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Chapter 12: Problem 32

Planet Vulcan. Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to \(\frac{2}{3}\) of the average orbit radius of Mercury. What would be the orbital period of such a planet? (Such a planet was once postulated, in part to explain the precession of Mercury's orbit. It was even given the name Vulcan, although we now have no evidence that it actually exists. Mercury's precession has been explained by general relativity.)

Short Answer

Expert verified
Vulcan's orbital period would be approximately 48 days.

Step by step solution

01

Identify Known Values

We know that Mercury's average orbital radius is approximately \( R_M \) and its orbital period is \( T_M = 88 \) days. The planet Vulcan would have an orbital radius \( R_V = \frac{2}{3}R_M \).
02

Apply Kepler's Third Law of Planetary Motion

Kepler's Third Law states \( \frac{T^2}{R^3} = \text{constant} \). For both Mercury and Vulcan, this gives us \( \frac{T_M^2}{R_M^3} = \frac{T_V^2}{R_V^3} \).
03

Substitute Known Values Into the Equation

Substitute \( R_V = \frac{2}{3}R_M \) into the equation: \[ \frac{T_M^2}{R_M^3} = \frac{T_V^2}{(\frac{2}{3}R_M)^3} \]. This simplifies to \( \frac{T_M^2}{R_M^3} = \frac{T_V^2}{\frac{8}{27} R_M^3} \).
04

Solve for Vulcan's Orbital Period \( T_V \)

Solving for \( T_V \), we find: \( T_V^2 = T_M^2 \cdot \frac{8}{27} \). Taking the square root, we obtain \( T_V = T_M \cdot \sqrt{\frac{8}{27}} \).
05

Calculate \( T_V \)

Substitute \( T_M = 88 \) days into the equation: \( T_V = 88 \times \sqrt{\frac{8}{27}} \). Calculate \( \sqrt{\frac{8}{27}} \approx 0.545 \). Thus, \( T_V \approx 88 \times 0.545 \approx 47.96 \) days.

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Key Concepts

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

Planetary Motion
Planetary motion refers to the way planets travel around the sun or any other star. This motion is governed by laws of physics, most notably, Kepler鈥檚 laws of planetary motion. These laws describe how the planets move in their orbits.
  • First Law (The Law of Orbits): States that planets move in elliptical orbits with the sun at one focus.
  • Second Law (The Law of Areas): Explains that a line joining a planet to the sun sweeps out equal areas in equal times.
  • Third Law (The Law of Periods): Connects the time it takes for a planet to orbit the sun with the size of its orbit. Specifically, it states that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit (\( \frac{T^2}{R^3} = \text{constant} \)).
This third law helps us understand how the orbital period changes with the distance from the star. In our exercise, it was used to determine the orbital period of the hypothetical Planet Vulcan.
Orbital Period
The orbital period is the time a planet takes to complete one full orbit around a star. For the planets in our solar system, this is measured in Earth days or years. It is a crucial aspect of planetary motion and is calculated using Kepler's Third Law.
Kepler's Third Law states that the orbital period (\(T\)) is related to the size of the orbit (\(R\)). Specifically, \(T^2 \propto R^3\). This means:
  • If the orbital radius doubles, the period increases by a factor of about 2.828 (since \((2^3)^{0.5} = 2.828\)).
  • Therefore, a smaller orbit means a shorter period, and vice versa.
In the case of the imaginary Planet Vulcan, the orbit was smaller than Mercury's, so its orbital period was shorter.
Precession of Mercury
The precession of Mercury refers to the slow rotation of the orbit of Mercury around the sun, which is different from the more common motion where planets stay roughly in the same orbit. Mercury's orbit is not fixed; it rotates or 'precesses' over time.
In the past, scientists puzzled over why Mercury鈥檚 orbit behaved in this way. The belief was that another planet, theorized as Vulcan, might exist within Mercury's orbit, affecting it through gravity. However, there was never any evidence of Vulcan's existence.
The mystery behind the precession was ultimately solved by Einstein's theory of General Relativity. The theory suggested that the effect arises due to the warping of spacetime around the sun. This explained the precession accurately without needing another planet.

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Most popular questions from this chapter

Titania, the largest moon of the planet Uranus, has \(\frac{1}{8}\) the radius of the earth and \(\frac{1}{100}\) the mass of the earth. (a) What is the acceleration due to gravity at the surface of Titania? (b) What is the average density of Titania? (This is less than the density of rock, which is one piece of evidence that Titania is made primarily of ice. \()\)

An \(8.00-\mathrm{kg}\) point mass and a \(15.0-\mathrm{kg}\) point mass are held in place 50.0 \(\mathrm{cm}\) apart. A particle of mass \(m\) is released from a point between the two masses 20.0 \(\mathrm{cm}\) from the \(8.00-\mathrm{kg}\) mass along the line connecting the two fixed masses. Find the magnitude and direction of the acceleration of the particle.

Binary Star-Different Masses. Two stars, with masses \(M_{1}\) and \(M_{2},\) are in circular orbits around their center of mass. The star with mass \(M_{1}\) has an orbit of radius \(R_{1}\) ; the star with mass \(M_{2}\) has an orbit of radius \(R_{2}\) . (a) Show that the ratio of the orbital radii of the two stars equals the reciprocal of the ratio of their masses, that is, \(R_{1} / R_{2}=M_{2} | M_{1},(b)\) Explain why the two stars have the same orbital period, and show that the period \(T\) is given by \(T=2 \pi\left(R_{1}+R_{2}\right)^{3 / 2} / \sqrt{G}\left(M_{1}+M_{2}\right) .\) (c) The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36.0 \(\mathrm{km} / \mathrm{s}\) . The second star, Beta, has an orbital speed of 12.0 \(\mathrm{km} / \mathrm{s}\) . The orbital period is 137 \(\mathrm{d}\) . What are the masses of each of the two stars? (d) One of the best candidates for a black hole is found in the binary system called A0620-0090. The two objects in the binary system are an orange star, V616 Monocerotis, and a compact object believed to be a black hole (Fig. \(12,22\) ). The orbital period of \(A 0620-0090\) is 7.75 hours, the mass of \(\mathrm{V} 616\) Monocerotis is estimated to be 0.67 times the mass of the sun, and the mass of the black hole is estimated to be 3.8 times the mass of the sun. Assuming that the orbits are circular, find the radius of each object's ortbit and the orbital speed of each object. Compare these answers to the orbital radius and orbital speed of the earth in its orbit around the sun.

Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is 400 \(\mathrm{km}\) above the earth's surface; at the high point, or apogee, it is 4000 \(\mathrm{km}\) above the earth's surface. (a) What is the period of the spacecraft's orbit the earth's surface. (a) What is the period of the spacecraft's orbit? (b) Using conservation of angular momentum, find the ratio of the spacecraft's speed at perigee to its speed at apogee. (c) Using conservation of energy, find the speed at perigee and the speed at apogee. (d) It is necessary to have the spacecraft escape from the earth completely. If the spacecraft's rockets are fired at perigee, by how much would the speed have to be increased to achieve this? What if the rockets were fired at apogee? Which point in the orbit is more efficient to use?

Binary Star-Equal Masses. Two identical stars with mass \(M\) orbit around their center of mass. Each orbit is circular and has radius \(R\) , so that the two stars are always on opposite sides of the circle. (a) Find the gravitational force of one star on the other. (b) Find the orbital speed of each star and the period of the orbit. (c) How much energy would be required to separate the two stars to infinity?
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