Problem 26 - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

The orbital periods for the satellites at 48,000 km and 64,000 km are approximately 12.3 days and 23.3 days, respectively.

Step 1: Understand the Problem

We are given three satellites orbiting Pluto. We need to find the orbital periods of the two newly discovered satellites, given the distance of their orbits. We already know the orbital distance and period of Pluto's large satellite, Charon.

Step 2: Use Kepler's Third Law

Kepler's Third Law asserts that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit:\[ T^2 = k imes a^3 \]where \(T\) is the orbital period, \(a\) is the semi-major axis (orbit radius), and \(k\) is a constant. Since we are considering objects orbiting the same primary body (Pluto), \(k\) remains constant. We will use this relation to infer the periods of the other satellites.

Step 3: Find k using Charon's information

Charon's semi-major axis \(a_C = 19,600 \, \text{km}\) and its orbital period \(T_C = 6.39 \, \text{days}\). So,\[ k = \frac{T_C^2}{a_C^3} \]Substitute \(T_C = 6.39\) days and \(a_C = 19,600\) km into the equation:\[ k = \frac{6.39^2}{(19600)^3} \]

Step 4: Calculate the period of the 48,000 km satellite

For the satellite at \(a_1 = 48,000 \, \text{km}\), we use the same value of \(k\):\[ T_1^2 = k \times a_1^3 \]Thus,\[ T_1 = \sqrt{k imes (48000)^3} \]Substitute the value of \(k\) obtained in Step 3 and calculate \(T_1\).

Step 5: Calculate the period of the 64,000 km satellite

Similar to the previous satellite, for the satellite at \(a_2 = 64,000 \, \text{km}\),\[ T_2^2 = k \times a_2^3 \]And,\[ T_2 = \sqrt{k \times (64000)^3} \]Substitute the value of \(k\) found earlier and compute \(T_2\).

Step 6: Solve for unknowns and interpret results

After substituting the value of \(k\) and solving for \(T_1\) and \(T_2\), simplify calculations to get the periods of the satellites in days. This step also checks consistency with Kepler's law and verifies that the periods correspond realistically to our initial conditions.

Key Concepts

Orbital Period

The orbital period of a satellite is the duration it takes to complete one full orbit around a celestial body. This is typically measured in days or years.
Understanding the orbital period is crucial because it gives insight into how fast or slow a satellite moves relative to its host planet or star.
To calculate the orbital period, we can use Kepler's Third Law, which relates the period squared to the semi-major axis of the orbit cubed:

• The formula: \( T^2 = k \times a^3 \)
• \(T\) is the orbital period
• \(a\) is the semi-major axis or the mean orbit distance
• \(k\) is a constant for all satellites of the same primary body

This means, if we know the semi-major axis and the constant, we can find the orbital period.
By comparing two satellites orbiting the same planet, as with Pluto's satellites, we use Charon's orbital data to find \(k\), then apply it to determine the periods of the newly discovered satellites.
My students often find it helpful to visualize this as a balance between a satellite's velocity and the gravitational pull balanced by the planet's distance, leading to a specific period.

Semi-Major Axis

The semi-major axis of an orbit is essentially half of the longest diameter of an elliptical orbit.
In simple terms, it's the average distance from the center of the orbit to the orbiting satellite.
This axis is central to understanding an orbit’s shape and size. When talking about circular orbits, the semi-major axis is just the radius.
For elliptical orbits, which are more common, it's a crucial part of Kepler's Third Law, as it influences the orbital period significantly. Consider the following aspects of the semi-major axis:

• Defining orbit size: The larger the semi-major axis, the larger the orbit.
• Relation to energy: A satellite with a larger semi-major axis has more orbital energy.

In context with the exercise, knowing the semi-major axes of Pluto's satellites allows us to use Kepler's Law efficiently.
It tasks us with comparing distances of the known satellite, Charon, to the others and finding the orbital periods based on proportional distances.

Natural Satellites

Natural satellites are bodies that orbit planets, dwarf planets or other non-stellar astronomical objects naturally.
They differ from artificial satellites, which are human-made and launched into orbit for various purposes.
In the case of Pluto, Charon is its largest known moon, followed by smaller moons like Hydra and Nix.
These satellites exhibit fascinating dynamics and help astronomers understand the gravitational influences and history of the celestial bodies they orbit. Key features include:

• Existence in stable orbits, thanks to gravitational forces.
• Rich history: Natural satellites often provide clues about their host planets’ formation and their own origin.

The complexity and diversity of natural satellites in our solar system remind us of the intricate gravitational tango celestial bodies engage in.
The locations of Pluto's new satellites, found further out than Charon, highlight how variances in distance can affect orbital periods and dynamics.
These discoveries aid in our understanding of how natural satellites interact with each other and their primary bodies.