kg}$

The magnitude of the gravitational force is $\boxed{\text{587.86 N}}$.

5.4.1
The weight of the parachutist when wearing suit A compared to suit B is EQUAL TO.
The problem states that the suits are of equal mass and the combined mass of the parachutist and suit is 60 kg for both. Weight depends only on mass and gravitational acceleration, which are the same in both cases.

5.4.2
He would hit the earth first with suit A.
Suit A experiences a very small frictional force, meaning less air resistance. Less air resistance results in a greater net downward force and thus a greater acceleration, causing him to reach the ground faster.

5.4.3
Yes, the mass of the parachutist affects his acceleration when wearing suit A.
Reason: The net force acting on the parachutist is the gravitational force minus the air friction.
Equation:

Since the air friction ($F_{friction}$) is not directly proportional to the mass ($m$) in the same way gravity is, a change in mass will alter the acceleration. For a given air friction, a larger mass will experience a smaller reduction in acceleration from $g$.

5.4.4
The gravitational acceleration will be SMALLER THAN.
Explanation: Gravitational acceleration ($g$) is inversely proportional to the square of the distance from the center of the Earth.
Equation:

Where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $r$ is the distance from the center of the Earth.
When jumping from 10 000 m, the distance $r$ from the center of the Earth is greater than when jumping from 3 630 m ($R + 10000 \text{ m} > R + 3630 \text{ m}$). Since $g$ is inversely proportional to $r^2$, a larger $r$ results in a smaller $g$.