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Chapter 8: Problem 27

Communications satellites are placed in circular orbits where they stay directly over a fixed point on the equator as the earth rotates. These are called geosynchronous orbits. The altitude of a geosynchronous orbit is \(3.58 \times 10^{7} \mathrm{m}\) (\(\sim 22,000\) miles). a. What is the period of a satellite in a geosynchronous orbit? b. Find the value of \(g\) at this altitude. c. What is the weight of a \(2000 \mathrm{kg}\) satellite in a geosynchronous orbit?

Short Answer

Expert verified
a. The period of a satellite in a geosynchronous orbit is 24 hours. b. The value of \(g\) at the given altitude can be calculated using the known values for the Earth's mass, the universal gravitational constant, and the distance from the center of the earth to the satellite. c. The weight of a 2000 kg satellite at the given altitude can be calculated using the gravitational acceleration value from part b and the formula \(F = ma\) where \(F\) is the force of gravity (the weight), \(m\) is the mass of the satellite, and \(a\) is the gravitational acceleration.

Step by step solution

01

Determining the period of a satellite in a geosynchronous orbit

By definition, a geosynchronous orbit matches the Earth's rotation. Since Earth completes one rotation in 24 hours, a satellite in a geosynchronous orbit also has a period of 24 hours.
02

Determining the value of \(g\) at the given altitude

Using the formula for gravitational force, we can express gravitational acceleration (g) at height \(r\) as \(GM / r^2\). The given altitude is 3.58 x 10^7 m, but since the altitude is from the earth's surface, we need to add this to the radius of the earth (about 6.37 x 10^6 m) to get the total distance from the center of the earth. Substituting \(G\) = 6.674 x 10^-11 N(m/kg)\^2, \(M\) = 5.972 x 10^24 kg, and \(r\) = 3.58 x 10^7 m + 6.37 x 10^6 m, we solve for \(g\).
03

Calculating the weight of the satellite in a geosynchronous orbit

Assuming the satellite's mass is \(m = 2000 \mathrm{kg}\), the weight of the satellite at that altitude can be calculated using \(F = ma\). The acceleration \(a\) is the acceleration due to gravity at the satellite's altitude we found in step 2.

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Key Concepts

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

Gravitational Acceleration
Gravitational acceleration, often represented as 'g', is a critical concept when discussing orbits. It is the acceleration on an object caused by the force of gravity from another object, typically a planet. The farther you are from the planet's center, the less gravitational acceleration you experience. This is why astronauts feel weightless in space; they are far enough from the Earth that the gravitational acceleration is much weaker.

When calculating gravitational acceleration at a specific altitude, such as that of a geosynchronous orbit, we use the formula \( g = \frac{GM}{r^2} \), where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(r\) is the distance from the Earth's center to the satellite. To find 'g' for a satellite in geosynchronous orbit, you need to add the Earth's radius to the altitude of the orbit before squaring this total distance.
Satellite Period
The period of a satellite is the time it takes to complete one full orbit around the Earth. In the case of geosynchronous satellites, this period is precisely 24 hours, matching the rotation period of the Earth.

To maintain a geosynchronous orbit, a satellite must travel at the exact speed that allows it to revolve around the Earth in 24 hours. If the satellite's period were any different, it wouldn't appear to hover over the same point on the Earth's surface. For satellites that are at other altitudes, we can determine the period using Kepler's third law, which relates the orbital period to the distance from the center of the earth.
Weight in Orbit
Normally, we calculate weight as the product of mass and the gravitational acceleration at the Earth's surface. But for objects in orbit, weight is the gravitational force they experience at their orbital distance.

Even in a geosynchronous orbit, a satellite is still under the influence of Earth's gravity, which is what keeps it in orbit. Weight in orbit isn't the same as what we feel on Earth's surface. To calculate the weight of a satellite in orbit, use the formula \( F = ma \), where 'F' is the force or weight, 'm' is the satellite's mass, and 'a' is the gravitational acceleration at the satellite's altitude. Despite common misconceptions, a satellite in orbit is not weightless; it simply experiences much less gravitational force than on the surface. This is crucial for keeping the satellite in a stable orbit without it plummeting back to Earth or drifting into space.

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Most popular questions from this chapter

\(A 200\) g block on a 50 -cm-long string swings in a circle on a horizontal, frictionless table at 75 rpm. a. What is the speed of the block? b. What is the tension in the string?

A \(4.0 \times 10^{10} \mathrm{kg}\) asteroid is heading directly toward the center of the earth at a steady \(20 \mathrm{km} / \mathrm{s}\). To save the planet, astronauts strap a giant rocket to the asteroid perpendicular to its direction of travel. The rocket generates \(5.0 \times 10^{9} \mathrm{N}\) of thrust. The rocket is fired when the asteroid is \(4.0 \times 10^{6} \mathrm{km}\) away from earth. You can ignore the earth's gravitational force on the asteroid and their rotation about the sun. a. If the mission fails, how many hours is it until the asteroid impacts the earth? b. The radius of the earth is \(6400 \mathrm{km}\). By what minimum angle must the asteroid be deflected to just miss the earth? c. The rocket fires at full thrust for 300 s before running out of fuel. Is the earth saved?

A \(5.0 \mathrm{g}\) coin is placed \(15 \mathrm{cm}\) from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of \(\mu_{3}=0.80\) and \(\mu_{\mathrm{x}}=0.50 .\) The turntable very slowly speeds up to 60 rpm. Does the coin slide off?

A heavy ball with a weight of \(100 \mathrm{N}(\mathrm{m}=10.2 \mathrm{kg})\) is hung from the ceiling of a lecture hall on a 4.5 -m-long rope. The ball is pulled to one side and released to swing as a pendulum, reaching a speed of \(5.5 \mathrm{m} / \mathrm{s}\) as it passes through the lowest point. What is the tension in the rope at that point?

A satellite orbiting the moon very near the surface has a period of 110 min. What is the moon's acceleration due to gravity? Astronomical data are inside the back cover of the book.
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