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Chapter 5: Problem 14

For communications satellites to work properly, they should appear from the surface of the Earth to remain stationary. That is, they should orbit the Earth exactly once each day. For any satellite, the period \(P\) (the length of time required to complete an orbit) is determined by its mean distance \(A\) from the center of the Earth. For a satellite of negligible mass, \(P\) and \(A\) are related by a power function \(A=c P^{2 / 3}\). a. The moon is 239,000 miles from the center of the Earth and has a period of about 28 days. How high above the center of the Earth should a geostationary satellite be? (Hint: You want the distance \(A\) for a satellite with period \(\frac{1}{28}\) that of the moon. The homogeneity property of power functions is applicable.) b. The radius of the Earth is about 3963 miles. How high above the surface of the Earth should a geostationary satellite be? \({ }^{14}\)

Short Answer

Expert verified
A geostationary satellite should be approximately 22,245 miles above Earth's surface.

Step by step solution

01

Understand the Relationship

We are given a power function that relates the mean distance \(A\) from the center of the Earth to the period \(P\) of the satellite: \(A = cP^{2/3}\). This implies that if we know the period of two satellites, we can find the constant \(c\) and then use it to find other distances given a new period.
02

Find the Constant 'c' Using the Moon's Values

We know that the moon's distance \(A_m = 239,000\) miles and its period \(P_m = 28\) days. We can solve for the constant \(c\) using these values: \(239,000 = c(28)^{2/3}\). We compute \((28)^{2/3}\) and then solve for \(c\):\[c = \frac{239,000}{28^{2/3}}.\]
03

Calculate 'A' for Geostationary Satellite

A geostationary satellite needs to orbit the Earth once per day, so its period \(P_s = 1\) day. We use the previously found constant \(c\) to find its distance \(A_s\): \[A_s = c(1)^{2/3} = c.\]Since \(c = \frac{239,000}{28^{2/3}}\), substitute it in:\[A_s = \frac{239,000}{28^{2/3}}.\]
04

Convert Distance Above Earth's Surface

The mean distance \(A_s\) we found is the distance from the center of the Earth, not the surface. The Earth's radius is given as 3,963 miles. We subtract it from \(A_s\) to find the satellite's height above Earth's surface:\[\text{Height above surface} = A_s - 3,963.\]
05

Solve for Exact Values

Calculate the value:1. Compute \((28)^{2/3}\).2. Calculate \(c = \frac{239,000}{28^{2/3}}\).3. With \(c\) known, \(A_s = c\).4. Finally compute \(A_s - 3,963\) to find the height above Earth's surface.

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Key Concepts

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

Orbital Dynamics
Understanding the concept of orbital dynamics is crucial for determining where a satellite should be placed in relation to Earth. Essentially, it involves the study of how objects move in space under the influence of gravity and other forces. When we think about satellites, especially geostationary ones, these are objects that need to maintain a consistent position relative to a point on the Earth's surface.
In simple terms, this means they orbit in synchronization with the Earth's rotation. A geostationary satellite revolves around the Earth once a day, matching the Earth's rotating period so that it appears motionless to an observer on the ground. This synchronization allows these types of satellites to consistently monitor or communicate with the same area of the Earth's surface.
We'll use principles of physics, including the effects of gravitational pull and the required velocity for achieving a stable orbit, to determine precise orbital altitude. These principles ensure that the satellite stays in its intended path without drifting.
Power Function
A power function is a mathematical expression where one variable changes in proportion to a power of another. In this case, the mean distance from the Earth's center, denoted as \( A \), is related to the satellite period, \( P \), by the formula \( A = cP^{2/3} \). This particular function helps us deduce the satellite's orbital path based on its revolution time around the Earth.
This power function gives us the ability to calculate the mean distance \( A \) by knowing the period \( P \) and a constant \( c \). In satellite terms, this relationship allows us to find how far away from the Earth's center a satellite needs to be to achieve a specific orbital period, thereby ensuring the satellite stays in the correct orbit for its intended purpose.
Homogeneity Property of Power Functions
The homogeneity property of power functions states that these functions maintain proportional relationships even when scaled. This is particularly useful when we want to compare distances and periods between different satellites, such as between the moon and a geostationary satellite.
By using the homogeneity property, if we scale the period of one object (say the moon), we can predict the scale required for the distance of another object with a different period. For example, with the moon having an orbital period of 28 days and geostationary satellites needing just 1 day, we can use this property to calculate how these differences in period scale affect their respective distances from the Earth's center.
This concept simplifies complex calculations, giving us a scalable approach to understanding varying distances and periods involved in satellite orbital dynamics.
Mean Distance from Earth's Center
Mean distance from the Earth's center \( A \) is an important measurement for satellites because it represents how far a satellite is from the point of gravitational attraction, which is the center of the Earth. The shorter or longer this mean distance, the faster or slower a satellite needs to be to maintain its orbit while balancing the gravity.
In geostationary satellite calculations, we start with the mean distance from the Earth's center and subtract the Earth's radius to get the distance above the surface. Thus, knowing the mean distance helps in placing the satellite at the optimal height for consistent communication or observation.
  • First, calculate the mean distance \( A \) using the power function relation \( A = cP^{2/3} \).
  • Next, subtract the Earth's radius (3963 miles in this case) to determine the satellite's actual altitude above sea level.
By understanding the mean distance, we ensure accurate satellite positioning for maximum efficiency and minimal energy use.

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Most popular questions from this chapter

The table on the following page shows the number, in millions, graduating from high school in the United States for selected years. \({ }^{48}\) $$ \begin{array}{|c|c|} \hline \text { Year } & \text { Number } \\ \hline 1986 & 2.79 \\ \hline 1987 & 2.65 \\ \hline 1989 & 2.45 \\ \hline 1990 & 2.36 \\ \hline 1991 & 2.28 \\ \hline 1992 & 2.40 \\ \hline 1994 & 2.52 \\ \hline 1996 & 2.66 \\ \hline 1998 & 2.80 \\ \hline \end{array} $$ a. Make a plot of the data and explain why a linear model is not appropriate. b. Use regression to find a linear model for the years 1986 through 1991 . (In this part and the next, round regression line parameters to three decimal places.) c. Use regression to find a linear model for the years 1992 through \(1998 .\) d. Write a formula for a model of the number, in millions, graduating as a piecewise-defined function using the linear models from part \(b\) and part c. e. Make a graph of the formula you found in part \(\mathrm{d}\). f. The number graduating in 1993 was \(2.34 \mathrm{mil}-\) lion. On the basis of your graph in part e, determine whether this is more or less than would be expected from your formula.

In southwest Georgia, the average pine pulpwood prices vary predictably over the course of the year, primarily because of weather. From 1993 through 1997, prices followed a similar pattern. In the first quarter of each year, the average price \(P\) was \(\$ 18.50\) per ton. It decreased at a steady rate to \(\$ 14\) in the second quarter and then increased at a steady rate up to \(\$ 18\) by the fourth quarter. \({ }^{51}\) a. Sketch a graph of pulpwood prices as a function of the quarter in the year. b. What type of function is \(P\) from the first to the second quarter? c. What formula for price \(P\) as a function of \(t\), the quarter, describes the price from the first to the second quarter? d. What type of function is \(P\) from the second to the fourth quarter? e. What formula for price \(P\) as a function of \(t\), the quarter, describes the price from the second to the fourth quarter? f. Write a formula for price \(P\) throughout the year as a piecewise-defined function of \(t\), the quarter.

By comparing the surface area of a sphere with its volume and assuming that air resistance is proportional to the square of velocity, it is possible to make a heuristic argument to support the following premise: For similarly shaped objects, terminal velocity varies in proportion to the square root of length. Expressed in a formula, this is $$ T=k L^{0.5}, $$ where \(L\) is length, \(T\) is terminal velocity, and \(k\) is a constant that depends on shape, among other things. This relation can be used to help explain why small mammals easily survive falls that would seriously injure or kill a human. a. A 6-foot man is 36 times as long as a 2-inch mouse (neglecting the tail). How does the terminal velocity of a man compare with that of a mouse? b. If the 6-foot man has a terminal velocity of 120 miles per hour, what is the terminal velocity of the 2 -inch mouse? c. Neglecting the tail, a squirrel is about 7 inches long. Again assuming that a 6-foot man has a terminal velocity of 120 miles per hour, what is the terminal velocity of a squirrel?

Physiologists have discovered that steady-state oxygen consumption (measured per unit of mass) in a running animal increases linearly with increasing velocity. The slope of this line is called the cost of transport of the animal, since it measures the energy required to move a unit mass 1 unit of distance. Table \(5.10\) gives the weight \(W\), in grams, and the cost of transport \(C\), in milliliters of oxygen per gram per kilometer, of seven animals. \({ }^{35}\) a. Judging on the basis of the table, does the cost of transport generally increase or decrease with increasing weight? Are there any exceptions to this trend? b. Make a plot of \(\ln C\) against \(\ln W\). c. Find a formula for the regression line of \(\ln C\) against \(\ln W\), and add this line to the plot you found in part \(b\). d. The cost of transport for a 20,790-gram emperor penguin is about \(0.43\) milliliter of oxygen per gram per kilometer. Use your plot in part c to compare this with the trend for cost of transport versus weight in the table. Does this confirm the stereotype of penguins as awkward waddlers? e. Find a formula that models \(C\) as a power function of \(W\). $$ \begin{array}{|l|c|c|} \hline \text { Animal } & \text { Weight } W & \begin{array}{c} \text { Cost of } \\ \text { transport } C \end{array} \\ \hline \text { White mouse } & 21 & 2.83 \\ \hline \text { Kangaroo rat } & 41 & 2.01 \\ \hline \text { Kangaroo rat } & 100 & 1.13 \\ \hline \text { Ground squirrel } & 236 & 0.66 \\ \hline \text { White rat } & 384 & 1.09 \\ \hline \text { Dog } & 2600 & 0.34 \\ \hline \text { Dog } & 18,000 & 0.17 \\ \hline \end{array} $$

Ecologists have studied how a population's intrinsic exponential growth rate \(r\) is related to the body weight \(W\) for herbivorous mammals. \({ }^{26}\) In Table \(5.2, W\) is the adult weight measured in pounds, and \(r\) is growth rate per year. $$ \begin{array}{|l|c|c|} \hline \text { Animal } & \text { Weight } W & r \\ \hline \text { Short-tailed vole } & 0.07 & 4.56 \\ \hline \text { Norway rat } & 0.7 & 3.91 \\ \hline \text { Roe deer } & 55 & 0.23 \\ \hline \text { White-tailed deer } & 165 & 0.55 \\ \hline \text { American elk } & 595 & 0.27 \\ \hline \text { African elephant } & 8160 & 0.06 \\ \hline \end{array} $$ a. Make a plot of \(\ln r\) against \(\ln W\). Is it reasonable to model \(r\) as a power function of \(W\) ? b. Find a formula that models \(r\) as a power function of \(W\), and draw a graph of this function.
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