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Chapter 4: Problem 62

A satellite in a circular orbit 1250 kilometers above Earth makes one complete revolution every 110 minutes. Assuming that Earth is a sphere of radius 6378 kilometers, what is the linear speed (in kilometers per minute) of the satellite?

Short Answer

Expert verified
The linear speed of the satellite is 436.78 kilometers per minute.

Step by step solution

01

Understanding the Concept

The total distance covered by the satellite is actually the circumference of the circular path which can be determined by using the formula \(2\pi r\). The radius is the sum of Earth's radius and the altitude of the satellite above Earth's surface.
02

Calculate the radius

The radius (r) of the orbit is the sum of earth鈥檚 radius and the height of the satellite above the earth, which is \(6378km + 1250km = 7628km\)
03

Calculate the Circumference

The total distance covered by the satellite in one revolution (the circumference of the circular path) is given by the formula \(C = 2\pi r\), Substituting the given values, \(C = 2 \times \pi \times 7628km = 47945.57km\)
04

Calculate the Speed

Speed is distance per time. The given time is 110 minutes. Therefore, the speed of the satellite is \(\frac{47945.57km}{110min} = 436.78km/min\)

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Key Concepts

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

Circular Orbit
When we talk about a satellite鈥檚 movement around the Earth, the term 'circular orbit' is used to describe a scenario where a satellite moves in a path that maintains a constant distance from the planet鈥檚 center. This path is an impeccable circle and is a simplified version of real satellite paths, which can sometimes be more elliptical due to various gravitational forces. The ideal circular orbit provides a useful model as it lets us apply uniform physics equations, such as those for calculating speed, without having to account for varying distances.
Circumference of a Circle
The circumference of a circle is the distance around its edge. It's a fundamental concept in geometry and is especially pertinent when we discuss orbits because the satellite's path around the Earth forms a circular loop. The formula for the circumference is given by \(C = 2\pi r\), where \(C\) represents the circumference and \(r\) is the radius of the circle. In the context of satellite movement, knowing the circumference allows us to determine how far the satellite travels in one full revolution around the Earth.
Radius of the Earth
The Earth's radius is the distance from the center of the planet to its surface. It鈥檚 a pivotal measurement in calculating the orbits of satellites. The current accepted average radius is 6378 kilometers. However, it is not sufficient to use this number alone when calculating satellite orbits because satellites do not skim the surface of Earth鈥攖hey orbit at varying altitudes above it. This means the radius of the satellite's orbit is the sum of the Earth鈥檚 radius and the satellite's altitude above Earth, providing us with the effective radius required for calculating orbital parameters like the circumference of the orbit.
Satellite Revolution Time
Satellite revolution time refers to the period it takes for a satellite to complete one full orbit around the Earth. This parameter is essential because it directly affects the satellite's speed. In our exercise, the satellite completes its revolution in 110 minutes, which is a known fixed time interval. By understanding revolution time in conjunction with the orbit's circumference, we can calculate the linear speed of the satellite, which is the distance traveled in a given time frame.
Orbital Mechanics
Orbital mechanics is the field of study concerned with the motion of objects in space under the influence of gravitational forces. The formulas and principles used in our step-by-step solution fall under the umbrella of orbital mechanics. When calculating the speed and orbits of satellites, we must apply these principles to predict and describe the satellite's behavior. This encompasses both the satellite's speed along its path and its position at various times, which is critical for tasks such as communication and Earth observations.

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Most popular questions from this chapter

The numbers of hours \(H\) of daylight in Denver, Colorado, on the 15 th of each month are: \(1(9.67), 2(10.72), 3(11.92), 4(13.25)\) \(5(14.37), \quad 6(14.97), \quad 7(14.72), \quad 8(13.77), \quad 9(12.48)\) \(10(11.18), \quad 11(10.00), \quad 12(9.38) . \quad\) The month is represented by \(t,\) with \(t=1\) corresponding to January. A model for the data is $$H(t)=12.13+2.77 \sin \left(\frac{\pi t}{6}-1.60\right)$$. (a) Use a graphing utility to graph the data points and the model in the same viewing window. (b) What is the period of the model? Is it what you expected? Explain. (c) What is the amplitude of the model? What does it represent in the context of the problem? Explain.

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