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Chapter 5: Problem 36

A satellite circles the earth in an orbit whose radius is twice the earth's radius. The earth's mass is \(5.98 \times 10^{24} \mathrm{kg}\), and its radius is \(6.38 \times 10^{6} \mathrm{m}\). What is the period of the satellite?

Short Answer

Expert verified
The satellite's period is approximately 8.91 hours.

Step by step solution

01

Identify the Required Formula

To find the period of a satellite, we use the formula for the orbital period: \[ T = 2\pi \sqrt{\frac{r^3}{GM}} \] where \(T\) is the period, \(r\) is the orbital radius, \(G\) is the gravitational constant \(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\), and \(M\) is the Earth's mass.
02

Calculate the Orbital Radius

Since the satellite's orbit radius is twice the Earth's radius, calculate it as follows:\[ r = 2 \times 6.38 \times 10^6 \, \text{m} = 1.276 \times 10^7 \, \text{m} \]
03

Substitute Known Values into the Formula

Substitute the known values into the orbital period formula:\[ T = 2\pi \sqrt{\frac{(1.276 \times 10^7 \, \text{m})^3}{6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \times 5.98 \times 10^{24} \, \text{kg}}} \]
04

Calculate the Expression Inside the Square Root

Calculate the value inside the square root:\[ \frac{(1.276 \times 10^7)^3}{6.674 \times 10^{-11} \times 5.98 \times 10^{24}} \approx 2.5987 \times 10^7 \]
05

Calculate the Orbital Period

Calculate the period by evaluating the expression:\[ T = 2\pi \sqrt{2.5987 \times 10^7} \approx 2\pi \times 5097.3 \approx 32088 \, \text{seconds} \].
06

Convert Period to Hours

Convert the period from seconds to hours by dividing by 3600 (the number of seconds in an hour):\[ T = \frac{32088}{3600} \approx 8.91 \, \text{hours} \]

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Key Concepts

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

Satellite Motion
Satellites move in space due to the gravitational pull of the Earth, which keeps them in orbit. This motion is known as satellite motion. A satellite maintains its orbit because its forward motion balances the gravitational pull from Earth. If a satellite moves too quickly, it might escape Earth's gravity and drift into space. Conversely, if it moves too slowly, it might fall back to Earth. Understanding satellite motion is crucial for calculating orbits, ensuring communication signals, and monitoring climate and weather patterns. The law of universal gravitation plays a vital role, as gravity is the main force acting upon a satellite in orbit.
Gravitational Constant
The gravitational constant, denoted as G, is a key component in the calculation of gravitational forces. It appears in Newton's law of universal gravitation and defines the strength of the gravitational force between two masses. The accepted value of G is \(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\). This constant helps us calculate the gravitational force that keeps the satellite moving along its orbit. Without G, we wouldn't be able to predict how celestial objects interact or how the force of gravity behaves across different scenarios.
Earth's Radius
The Earth's radius is a fundamental value in orbital calculations. It provides a reference point for determining a satellite's orbital radius when no altitude information is given. The average radius of Earth is approximately \(6.38 \times 10^6 \, \text{m}\). Knowing this helps in various scientific calculations and is especially important in understanding Earth's dimensions and its influence on nearby celestial objects. When calculating orbits, it's essential to note whether given values, such as a satellite's radius, are from the center of Earth or above its surface.
Orbital Radius Calculation
Calculating the orbital radius involves determining the distance from the center of the Earth to the satellite. For our specific problem, the orbital radius is defined as twice the Earth's radius. Therefore, to find the orbital radius, simply multiply the Earth's radius by two: \[ r = 2 \times 6.38 \times 10^6 \, \text{m} = 1.276 \times 10^7 \, \text{m} \]This calculation is crucial for the accurate application of formulas like the orbital period formula, which requires the orbital radius to find out how long a satellite takes to complete one orbit around Earth.

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Most popular questions from this chapter

A motorcycle is traveling up one side of a hill and down the other side. The crest of the hill is a circular arc with a radius of \(45.0 \mathrm{m}\). Determine the maximum speed that the cycle can have while moving over the crest without losing contact with the road.

Two cars are traveling at the same speed of \(27 \mathrm{m} / \mathrm{s}\) on a curve that has a radius of \(120 \mathrm{m}\). Car \(\mathrm{A}\) has a mass of \(1100 \mathrm{kg}\), and car \(\mathrm{B}\) has a mass of \(1600 \mathrm{kg} .\) Find the magnitude of the centripetal acceleration and the magnitude of the centripetal force for each car.

A child is twirling a \(0.0120-\mathrm{kg}\) plastic ball on a string in a horizontal circle whose radius is \(0.100 \mathrm{m}\). The ball travels once around the circle in 0.500 s. (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, does the centripetal force double? If not, by what factor does the centripetal force increase?

In a skating stunt known as crack-the-whip, a number of skaters hold hands and form a straight line. They try to skate so that the line rotates about the skater at one end, who acts as the pivot. The skater farthest out has a mass of \(80.0 \mathrm{kg}\) and is \(6.10 \mathrm{m}\) from the pivot. He is skating at a speed of \(6.80 \mathrm{m} / \mathrm{s}\). Determine the magnitude of the centripetal force that acts on him.

Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the planets are determined to be \(43.3 \mathrm{km} / \mathrm{s}\) and \(58.6 \mathrm{km} / \mathrm{s} .\) The slower planet's orbital period is 7.60 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?
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