Question

1 (a) the moon has a mass of \\(7.35 \times 10^{22}\text{ kg}\\) and can be considered to travel around the earth in a circular orbit of radius \\(3.84 \times 10^8\text{ m}\\). the time to complete one orbit is \\(2.36 \times 10^6\text{ s}\\).

calculate

(i) the speed of the moon in this orbit,

speed of the moon = ............................................................ \\(\text{ms}^{-1}\\) 2

(ii) the kinetic energy of the moon in this orbit,

kinetic energy of the moon = ............................................................ \\(\text{j}\\) 2

(iii) the gravitational potential energy of the moon in this orbit.

the mass of the earth is \\(5.98 \times 10^{24}\text{ kg}\\).

potential energy of the moon = ............................................................ \\(\text{j}\\) 2

Explanation:

Calculate orbital speed

We find the speed \(v\) using the circular orbit circumference and period.

$$ v = \frac{2\pi r}{T} $$

Given:

• Radius \(r = 3.84 \times 10^8\text{ m}\)
• Period \(T = 2.36 \times 10^6\text{ s}\)

$$ v = \frac{2\pi (3.84 \times 10^8)}{2.36 \times 10^6} \approx 1.022 \times 10^3\text{ m s}^{-1} $$

Calculate kinetic energy

We calculate the kinetic energy \(E_k\) using the mass and speed.

$$ E_k = \frac{1}{2} m v^2 $$

Given:

• Mass of Moon \(m = 7.35 \times 10^{22}\text{ kg}\)
• Speed \(v \approx 1.022 \times 10^3\text{ m s}^{-1}\)

$$ E_k = \frac{1}{2} (7.35 \times 10^{22}) (1.022 \times 10^3)^2 \approx 3.84 \times 10^{28}\text{ J} $$

Calculate gravitational potential energy

We calculate the potential energy \(E_p\) using Newton's law of gravitation.

$$ E_p = -\frac{G M m}{r} $$

Given:

• Gravitational constant \(G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}\)
• Mass of Earth \(M = 5.98 \times 10^{24}\text{ kg}\)
• Mass of Moon \(m = 7.35 \times 10^{22}\text{ kg}\)
• Distance \(r = 3.84 \times 10^8\text{ m}\)

$$ E_p = -\frac{(6.67 \times 10^{-11}) (5.98 \times 10^{24}) (7.35 \times 10^{22})}{3.84 \times 10^8} \approx -7.63 \times 10^{28}\text{ J} $$

Answer:

Question 1(a)(i)

speed of the Moon = \(1.02 \times 10^3\text{ m s}^{-1}\)

Question 1(a)(ii)

kinetic energy of the Moon = \(3.84 \times 10^{28}\text{ J}\)

Question 1(a)(iii)

potential energy of the Moon = \(-7.63 \times 10^{28}\text{ J}\)