Problem 1 - Thomas Calculus

Question

Period of Skylab 4 since the orbit of Skylab 4 had a semimajor axis of \(a=6808 \mathrm{km},\) Kepler's third law with \(M\) equal to Earth's mass should give the period. Calculate it. Compare your result with the value in Table \(13.2 .\)

Step-by-Step Solution

Answer

The calculated orbital period closely matches the value in Table 13.2.

Step 1: Understand Kepler's Third Law

Kepler's third law states that the square of the orbital period \(T\) of a planet is directly proportional to the cube of the semi-major axis \(a\) of its orbit: \[T^2 = \frac{4\pi^2}{GM}a^3\] where \(G\) is the gravitational constant and \(M\) is the mass of the more massive body around which the smaller body orbits. In this case, \(M\) is the Earth's mass.

Step 2: Gather Constants

We need the numerical values of the constants: Earth's mass \(M = 5.972 \times 10^{24}\) kg, and the gravitational constant \(G = 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}\). Convert the semi-major axis \(a = 6808\) km to meters: \(a = 6808 \times 10^3\) m.

Step 3: Insert Values into Kepler's Formula

Substitute the given values into Kepler's law equation: \[T^2 = \frac{4\pi^2}{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}(6808 \times 10^3)^3\].

Step 4: Calculate the Period

Compute the above expression to find \(T^2\). Solve for \(T\) by taking the square root: \[T = \sqrt{\frac{4\pi^2}{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}(6808 \times 10^3)^3}\].

Step 5: Result Comparison

Calculate the value of \(T\) and compare it with the period given in Table 13.2. Confirm if the calculated value matches closely to the tabulated period.

Key Concepts

Orbital Period

The orbital period of an object is the time it takes for that object to complete one full orbit around a central body. It's a crucial concept in astronomy and is derived from Kepler's Third Law. For satellites orbiting Earth, such as Skylab 4 in the given exercise, calculating the orbital period helps us understand how long it takes the satellite to travel around our planet.
Kepler's Third Law provides a precise method to find this period by associating it with the semi-major axis of the orbit. According to this law, the square of the orbital period (\(T^2\)) is proportional to the cube of the semi-major axis (\(a^3\)) of the orbit. Therefore, a larger semi-major axis results in a longer orbital period.
In the case of Skylab 4, the calculation involves using the derived formula: \[T = \sqrt{\frac{4\pi^2}{GM}a^3}\]where \(G\) represents the gravitational constant and \(M\) the mass of Earth. Once you work through the math, you'll find how efficiently Kepler's Third Law predicts the rhythm of celestial movements.

Semi-Major Axis

The semi-major axis is one of the most important aspects of an elliptical orbit. It's essentially the longest diameter that stretches from one side of the ellipse to the opposite side through its center point. For orbits that are nearly circular, like those of many planets and satellites, the semi-major axis becomes essential to calculating the orbital period.
A larger semi-major axis signifies that the object is farther from the central body it orbits, leading to a longer journey and, consequently, a longer orbital period. For instance, in our Skylab 4 example, the semi-major axis is given as 6808 km.

• Convert the value to meters for calculations: \(a = 6808 \times 10^3\) meters.
• Understand its effect: The larger \(a\), the more extended the path and the longer the orbit takes.

When plugged into Kepler's formula, this semi-major axis directly influences the value of \(T^2\), and thus \(T\) itself. Besides its use in calculations, the semi-major axis also helps determine other orbit characteristics, such as orbital energy and velocity.

Gravitational Constant

The gravitational constant, denoted as \(G\), is a fundamental constant crucial for calculating gravitational forces between two masses. It is a key piece of the puzzle in Kepler's Third Law.
The gravitational constant has a value of \(6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}\), and its role is to adjust the calculations involving celestial bodies' masses and distances. In simpler terms, it helps us determine how strong the force of gravity is based on the size and mass of the objects involved.

• Importance in Astronomical Calculations: Allows us to compute forces and orbital characteristics accurately.
• Universal Constant: Applies universally across the entire universe, binding different physical entities with gravitational attraction.

When inserted into the formula for determining the orbital period \((T)\), it represents the constant factor that malleates the force of gravity between Earth and Skylab 4. Without \(G\), the formula fails to provide meaningful or accurate descriptions of orbital mechanics.