Question

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Explanation:

Identify missing values from previous parts

The table refers to "Answer from (a)(iii)" and "Answer from (a)(ii)" for a circular orbit of radius \(3.84 \times 10^8\text{ m}\).
For a circular orbit of the Moon around the Earth:

• The mass of the Earth is \(M \approx 5.97 \times 10^{24}\text{ kg}\).
• The mass of the Moon is \(m \approx 7.35 \times 10^{22}\text{ kg}\).
• The orbital radius is \(r = 3.84 \times 10^8\text{ m}\).
• Gravitational Potential Energy is:

$$E_p = -\frac{GMm}{r} \approx -\frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22}}{3.84 \times 10^8} \approx -7.63 \times 10^{28}\text{ J}$$

• For a circular orbit, the Kinetic Energy is:

$$E_k = \frac{GMm}{2r} = -\frac{1}{2}E_p \approx 3.82 \times 10^{28}\text{ J}$$

• The total energy \(E\) is:

$$E = E_p + E_k \approx -3.82 \times 10^{28}\text{ J}$$

Complete the energy table

Using the conservation of total mechanical energy:

• The total energy is constant during the orbit: \(E = -3.82 \times 10^{28}\text{ J}\) (or more precisely, from the circular orbit values, \(E = -3.81 \times 10^{28}\text{ J}\) to \( -3.82 \times 10^{28}\text{ J}\)).
• At \(r = 3.56 \times 10^8\text{ m}\):
• \(E_p = -8.24 \times 10^{28}\text{ J}\)
• \(E = -3.82 \times 10^{28}\text{ J}\)
• \(E_k = E - E_p = -3.82 - (-8.24) = 4.42 \times 10^{28}\text{ J}\)
• At \(r = 4.07 \times 10^8\text{ m}\):
• \(E_p = -\frac{GMm}{r} = -8.24 \times \frac{3.56}{4.07} \approx -7.21 \times 10^{28}\text{ J}\)
• \(E = -3.82 \times 10^{28}\text{ J}\)
• \(E_k = E - E_p = -3.82 - (-7.21) = 3.39 \times 10^{28}\text{ J}\)

Calculate the maximum speed of the Moon

The maximum speed occurs at the closest approach (perigee), where the Kinetic Energy is at its maximum:

• \(E_{k,\text{max}} = 4.42 \times 10^{28}\text{ J}\)
• Using the formula for kinetic energy:

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

• Solving for \(v_{\text{max}}\):

$$v_{\text{max}} = \sqrt{\frac{2 E_{k,\text{max}}}{m}} = \sqrt{\frac{2 \times 4.42 \times 10^{28}}{7.35 \times 10^{22}}} \approx 1.10 \times 10^3\text{ m s}^{-1}$$

Answer:

Question 1

The completed table showing the energy values (in units of \(10^{28}\text{ J}\)):

distance from Earth / \(10^8\text{ m}\)	gravitational potential energy / \(10^{28}\text{ J}\)	total energy / \(10^{28}\text{ J}\)	kinetic energy / \(10^{28}\text{ J}\)
3.84	\(-7.63\)	\(-3.82\)	\(3.81\)
4.07	\(-7.21\)	\(-3.82\)	\(3.39\)

Question 2

The maximum speed of the Moon is:
maximum speed = <blank>\(1.10 \times 10^3\)</blank> \(\text{m s}^{-1}\)