/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 108 Halley's comet travels in an elo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Halley's comet travels in an elongated elliptic orbit for which the minimum distance from the sun is approximately \(\frac{1}{2} r_{f},\) where \(r_{E}=150 \times 10^{6} \mathrm{km}\) is the mean distance from the sun to the earth. Knowing that the periodic time of Halley's comet is about 76 years, determine the maximum distance from the sun reached by the comet.

Short Answer

Expert verified
Maximum distance is approximately \(5343 \times 10^6\) km.

Step by step solution

01

Understand the Problem

We need to find the maximum distance that Halley's comet reaches from the sun, given that its minimum distance is half the Earth's mean distance to the sun and its orbital period is 76 years. Halley's comet has an elliptical orbit, where one focus is the sun.
02

Apply Kepler's Third Law

Kepler's Third Law states that the square of the period of orbit (\(T\)) is proportional to the cube of the semi-major axis (\(a\)) of the orbit. Mathematically, this is expressed as \(T^2 = k \cdot a^3\), where \(k\) is a constant. For objects orbiting the same body, \(k\) remains the same, so we can set up a relative equation using Earth's orbit:\[\left(\frac{T_h}{T_E}\right)^2 = \left(\frac{a_h}{a_E}\right)^3\]where \(T_h = 76\) years, \(T_E = 1\) year (Earth's orbital period), and \(a_E = r_E\).
03

Solve for Semi-Major Axis of Halley's Comet

Using the equation from Step 2:\[\left(\frac{76}{1}\right)^2 = \left(\frac{a_h}{150 \times 10^6}\right)^3\]Calculate:\[76^2 = a_h^3 / (150 \times 10^6)^3\]\[a_h = (76^2)^{1/3} \times 150 \times 10^6\]
04

Calculate Semi-Major Axis

Now we calculate \((76^2)^{1/3}\):\[76^2 = 5776\]\[(5776)^{1/3} \approx 18.06\]Substitute into the equation for \(a_h\):\[a_h \approx 18.06 \times 150 \times 10^6 \text{ km}\]
05

Find the Maximum Distance (Aphelion)

The orbit is elliptical, so maximum distance (aphelion) \(r_a\) is calculated by:\(r_a = 2a - r_p\), where \(r_p = \frac{1}{2}r_E = \frac{1}{2} \times 150 \times 10^6 \) km.Calculate \(r_a\):\[r_a = 2 \times 18.06 \times 150 \times 10^6 - \frac{1}{2} \times 150 \times 10^6\]
06

Simplify and Solve for Aphelion

\(r_a = 2 \times 18.06 \times 150 \times 10^6 - 75 \times 10^6\)\[r_a = (2 \times 18.06 \times 150 - 75) \times 10^6\]Calculating:\[r_a = (5418 - 75) \times 10^6\]\[r_a \approx 5343 \times 10^6 \text{ km}\]
07

Conclusion

The maximum distance from the sun that Halley's comet reaches is approximately \(5343 \times 10^6\) km.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elliptical Orbits
An elliptical orbit is the path followed by celestial objects, such as planets and comets, around a larger body like the sun. Unlike circular orbits, which have a constant distance from the center, elliptical orbits have varying distances. This is because ellipses are oval-shaped and have two foci. For objects in our solar system, the sun is located at one of these foci.

Elliptical orbits are significant because they help us understand how objects like Halley's comet move through space. The shape of an ellipse is described by its semi-major axis, the longest diameter, and the eccentricity which determines how stretched out the ellipse is. In an elliptical orbit:
  • The closest point to the sun is called the perihelion.
  • The farthest point from the sun is called the aphelion.
These distinct points cause objects to speed up when they are near the sun and slow down when they are farther away. This characteristic motion is predicted by Kepler's laws of planetary motion and is crucial for predicting the positions of objects in space over time.
Orbital Period
The orbital period is the time a celestial object takes to complete one full orbit around another body. For instance, Earth has an orbital period of approximately one year as it travels around the sun. Kepler's laws tell us that the orbital period is related to the size of the orbit. Specifically, Kepler's Third Law states that the square of the orbital period (\(T^2\)) is proportional to the cube of the semi-major axis of the orbit (\(a^3\)).

This relationship is important for understanding how different objects move through space. For Halley's comet, with an orbital period of 76 years, we used this principle to find out more about its orbit. Given Earth's orbital period of one year as a benchmark, we compare Halley's longer period to deduce information about its larger orbit. This relationship allows astronomers to calculate either the period or the size of the orbit if one of the values is known. For students learning this, it’s key to remember Kepler's formula:
  • \[T^2 = k imes a^3\]
where \(k\) is the constant of proportionality for bodies orbiting the same sun.
Semi-Major Axis
The semi-major axis is a vital element in describing the size of an elliptical orbit. It is half of the longest diameter of the ellipse. This measurement helps us understand the scale and size of an orbit, which influences the orbital period of an object. For example, Halley's comet has a semi-major axis that is significantly larger than Earth's, explaining its much longer orbital period.

To find the semi-major axis of Halley's comet, we applied Kepler's Third Law. By knowing its period and the constant derived from Earth's orbit, we calculated that Halley's orbit is approximately 18 times that of Earth's mean distance from the sun. This was done by solving the equation:
  • \[a_h = (T_h^2)^{1/3} \times a_E\]
where \(T_h\) is the orbital period of Halley's Comet and \(a_E\) is the Earth's semi-major axis. The semi-major axis not only helps determine the period of orbit but is also crucial in calculating the farthest and nearest points an object reaches in its orbit, known as the aphelion and perihelion. Understanding this concept helps in predicting the behavior of celestial bodies throughout their orbits.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that the angular momentum per unit mass \(h\) of a satellite describing an elliptic orbit of semimajor axis \(a\) and eccentricity \(\varepsilon\) about a planet of mass \(M\) can be expressed as $$h=\sqrt{G M a\left(1-\varepsilon^{2}\right)}$$

To unload a bound stack of plywood from a truck, the driver first tilts the bed of the truck and then accelerates from rest. Knowing that the coefficients of friction between the bottom sheet of plywood and the bed are \(\mu_{s}=0.40\) and \(\mu_{k}=0.30\), determine (a) the smallest acceleration of the truck which will cause the stack of plywood to slide, ( \(b)\) the acceleration of the truck which causes corner \(A\) of the stack to reach the end of the bed in 0.9 s.

To transport a series of bundles of shingles \(A\) to a roof, a contractor uses a motor-driven lift consisting of a horizontal platform \(B C\) which rides on rails attached to the sides of a ladder. The lift starts from rest and initially moves with a constant acceleration \(a_{1}\) as shown. The lift then decelerates at a constant rate \(a_{2}\) and comes to rest at \(D,\) near the top of the ladder. Knowing that the coefficient of static friction between a bundle of shingles and the horizontal platform is 0.30 , determine the largest allowable acceleration \(\mathbf{a}_{1}\) and the largest allowable deceleration \(\mathbf{a}_{2}\) if the bundle is not to slide on the platform.

The Clementine spacecraft described an elliptic orbit of minimum altitude \(h_{A}=400 \mathrm{km}\) and maximum altitude \(h_{B}=2940 \mathrm{km}\) above the surface of the moon. Knowing that the radius of the moon is \(1737 \mathrm{km}\) and that the mass of the moon is 0.01230 times the mass of the earth, determine the periodic time of the spacecraft.

A small 250 -g collar \(C\) can slide on a semicircular rod which is made to rotate about the vertical \(A B\) at a constant rate of 7.5 rad/s. Determine the three values of \(\theta\) for which the collar will not slide on the rod, assuming no friction between the collar and the rod.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.