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Chapter 13: Problem 32

In March \(2006,\) two small satellites were discovered orbiting Pluto, one at a distance of \(48,000 \mathrm{~km}\) and the other at \(64,000 \mathrm{~km}\). Pluto already was known to have a large satellite Charon, orbiting at \(19,600 \mathrm{~km}\) with an orbital period of 6.39 days. Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites without using the mass of Pluto.

Short Answer

Expert verified
The orbital period of the first small satellite (48,000 km away from Pluto) is approximately 24.56 days, and the orbital period of the second small satellite (64,000 km away from Pluto) is approximately 39.22 days.

Step by step solution

01

Understand Kepler's Third Law

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, this can be written as \(T^2 \propto r^3\), where T is the period and r is the radius (or semi-major axis). We are to find the periods of the two small satellites, given the period and the radius of Charon's orbit around Pluto.
02

Formulate the ratios

Given that the periods of orbit and semi-major axis lengths are proportional, we can set up a ratio comparing the known satellite (Charon) and each of the unknown satellites in turn. This will look like this: \(\frac{T_{Charon}^2}{r_{Charon}^3} = \frac{T_{SmallSatellite}^2}{r_{SmallSatellite}^3}\). We know the values for Charon and we also know the semi-major axis lengths/radius distances for the smaller satellites.
03

Solve for periods of the smaller satellites

Rearranging the above equation gives: \(T_{SmallSatellite} = \sqrt{\frac{T_{Charon}^2 * r_{SmallSatellite}^3}{r_{Charon}^3}}\). First calculate for the satellite at 48000 km: \(T_{SmallSatellite1} = \sqrt{\frac{(6.39days)^2 * (48000km)^3}{(19600km)^3}}\), which simplifies to approximately 24.56 days. The same method is used for the satellite at 64000 km: \(T_{SmallSatellite2} = \sqrt{\frac{(6.39days)^2 * (64000km)^3}{(19600km)^3}}\), which simplifies to approximately 39.22 days.

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Key Concepts

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

Orbital Period
The orbital period of a celestial object, such as a planet or moon, is the time it takes to complete one full orbit around another body. For instance, Earth's orbital period around the Sun is approximately 365.25 days, which defines one year. In the example of the two satellites orbiting Pluto, we utilized the known orbital period of Charon to determine the periods of the newly discovered satellites.

We employed Kepler's Third Law, which builds a relation between the orbital period and the distance from the central body, to complete our calculation. By understanding this law, we were able to predict the durations of the satellites' orbits without needing the mass of Pluto. This principle is critical as it simplifies the complex task of determining the orbital characteristics of celestial bodies by using simple proportional relationships, a corner stone in the study of astronomy.
Semi-Major Axis
The semi-major axis is essentially half of the longest diameter of an elliptical orbit. This value is crucial when describing orbital shapes and sizes, and it plays a pivotal role in determining the orbital period of an object according to Kepler's Third Law. In an orbit, the semi-major axis is the average distance between the orbiting object and the central object, providing a practical way to express the size of the orbit.

In our Pluto's satellites problem, the radius given can be considered as the semi-major axis since orbits are typically not perfectly circular. Using the distances provided (48,000 km and 64,000 km), along with the well-known distance for Charon (19,600 km), we implemented Kepler's law to uncover the unknown orbital periods, demonstrating the practical application of the semi-major axis in orbital mechanics.
Proportional Relationships in Astronomy
Astronomy often makes use of proportional relationships to relate various celestial phenomena, which makes calculations regarding the vast expanse of space more manageable. Kepler's laws, specifically the third law, provide one of the most clear-cut examples of these relationships, directly relating the cube of the semi-major axis to the square of the orbital period.

This principle allows astronomers to predict and compare the behavior of objects within a system relative to each other, as demonstrated in our Pluto example. By comparing the known ratio from Charon's orbit to those of the smaller satellites, we were able to determine the orbital periods without the need for complex gravitational equations. These proportional relationships form the bedrock of classical astronomy, enabling scientists to map the heavens with remarkable precision well before the age of modern telescopes and spacecraft.

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Most popular questions from this chapter

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?

A planet orbiting a distant star has radius \(3.24 \times 10^{6} \mathrm{~m}\). The escape speed for an object launched from this planet's surface is \(7.65 \times 10^{3} \mathrm{~m} / \mathrm{s}\). What is the acceleration due to gravity at the surface of the planet?

A uniform, solid, \(1000.0 \mathrm{~kg}\) sphere has a radius of \(5.00 \mathrm{~m}\). (a) Find the gravitational force this sphere exerts on a \(2.00 \mathrm{~kg}\) point mass placed at the following distances from the center of the sphere: (i) \(5.01 \mathrm{~m}\), (ii) \(2.50 \mathrm{~m}\). (b) Sketch a qualitative graph of the magnitude of the gravitational force this sphere exerts on a point mass \(m\) as a function of the distance \(r\) of \(m\) from the center of the sphere. Include the region from \(r=0\) to \(r \rightarrow \infty\)

Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chicf scientist-engineer, you make the following measurements: A \(2.50 \mathrm{~kg}\) stone thrown upward from the ground at \(12.0 \mathrm{~m} / \mathrm{s}\) returns to the ground in \(4.80 \mathrm{~s} ;\) the circumference of Mongo at the equator is \(2.00 \times 10^{5} \mathrm{~km} ;\) and there is no appreciable atmosphere on Mongo. The starship commander, Captain Confusion, asks for the following information: (a) What is the mass of Mongo? (b) If the Aimless Wanderer goes into a circular orbit \(30,000 \mathrm{~km}\) above the surface of Mongo. how many hours will it take the ship to complete one orbit?

(a) Calculate how much work is required to launch a spacecraft of mass \(m\) from the surface of the earth (mass \(m_{E}\), radius \(R_{\mathrm{E}}\) ) and place it in a circular low earth orbit - that is, an orbit whose altitude above the earth's surface is much less than \(R_{\mathrm{E}}\). (As an example, the International Space Station is in low earth orbit at an altitude of about \(400 \mathrm{~km},\) much less than \(R_{\mathrm{E}}=6370 \mathrm{~km} .\) ) Ignore the kinetic energy that the spacecraft has on the ground due to the earth's rotation. (b) Calculate the minimum amount of additional work required to move the spacecraft from low earth orbit to a very great distance from the earth. Ignore the gravitational effects of the sun, the moon, and the other planets. (c) Justify the statement "In terms of energy, low earth orbit is halfway to the edge of the universe."
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