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If the average speed of an orbiting space shuttle is \(27800 \mathrm{km} / \mathrm{h},\) determine the time required for it to circle Earth. Assume that the shuttle is orbiting about \(320.0 \mathrm{km}\) above Earth's surface, and that Earth's radius is \(6380 \mathrm{km}\)

Short Answer

Expert verified
To find the time taken by the shuttle to circle the Earth, first calculate the total circumference that the shuttle needs to travel. Convert the speed of the shuttle to the same units as the distance. Substituting these values in the time formula, the time required for the shuttle to circle Earth can be determined.

Step by step solution

01

Determine the total distance the shuttle travels

The total distance is the circumference of the circle formed by the shuttle's orbit. This can be calculated using the formula: \(C=2\pi r\) where \(r\) is the radius of the circle, which is the sum of the radius of the earth and the distance the shuttle is away from the earth's surface. So, \(r=6380 \mathrm{km} + 320.0 \mathrm{km} = 6700\ \mathrm{km}\). Therefore, \(C = 2\pi \times 6700 \mathrm{km}\).
02

Convert speed to the same units

The speed of the shuttle is given in km/hr, while we are asked to find the time in seconds. To make the units compatible, we need to convert the speed from km/hr to km/s. This conversion can be done using the conversion ratio \(1\hr / 3600\sec\). Therefore, the speed in km/s is \(27800 \mathrm{km/hr} \times 1 \hr/3600\ \mathrm{s} = 7.72\ \mathrm{km/s}\)
03

Calculate Time

Using the formula \(speed = \frac{distance}{time}\), time can be determined by rearranging the formula to \(time = \frac{distance}{speed}\). Substituting the distance as the circumference calculated in step 1 and the speed in km/s from step 2, the time required the shuttle to circle Earth can be found: \(time = \frac{C}{speed}\)

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Key Concepts

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

Understanding the Circumference of an Orbit
The circumference of an orbit plays a fundamental role in understanding satellite motion, including that of a space shuttle in orbit around Earth. In simple terms, the circumference represents the boundary of a circle and the path along which a satellite travels while orbiting a celestial body, such as Earth. By using the formula \(C = 2\pi r\), where \(C\) is the circumference and \(r\) is the combined radius of Earth and the altitude of the satellite, we can calculate the total distance the shuttle will cover in one complete orbit. Specifically, for a shuttle orbiting 320 km above Earth's surface, with Earth's radius being 6380 km, the radius of the orbit is the sum: 6380 km + 320 km = 6700 km. Thus, the circumference of the orbit is \(C = 2\pi \times 6700 \text{km}\).
In our daily lives, we rarely deal with circumferences of such magnitude, making it an intriguing concept when studying the vastness of space travel and the physics behind it. It is crucial to acclimate ourselves to think in terms of astronomical distances when dealing with orbital mechanics.
Converting Average Speed for Orbital Mechanics
To assess the motion of an orbiting shuttle, students must understand how to convert average speed into different units. The average speed of the shuttle is initially provided in kilometers per hour (km/h). In the realm of orbital physics, time is often measured in seconds, necessitating the conversion of speed into kilometers per second (km/s). This conversion is essential for compatibility with other formulas and is accomplished by dividing the speed by the number of seconds in an hour (\(3600 \text{s}\)).
Using the conversion formula, \(27800 \text{km/h} \times \frac{1 \text{h}}{3600 \text{s}} = 7.72 \text{km/s}\), we transmute the given speed to a usable form for further calculations. This step is vital as it aligns our speed units with the time units we aim to calculate. Remember, understanding how to convert units is a key skill in physics and enables students to navigate different measurement systems used in scientific equations.
Calculating the Orbital Period of a Space Shuttle
The orbital period of a space shuttle refers to the time it takes to complete one full orbit around Earth. It is a critical concept for determining the frequency at which satellites pass overhead and for synchronizing various operations in space. By reworking the simple speed formula, where speed equals distance divided by time (\(speed = \frac{distance}{time}\)), we can solve for time (\(time = \frac{distance}{speed}\)) to find the orbital period.
Given the shuttle's average speed in km/s and the circumference of its orbit, the orbital period is the ratio of these two quantities. Hence, for the shuttle orbiting at a speed of 7.72 km/s, with a circumference of \(2\pi \times 6700 \text{km}\), the calculation yields the time it takes for one complete revolution around Earth. This period is essential for mission planning, including when to communicate with the shuttle and how to manage its trajectory. Mastery of this calculation is useful for careers in aerospace, astrophysics, and various engineering disciplines where precision and timing are of the essence.

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Most popular questions from this chapter

A train travels between stations 1 and \(2,\) as shown below. The engineer of the train is instructed to start from rest at station 1 and accelerate uniformly between points \(A\) and \(B\), then coast with a uniform velocity between points \(B\) and \(C\), and finally accelerate uniformly between points \(C\) and \(D\) until the train stops at station \(2 .\) The distances \(A B, B C\), and \(C D\) are all equal, and it takes 5.00 min to travel between the two stations. Assume that the uniform accelerations have the same magnitude, even when they are opposite in direction. a. How much of this 5.00 min period does the train spend between points \(A\) and \(B ?\) b. How much of this 5.00 min period does the train spend between points \(B\) and \(C ?\) c. How much of this 5.00 min period does the train spend between points \(C\) and \(D ?\)

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