Question

Halley's comet orbits the Sun with a period of 75.3 yr. a) Find the semimajor axis of the orbit of Halley's comet in astronomical units ( \(1 \mathrm{AU}\) is equal to the semimajor axis of the Earth's orbit). b) If Halley's comet is 0.586 AU from the Sun at perihelion, what is its maximum distance from the Sun, and what is the eccentricity of its orbit?

Short answer

Question: Calculate the semimajor axis, aphelion, and eccentricity of Halley's comet, given its orbital period of 75.3 years and perihelion distance of 0.586 AU. Answer: The semimajor axis of Halley's comet is 17.9 AU, its maximum distance from the Sun (aphelion) is 35.214 AU, and its orbital eccentricity is 0.967.

Step-by-step solution

Convert the orbital period to seconds and find the semimajor axis

Given that the orbital period of Halley's comet is 75.3 years, we'll first convert that to seconds: \(75.3 \,\text{yr} \times \frac{365\, \text{days}}{1\, \text{yr}} \times \frac{24\, \text{hours}}{1\, \text{day}} \times \frac{60\, \text{minutes}}{1\, \text{hour}} \times \frac{60\, \text{seconds}}{1\, \text{minute}} = 2376729600\,\text{seconds}\) Now, we'll use Kepler's Third Law of Planetary Motion, which states that the square of the period of a planet (or comet) is proportional to the cube of the semimajor axis of its orbit: \(\frac{T^2}{a^3} = C\) Where \(T\) is the orbital period in seconds, \(a\) is the semimajor axis in astronomical units, and \(C\) is a constant. Since the Earth's orbital period is 1 year, we can express the constant as: \(C = \frac{T_E^2}{a_E^3} = \frac{(1\, \text{year})^2}{(1\, \mathrm{AU})^3}\) Now, we'll find the semimajor axis (\(a\)) of Halley's comet: \(\frac{T_H^2}{a_H^3} = C\) \(a_H = \sqrt[3]{\frac{T_H^2}{C}} = \sqrt[3]{\frac{(2376729600\, \text{seconds})^2}{(1\, \text{year})^2 / (1\, \mathrm{AU})^3}} = 17.9\, \mathrm{AU}\) So, the semimajor axis of Halley's comet is 17.9 AU.

Calculate the maximum distance (aphelion) from the Sun

Given the perihelion distance (\(q\)) of 0.586 AU, we can calculate the aphelion distance (\(Q\)) using the semimajor axis and the formula: \(a = \frac{q + Q}{2}\) Now, we'll solve for \(Q\): \(Q = 2a - q = 2(17.9\, \mathrm{AU}) - 0.586\, \mathrm{AU} = 35.214\, \mathrm{AU}\) Therefore, Halley's comet's maximum distance from the Sun (aphelion) is 35.214 AU.

Calculate the eccentricity of the orbit

We can find the eccentricity (\(e\)) of Halley's comet's orbit using the formula: \(e = \frac{Q - q}{Q + q}\) Plugging in the values for \(Q\) and \(q\), we get: \(e = \frac{35.214\, \mathrm{AU} - 0.586\, \mathrm{AU}}{35.214\, \mathrm{AU} + 0.586\, \mathrm{AU}} = 0.967\) So, the eccentricity of Halley's comet's orbit is 0.967. In conclusion, we found that Halley's comet has a semimajor axis of 17.9 AU, a maximum distance from the Sun (aphelion) of 35.214 AU, and an orbital eccentricity of 0.967.