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Chapter 12: Problem 48

Halley's comet orbits the Sun with a period of 76.2 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 \(0.56 \mathrm{AU}\) from the Sun at perihelion, what is its maximum distance from the Sun, and what is the eccentricity of its orbit?

Short Answer

Expert verified
Answer: The semimajor axis of Halley's comet's orbit is approximately 17.9 AU, the maximum distance (aphelion) from the Sun is approximately 35.24 AU, and the eccentricity of the orbit is approximately 0.97.

Step by step solution

01

Use Kepler's third law to find the semimajor axis of Halley's comet's orbit

Kepler's third law states that \((\frac{a}{1 AU})^3 = (\frac{T}{1 yr})^2\), where \(a\) is the semimajor axis of the orbit, \(T\) is the period, \(1 AU\) is the astronomical unit (the semimajor axis of Earth's orbit around the Sun), and \(1 yr\) is one Earth year. Here, \(T = 76.2 yr\). Substituting the values and solving for \(a\), we get: \((\frac{a}{1 AU})^3 = (\frac{76.2 yr}{1 yr})^2\) \(a^3 = (76.2^2) (1 AU)^3\) \(a = \sqrt[3]{(76.2)^2 (1 AU)^3}\)
02

Calculate the semimajor axis of Halley's comet's orbit

Now, we can plug in the values and solve for the semimajor axis \(a\): \(a = \sqrt[3]{(76.2)^2 (1 AU)^3}\) \(a = \sqrt[3]{5800.84 (1 AU)^3}\) \(a \approx 17.9 AU\) So, the semimajor axis of Halley's comet is approximately \(17.9 AU\).
03

Find the maximum distance (aphelion) of Halley's comet from the Sun

We are given that the distance from the Sun at perihelion is \(0.56 AU\). Let \(q\) be the perihelion distance and \(Q\) be the aphelion distance. We know that \(a = \frac{q+Q}{2}\), so we can solve for \(Q\): \(Q = 2a - q\) \(Q = 2(17.9 AU) - 0.56 AU\) \(Q \approx 35.24 AU\) The maximum distance (aphelion) of Halley's comet from the Sun is approximately \(35.24 AU\).
04

Find the eccentricity of Halley's comet's orbit

To find the eccentricity \(e\) of Halley's comet's orbit, we use the formula: \(e = \frac{Q-q}{Q+q}\) Plugging in the values for \(Q\) and \(q\), we get: \(e = \frac{35.24 AU - 0.56 AU}{35.24 AU + 0.56 AU}\) \(e = \frac{34.68 AU}{35.8 AU}\) \(e \approx 0.97\) The eccentricity of Halley's comet's orbit is approximately \(0.97\).

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