Problem 17 - Thomas Calculus

Question

Mass of Jupiter Io is one of the moons of Jupiter. It has a semi major axis of \(0.042 \times 10^{10} \mathrm{m}\) and an orbital period of 1.769 days. Use these data to estimate the mass of Jupiter.

Step-by-Step Solution

Answer

Jupiter's estimated mass is approximately \(1.898 \times 10^{27} \text{ kg}\).

Step 1: Convert units

First, we need to convert the orbital period from days to seconds because we will use it in the formula involving gravitational force, where time is measured in seconds.1 day = 24 hours, 1 hour = 60 minutes, 1 minute = 60 seconds. So,\(1.769 \text{ days} = 1.769 \times 24 \times 60 \times 60 \text{ seconds} \)Calculate this multiplication to get the period in seconds.

Step 2: Apply Kepler's Third Law

Kepler's Third Law can be written in terms of the masses and radius of orbits as \(T^2 = \frac{4 \pi^2}{G} \cdot \frac{a^3}{M_{Jupiter}}\), where \(T\) is the period, \(a\) is the semi-major axis of the orbit, and \(G\) is the gravitational constant \(6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2\). Rearrange this formula to solve for the mass of Jupiter, \(M_{Jupiter}\):\(M_{Jupiter} = \frac{4 \pi^2 a^3}{G T^2}\).

Step 3: Calculate the mass of Jupiter

Substitute the known values into the rearranged formula: \(a = 0.042 \times 10^{10} \text{ m}\) and \(T\) in seconds from Step 1. Calculate the cube of the semi-major axis \(a^3\) and the square of the period \(T^2\). Then, use these to find \(M_{Jupiter}\):\[M_{Jupiter} = \frac{4 \pi^2 (0.042 \times 10^{10})^3}{6.674 \times 10^{-11} (1.769 \times 24 \times 60 \times 60)^2}\]

Step 4: Evaluate the expression

Use a calculator to evaluate the expression from Step 3. Take care to correctly execute the operations one step at a time, respecting the order of operations found within nested calculations (e.g., calculate exponents first, then multiplications and divisions). This calculation will give you the approximate mass of Jupiter in kilograms.

Key Concepts

Orbital Period Conversion

Converting the orbital period from days to seconds is crucial because many formulas in physics, especially those dealing with gravitational forces, use seconds as the standard unit of time. To convert from days to seconds:

• First, recognize that 1 day equals 24 hours.
• One hour equals 60 minutes.
• Each minute consists of 60 seconds.

To find the total number of seconds in 1.769 days, we perform the multiplication:\[1.769 \times 24 \times 60 \times 60\]This will give us the orbital period in seconds, ensuring compatibility with the gravitational constant, which is also typically expressed in seconds. This conversion forms an essential first step in applying Kepler's Third Law effectively for calculations involving planetary dynamics.

Mass of Jupiter

Kepler’s Third Law provides a relationship between the orbital period and the semi-major axis of an orbit. Specifically, it allows us to find the mass of a central astronomical object, such as Jupiter in this instance. In a simplified form, Kepler’s Third Law can be expressed as:\[T^2 = \frac{4 \pi^2}{G} \cdot \frac{a^3}{M}\]where

• \(T\) is the orbital period in seconds.
• \(a\) is the semi-major axis.
• \(G\) is the gravitational constant, approximately \(6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2\).
• \(M\) is the mass of the central object (Jupiter in this exercise).

To determine the mass of Jupiter, rearrange the equation to solve for \(M\):\[M_{Jupiter} = \frac{4 \pi^2 a^3}{G T^2}\]By substituting the values we obtain for \(a\) and \(T\) after conversion, you can calculate Jupiter's mass. It’s essential to handle this equation with care, especially when dealing with the large values associated with celestial mechanics.

Semi-major Axis Calculation

The semi-major axis is a vital part of understanding planetary orbits. It represents the longest radius of an elliptical orbit, like those planets and moons often have. In this problem, Io’s orbit around Jupiter benefits from knowing that its semi-major axis is provided in meters:
\(a = 0.042 \times 10^{10} \text{ m}\).
To correctly use this value in the Kepler’s Third Law formula, you need to cube it:\[a^3 = (0.042 \times 10^{10})^3\]This cubing accounts for the three-dimensional space the orbiting body traverses over each complete orbit.

Successful calculation of this value is crucial since it factor significantly into determining the mass of Jupiter. The interplay between the orbital characteristics of Io and Jupiter’s mass showcases the fascinating complexities of orbital mechanics.