Problem 2 - Computational Physics

Question

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?

Step-by-Step Solution

Answer

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 1: 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.

Step 2: 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 \[ F_c = \frac{m v^2}{R+h} \].

Step 3: 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 \[ v^2 = \frac{G M}{R+h} \].

Step 4: 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:\[ \left(\frac{2 \pi (R+h)}{T}\right)^2 = \frac{G M}{R+h} \].

Step 5: Simplify the Expression

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

Step 6: 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,\[ h = \left( \frac{G M T^2}{4 \pi^2} \right)^{1/3} - R \].

Step 7: 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 \(T\).

Step 8: 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.

Step 9: 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.

Step 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.

Key Concepts

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.