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Chapter 2: Problem 2

Altitude of a satellite A satellite is to be launched into a circular orbit around the Earth so that it orbits the planet once every \(T\) seconds. a) Show that the altitude \(h\) above the Earth's surface that the satellite must have is $$ h=\left(\frac{G M T^{2}}{4 \pi^{2}}\right)^{1 / 3}-R $$ where \(G=6.67 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\) is Newton's gravitational constant, \(M=\) \(5.97 \times 10^{24} \mathrm{~kg}\) is the mass of the Earth, and \(R=6371 \mathrm{~km}\) is its radius. b) Write a program that asks the user to enter the desired value of \(T\) and then calculates and prints out the correct altitude in meters. c) Use your program to calculate the altitudes of satellites that orbit the Earth once a day (so-called "geosynchronous" orbit), once every 90 minutes, and once every 45 minutes. What do you conclude from the last of these calculations? d) Technically a geosynchronous satellite is one that orbits the Earth once per sidereal day, which is \(23.93\) hours, not 24 hours. Why is this? And how much difference will it make to the altitude of the satellite?

Short Answer

Expert verified
Formulate altitude with given inputs using \( h=\left(\frac{G M T^{2}}{4 \pi^{2}}\right)^{1 / 3}-R \). Program altitude calculations based on orbital periods of 24 hours, 90 minutes, and 45 minutes show impracticality for the 45-minute orbit. Geosynchronous satellites align with Earth's sidereal day.

Step by step solution

01

Understand the Given Formula

The formula for the altitude of a satellite is given as \[ h=\left(\frac{G M T^{2}}{4 \pi^{2}}\right)^{1 / 3}-R \]To show this derivation, we need to use Newton’s law of gravitation and the centripetal force that maintains the satellite’s circular orbit.
02

Apply Newton’s Law of Gravitation and Centripetal Force

The gravitational force acting on the satellite is given by\[ F_g = \frac{G M m}{(R+h)^2} \]where \(m\) is the mass of the satellite. The centripetal force required to keep the satellite in circular orbit is \[ F_c = \frac{m v^2}{R+h} \].
03

Equalize Gravitational and Centripetal Forces

For the satellite to stay in orbit, the gravitational force must equal the centripetal force:\[ \frac{G M m}{(R+h)^2} = \frac{m v^2}{R+h} \]This simplifies to \[ v^2 = \frac{G M}{R+h} \].
04

Relate Orbital Velocity to Orbital Period

The orbital velocity \(v\) is also related to the period \(T\) by\[ v = \frac{2 \pi (R+h)}{T} \]Substitute this into the previous equation:\[ \left(\frac{2 \pi (R+h)}{T}\right)^2 = \frac{G M}{R+h} \].
05

Simplify the Expression

Simplify the equation to obtain\[ \frac{4 \pi^2 (R+h)^3}{T^2} = G M \]Solve for \((R+h)^3\):\[ (R+h)^3 = \frac{G M T^2}{4 \pi^2} \].
06

Solve for the Altitude \( h \)

Take the cube root of both sides to find\[ R+h = \left( \frac{G M T^2}{4 \pi^2} \right)^{1/3} \]Then,\[ h = \left( \frac{G M T^2}{4 \pi^2} \right)^{1/3} - R \].
07

Write the Program (b)

A simple program can be written in Python:```pythonimport mathG = 6.67e-11M = 5.97e24R = 6371e3def find_altitude(T): term = (G * M * T**2) / (4 * math.pi**2) h = term**(1/3) - R return hT = float(input('Enter the period T in seconds: '))altitude = find_altitude(T)print(f'The altitude is {altitude} meters')```This program calculates and prints the altitude of the satellite based on the input period \(T\).
08

Calculate Altitudes for Specific Periods (c)

To calculate the altitudes:- For geosynchronous orbit (24 hours):\[ T = 24 \times 3600 = 86400 \text{s} \]- For a 90-minute orbit:\[ T = 90 \times 60 = 5400 \text{s} \]- For a 45-minute orbit:\[ T = 45 \times 60 = 2700 \text{s} \]Using the program, these values can be input to get the corresponding altitudes.
09

Comparison and Conclusion from 45-minute Orbit (c)

Calculate using the program and compare the altitudes. The 45-minute orbit results in an altitude very close to the Earth's surface, likely within the atmosphere and hence not a stable orbit.
10

Understanding Geosynchronous Satellite (d)

A sidereal day is 23.93 hours, not 24 hours owing to Earth's rotation relative to distant stars. To calculate this difference in altitude, use \( T = 23.93 \times 3600 \) and compare it against the 24-hour calculation.

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Key Concepts

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

Newton's law of gravitation
Newton's law of gravitation is fundamental to understanding how satellites stay in orbit around the Earth. This law states that every point mass attracts every other point mass by a force acting along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as: \( F = \frac{G M_1 M_2}{r^2} \)where:
  • \( F \) is the gravitational force between two masses.
  • \( G \) is the gravitational constant \( (6.67 \times 10^{-11} \text{m}^3\text{kg}^{-1}\text{s}^{-2}) \).
  • \( M_1 \) and \( M_2 \) are the masses of the two objects.
  • \( r \) is the distance between the centers of the two masses.
This formula helps us understand the attractive force between Earth and a satellite, which is crucial for orbital mechanics.
Orbital mechanics
Orbital mechanics deals with the motions of objects in space, primarily under the influence of gravity. For a satellite to remain in a stable orbit around Earth, the gravitational force pulling the satellite towards Earth must equal the centripetal force needed to keep it moving in a circular path. The centripetal force is given by:\( F_c = \frac{m v^2}{r} \)where:
  • \( m \) is the mass of the satellite.
  • \( v \) is the orbital velocity.
  • \( r \) is the radius of the orbit (which is the Earth's radius plus the altitude of the satellite).
By equating the gravitational force with the centripetal force, we derive important formulas that help us calculate the altitude of a satellite for a given orbital period.
Programming with Python
Programming provides a great way to handle calculations for satellite orbits. In this case, Python can be used to create a program to find the altitude of a satellite given its orbital period. Python is excellent for scientific computing because of its simplicity and rich set of libraries. Here's a simplified version of a Python program to calculate satellite altitude:```pythonimport mathG = 6.67e-11M = 5.97e24R = 6371e3# Function to find satellite altitude def find_altitude(T): term = (G * M * T**2) / (4 * math.pi**2) h = term**(1/3) - R return h# Input and calculationT = float(input('Enter the period T in seconds: '))altitude = find_altitude(T)print(f'The altitude is {altitude} meters')```This program prompts the user for the orbital period \( T \) and then calculates the altitude \( h \) using the derived formula.
Geosynchronous satellite
A geosynchronous satellite orbits the Earth once per sidereal day, approximately 23.93 hours. This special orbit means the satellite appears to be stationary relative to the Earth's surface, making it ideal for communications, weather monitoring, and surveillance. The orbital period for such a satellite is 24 hours (or 86400 seconds). Converting this period into altitude can be done using the derived formula.However, it's important to note that the exact geosynchronous orbit accounts for the sidereal day, requiring a slight adjustment in the calculation to 23.93 hours (or approximately 86148 seconds). This minor change affects the altitude slightly, but enough to be significant for precise satellite operations.
Altitude calculation
Calculating the altitude of a satellite involves understanding the relationship between the orbital period and the radius of the orbit, which includes the Earth's radius and the altitude \( h \). The formula for the altitude given the orbital period \( T \) is:\[ h = \left( \frac{G M T^2}{4 \pi^2} \right)^{1/3} - R \]where:
  • \( T \) is the orbital period in seconds.
  • \( G \) is the gravitational constant (\(6.67 \times 10^{-11} \text{m}^3\text{kg}^{-1}\text{s}^{-2}\)).
  • \( M \) is the mass of the Earth (\(5.97 \times 10^{24} \text{kg}\)).
  • \( R \) is the Earth's radius (approx. 6371 km).
By plugging in the appropriate values for \( T \), you can determine the altitude of a satellite for different orbital periods, such as 24 hours, 90 minutes, and 45 minutes. For instance, a 45-minute orbit results in a very low altitude that would likely be unsustainable due to atmospheric drag.

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Most popular questions from this chapter

Catalan numbers The Catalan numbers \(C_{n}\) are a sequence of integers \(1,1,2,5,14,42,132 \ldots\) that play an important role in quantum mechanics and the theory of disordered systems. (They were central to Eugene Wigner's proof of the so- called semicircle law.) They are given by $$ C_{0}=1, \quad C_{n+1}=\frac{4 n+2}{n+2} C_{n} . $$ Write a program that prints in increasing order all Catalan numbers less than or equal to one billion.

Another ball dropped from a tower A ball is again dropped from a tower of height \(h\) with initial velocity zero. Write a program that asks the user to enter the height in meters of the tower and then calculates and prints the time the ball takes until it hits the ground, ignoring air resistance. Use your program to calculate the time for a ball dropped from a \(100 \mathrm{~m}\) high tower.

The semi-empirical mass formula In nuclear physics, the semi-empirical mass formula is a formula for calculating the approximate nuclear binding energy \(B\) of an atomic nucleus with atomic number \(Z\) and mass number \(A\) : $$ B=a_{1} A-a_{2} A^{2 / 3}-a_{3} \frac{Z^{2}}{A^{1 / 3}}-a_{4} \frac{(A-2 Z)^{2}}{A}+\frac{a_{5}}{A^{1 / 2}} $$ where, in units of millions of electron volts, the constants are \(a_{1}=15.8, a_{2}=18.3\), \(a_{3}=0.714, a_{4}=23.2\), and $$ a_{5}= \begin{cases}0 & \text { if } A \text { is odd, } \\ 12.0 & \text { if } A \text { and } Z \text { are both even, } \\ -12.0 & \text { if } A \text { is even and } Z \text { is odd. }\end{cases} $$ a) Write a program that takes as its input the values of \(A\) and \(Z\), and prints out the binding energy for the corresponding atom. Use your program to find the binding energy of an atom with \(A=58\) and \(Z=28\). (Hint: The correct answer is around \(500 \mathrm{MeV}\) ) b) Modify your program to print out not the total binding energy B, but the binding energy per nucleon, which is \(B / A\). c) Now modify your program so that it takes as input just a single value of the atomic number \(Z\) and then goes through all values of \(A\) from \(A=Z\) to \(A=3 Z\), to find the one that has the largest binding energy per nucleon. This is the most stable nucleus with the given atomic number. Have your program print out the value of \(A\) for this most stable nucleus and the value of the binding energy per nucleon d) Modify your program again so that, instead of taking \(Z\) as input, it runs through all values of \(Z\) from 1 to 100 and prints out the most stable value of \(A\) for each one. At what value of \(Z\) does the maximum binding energy per nucleon occur? (The true answer, in real life, is \(Z=28\), which is nickel.)

Suppose arrays a and b are defined as follows: $$ \begin{aligned} &\text { from numpy import array } \\ &a=\operatorname{array}([1,2,3,4], \text { int }) \\ &b=\operatorname{array}([2,4,6,8], \text { int }) \end{aligned} $$ What will the computer print upon executing the following lines? (Try to work out the answer before trying it on the computer.) a) print \((b / a+1)\) b) print \((b /(a+1))\) c) print(1/a)

Prime numbers The program in Example \(2.8\) is not a very efficient way of calculating prime numbers: it checks each number to see if it is divisible by any number less than it. We can develop a much faster program for prime numbers by making use of the following observations: a) A number \(n\) is prime if it has no prime factors less than \(n\). Hence we only need to check if it is divisible by other primes. b) If a number \(n\) is non-prime, having a factor \(r\), then \(n=r s\), where \(s\) is also a factor. If \(r \geq \sqrt{n}\) then \(n=r s \geq \sqrt{n} s\), which implies that \(s \leq \sqrt{n}\). In other words, any non-prime must have factors, and hence also prime factors, less than or equal to \(\sqrt{n}\). Thus to determine if a number is prime we have to check its prime factors only up to and including \(\sqrt{n}\)-if there are none then the number is prime. c) If we find even a single prime factor less than \(\sqrt{n}\) then we know that the number is non-prime, and hence there is no need to check any further- we can abandon this number and move on to something else. Write a Python program that finds all the primes up to ten thousand. Create a list to store the primes, which starts out with just the one prime number 2 in it. Then for each number \(n\) from 3 to 10000 check whether the number is divisible by any of the primes in the list up to and including \(\sqrt{n}\). As soon as you find a single prime factor you can stop checking the rest of them- you know \(n\) is not a prime. If you find no prime factors \(\sqrt{n}\) or less then \(n\) is prime and you should add it to the list. You can print out the list all in one go at the end of the program, or you can print out the individual numbers as you find them.
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