Problem 43 - Single Variable Calculus

Question

Period of a Communications Satellite According to Kepler's Third Law, the period \(T\) (in days) of a satellite moving in a circular orbit \(x\) mi above the surface of the earth is given by $$ T=0.0588\left(1+\frac{x}{3959}\right)^{5 / 2} $$ Suppose that a communications satellite is moving in a circular orbit \(22,000 \mathrm{mi}\) above the earth's surface. Because of friction, the satellite drops down to a new orbit \(21,500 \mathrm{mi}\) above the earth's surface. Estimate the decrease in the period of the satellite to the nearest one- hundredth hour.

Step-by-Step Solution

Answer

The decrease in the period of the satellite is approximately \(0.55\) hours when it drops to a new orbit of 21,500 mi above the Earth's surface.

Step 1: Calculate the initial period of the satellite

To calculate the initial period of the satellite, we'll use Kepler's Third Law equation with \(x = 22,000\). \(T = 0.0588\left(1+\frac{22000}{3959}\right)^{5 / 2}\) Now, calculate the resulting value for \(T\).

Step 2: Calculate the new period of the satellite after it drops down

Next, calculate the new period of the satellite after it drops down to an orbit of 21,500 mi above the earth's surface. Substitute \(x = 21,500\) into Kepler's Third Law equation. \(T = 0.0588\left(1+\frac{21500}{3959}\right)^{5 / 2}\) Now, calculate the resulting value for \(T\).

Step 3: Find the decrease in the period of the satellite

To find the decrease in the period, subtract the new period (Step 2) from the initial period (Step 1). Decrease in period = Initial period - New period

Step 4: Convert the decrease in period to the nearest one-hundredth hour

Since the period is given in days, we need to convert the decrease in period to hours. To do that, we'll multiply the decrease in period by 24 (as there are 24 hours in a day). Decrease in period (in hours) = Decrease in period (in days) * 24 Then round the result to the nearest one-hundredth hour.

Key Concepts

satellite orbit

A satellite orbit refers to the path that a satellite follows as it revolves around a larger celestial body, such as the Earth. Satellites, like our planets, follow the laws of celestial mechanics. These orbits can be circular or elliptical in shape.
The altitude of a satellite orbit directly influences its characteristics:

• Higher orbits result in longer periods.
• Lower orbits allow for shorter periods and closer proximity to the Earth.

In the case of the communications satellite mentioned in the problem, it initially moves in a circular orbit 22,000 miles above the Earth. After encountering friction, it drops to a new orbit 21,500 miles above the Earth. Satellite orbits are crucial for ensuring that the satellite can accomplish its designated mission, whether that’s for communications, navigation, or observation.

period estimation

Period estimation in orbital mechanics involves calculating how long it takes a satellite to complete one full orbit around the Earth. This is often referred to as the orbital period.
In the given exercise, Kepler's Third Law is applied in a simplified form to estimate the period (\(T\)) of a satellite in a circular orbit. Kepler's Third Law relates the orbital period of a planet or satellite to its distance from the celestial body it’s orbiting. It is expressed as:

• \[ T = 0.0588\left(1+\frac{x}{3959}\right)^{3/2} \]

where \(x\) is the altitude of the satellite above the Earth's surface.
In practical terms, this formula helps us quickly estimate how frictional forces affecting a satellite’s orbit can change the period.
By plugging in different orbital heights, one can find how much the satellite’s period decreases due to a lower orbit.

orbital mechanics

Orbital mechanics, also known as celestial mechanics, involves the study of the motion of objects in space under the influence of gravitational forces. This field uses mathematical models to predict and understand the trajectories of satellites and other space objects.
Kepler's laws are fundamental in this field, particularly when calculating specific elements such as distances, periods, and velocity.

• Kepler's First Law: Describes the shape of an orbit as an ellipse with the celestial body at one of its foci.
• Kepler's Second Law: States that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
• Kepler's Third Law: The square of an orbital period is proportional to the cube of the semi-major axis of its orbit.

These laws enable scientists to predict satellite positions, necessary for various applications such as GPS technology, television satellites, and other communication devices. Understanding these principles is essential to solve real-world engineering problems related to satellite design and operation.