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Chapter 13: Problem 28

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. Pluto already was known to have a large satellite Charon, orbiting at 19,600 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
For the first satellite, \( T_1 \approx 15.8 \) days; for the second satellite, \( T_2 \approx 24.8 \) 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 satellite is directly proportional to the cube of the semi-major axis of its orbit. In this case, we assume circular orbits, hence semi-major axis equals orbital radius. The formula is \( T^2 \propto r^3 \).
02

Establish Proportionality Relationship

Using Charon as a reference, we have for any satellite \( \frac{T^2}{r^3} = \frac{T_{Charon}^2}{r_{Charon}^3} \). Substituting the known values for Charon: \( T_{Charon} = 6.39 \text{ days} \) and \( r_{Charon} = 19,600 \text{ km} \).
03

Calculate Proportionality Constant for Charon

Calculate the constant for Charon: \( \frac{6.39^2}{19600^3} \). This constant applies to the other satellites due to their similar nature being unaffected by each other.
04

Set Up Equation for Satellite 1

For the first satellite with \( r_1 = 48000 \text{ km} \), set up the equation: \( T_1^2 = r_1^3 \times \frac{6.39^2}{19600^3} \).
05

Solve for the Orbital Period of Satellite 1

Calculate \( T_1 \) by solving \( T_1 = \sqrt{48000^3 \times \frac{6.39^2}{19600^3}} \).
06

Set Up Equation for Satellite 2

For the second satellite with \( r_2 = 64000 \text{ km} \), set up the equation: \( T_2^2 = r_2^3 \times \frac{6.39^2}{19600^3} \).
07

Solve for the Orbital Period of Satellite 2

Calculate \( T_2 \) by solving \( T_2 = \sqrt{64000^3 \times \frac{6.39^2}{19600^3}} \).

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Key Concepts

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

Understanding Orbital Period
The orbital period of a satellite is the time it takes for the satellite to complete one full orbit around its planet or celestial body. For instance, the Moon orbits Earth, and its orbital period is about 27.3 days. This concept is crucial to understanding how celestial bodies interact with each other. When calculating the orbital period, we usually refer to Kepler's Third Law. According to this law, there is a mathematical relationship between the period and the semi-major axis of the orbit. To be precise, the square of the orbital period ( \( T^2 \) ) is proportional to the cube of the radius ( \( r^3 \) ) of its orbit. This relationship helps us calculate the periods of satellites like those orbiting Pluto without directly needing the planet's mass. As in the example from the exercise, knowing Charon's orbital period and radius allowed us to determine the periods of other satellites around Pluto using their respective radii. r
This makes it easier to understand the dynamics of celestial bodies using fundamental physics without extensive data.
Satellite Orbits and Their Properties
Satellite orbits describe the path satellites take around their celestial bodies. An orbit's shape can be circular or elliptical, but in the exercise, we assumed circular orbits around Pluto. In a simpler sense, this means the gravitational pull between Pluto and its satellites is just right to keep them in motion along a circular path. Each satellite orbit is unique and depends on several factors such as:
  • The mass of the satellite.
  • The velocity at which it travels.
  • The gravitational influence of Pluto.
The greater the distance from the planet, the longer the orbital period 鈥� this means that the orbital radius directly affects how long it takes for a satellite to complete an orbit. Understanding these basic properties of satellite orbits helps in calculating orbital periods efficiently, as demonstrated in the problem.
Proportionality Constant in Kepler's Third Law
In Kepler's Third Law, the proportionality constant is a key factor that allows us to compare different satellite orbits around the same celestial body.This constant ( \( k \) ) is determined using a reference satellite's known orbital period and radius. In the exercise, Charon was used to find this constant because we had its orbital data. The formula is derived from:\[ k = \frac{T_{Charon}^2}{r_{Charon}^3} \] Once the proportionality constant is calculated, it remains constant for all satellites orbiting the same body. This applies as long as gravitational interactions do not affect the satellites differently. Using this constant, we simplified the calculations for the new satellites' orbital periods. With known radii, we used the constant to find the orbital periods efficiently, showcasing how this principle aids calculations in astronomy.

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Most popular questions from this chapter

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.980 m/s\(^2\) if the acceleration due to gravity at the surface has magnitude 9.80 m/s\(^2\)?

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.)

The dwarf planet Pluto has an elliptical orbit with a semimajor axis of 5.91 \(\times\) 10\(^{12}\) m and eccentricity 0.249. (a) Calculate Pluto's orbital period. Express your answer in seconds and in earth years. (b) During Pluto's orbit around the sun, what are its closest and farthest distances from the sun?

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.0 \(\times\) 10\(^4\) m/s when at a distance of 2.5 \(\times\) 10\(^{11}\) m from the center of the sun, what is its speed when at a distance of 5.0 \(\times\) 10\(^{10}\) m?

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