Question

Be sure to show all calculations clearly and state your final answers in complete sentences. Halley Orbit. Halley's comet orbits the Sun every 76.0 years and has an orbital eccentricity of 0.97 a. Find its average distance (semimajor axis). b. Halley's orbit is a very eccentric (stretched-out) ellipse, so that at perihelion it is only about 90 million \(\mathrm{km}\) from the Sun, compared to more than 5 billion \(\mathrm{km}\) at aphelion. Does Halley's comet spend most of its time near its perihelion distance, its aphelion distance, or halfway in between? Explain.

Short answer

Halley's comet's average distance from the Sun is about 17.8 AU, and it spends most time near aphelion.

Step-by-step solution

Identify given values and constants

We are given that Halley's comet orbits the Sun every \( T = 76.0 \) years and has an orbital eccentricity \( e = 0.97 \). We also know the gravitational constant \( G = 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \) and the mass of the Sun \( M = 1.989 \times 10^{30} \text{ kg} \). We need to find its average distance from the Sun (semimajor axis \( a \)).

Calculate the semimajor axis using Kepler's Third Law

Kepler's Third Law states that \( T^2 = \frac{4\pi^2}{GM} a^3 \). Rearranging for \( a \), we get:\[a = \left( \frac{T^2 GM}{4\pi^2} \right)^{1/3}\]Substitute the given values (converting \( T \) from years to seconds) and solve for \( a \).

Convert orbital period to seconds

Since \( T \) is given in years, we convert it to seconds:\[\ T = 76.0 \times 365.25 \times 24 \times 3600 \text{ s} \ = 2.397 \times 10^9 \text{ s}.\]

Substitute values and solve for semimajor axis

Using the formula for \( a \):\[a = \left( \frac{(2.397 \times 10^9)^2 \times (6.674 \times 10^{-11}) \times (1.989 \times 10^{30})}{4\pi^2} \right)^{1/3}\]Solve for \( a \) to find that \( a \approx 2.67 \times 10^{12} \) meters or \( \approx 17.8 \text{ AU} \) (astronomical units).

Analyze the orbit characteristics

Halley's orbit has a very high eccentricity of 0.97, meaning the orbit is highly elliptical. The perihelion distance is about 90 million km, and aphelion is more than 5 billion km. In highly elliptical orbits, objects move faster when near the perihelion and spend more time at greater distances. Thus, Halley's comet spends most of its time near the aphelion distance.

Key concepts

orbital eccentricity

Orbital eccentricity is a measure of how much an orbit deviates from being circular. For an orbit, the eccentricity \( e \) can vary from 0 to 1.
- When \( e = 0 \), the orbit is perfectly circular.
- When \( e = 1 \), the orbit is parabolic, indicating the object is not bound to the central body.
Halley's comet has an eccentricity of 0.97. This means its orbit is highly stretched out, resembling an elongated ellipse. The high eccentricity indicates that there is a significant difference between its closest and farthest distance from the Sun, known as the perihelion and aphelion distances, respectively. A high eccentricity like this also affects the speed of the comet, causing it to move much faster when close to the Sun and slower when it is farther away.

semimajor axis

The semimajor axis (\( a \)) of an orbit is essentially the average distance between the orbital object and the body it is orbiting. It is the longest diameter of an elliptical orbit, stretching from one edge to the opposite edge via the center.
For elliptical orbits like that of Halley's comet, the semimajor axis provides essential information about the size of the orbit. The size of the semimajor axis can be determined using Kepler's Third Law if the period of the orbit and other constants are known. In the case of Halley's comet, the semimajor axis is approximately 17.8 astronomical units (AU), which is derived from its orbital period and high eccentricity. This significant distance indicates how the comet sweeps vast ranges across its path as it speeds through its eccentric orbit.

Kepler's Third Law

Kepler's Third Law of Planetary Motion is an essential law relating the motion of orbiting bodies around a star. This law states that the square of the orbital period (\( T \)) is proportional to the cube of the semimajor axis \( a \), of the object's orbit. Mathematically, it is expressed as: \( T^2 = \frac{4\pi^2}{GM} a^3 \), where \( G \) is the gravitational constant and \( M \) is the mass of the primary body, like the Sun in our solar system.
This relationship helps in determining the semimajor axis of an object if its orbital period is known, and vice versa.
For Halley's comet, the application of Kepler’s Third Law allows us to calculate its semimajor axis by factoring in constants like the gravitational constant and the mass of the Sun, with its known orbital period of 76 years.

perihelion and aphelion distances

Perihelion and aphelion are terms used to describe the closest and farthest points along an orbit concerning the sun. For an object in an elliptical orbit like Halley's comet:
- The **perihelion distance** is the shortest distance from the Sun, where the object travels the fastest due to the increased gravitational pull. For Halley's comet, this distance is about 90 million kilometers.
- The **aphelion distance** is the farthest point from the Sun, where the object moves slower and spends most of its time. Halley's comet's aphelion distance is more than 5 billion kilometers.
Due to its high orbital eccentricity of 0.97, Halley's comet spends the majority of its time closer to the aphelion point. The reason for this is the slower speed at greater distances due to weaker gravitational effects, causing it to linger far from the Sun for a more extended period before returning to the perihelion swiftly.