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Ch 13: Gravitation
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 13, Problem 26a

On July 15, 2004, NASA launched the Aura spacecraft to study the earth's climate and atmosphere. This satellite was injected into an orbit 705 km above the earth's surface. Assume a circular orbit. How many hours does it take this satellite to make one orbit?

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First, understand that the satellite is in a circular orbit around the Earth. To find the time it takes to complete one orbit, we need to calculate the orbital period.
The orbital period can be found using Kepler's Third Law, which relates the period of orbit to the radius of the orbit. The formula is: T=2πr3GM, where r is the radius of the orbit, G is the gravitational constant, and M is the mass of the Earth.
Calculate the radius of the orbit. The radius is the sum of the Earth's radius and the altitude of the satellite. Earth's average radius is approximately 6371 km, so the orbital radius r is 6371+705 km.
Substitute the values into the formula. Use G = 6.674×10^-11 m3 kg-1 s-2 and M = 5.972×10^24 kg.
Convert the orbital period from seconds to hours by dividing by 3600 seconds per hour. This will give you the time it takes for the satellite to complete one orbit in hours.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Circular Orbit

A circular orbit is a path followed by an object around a celestial body where the distance from the center of the body remains constant. In this context, the satellite maintains a constant altitude of 705 km above Earth's surface, implying a uniform circular motion. Understanding circular orbits is crucial for calculating orbital periods and velocities.
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Gravitational Force

Gravitational force is the attractive force between two masses, such as a satellite and Earth. It provides the necessary centripetal force to keep the satellite in orbit. The balance between gravitational pull and the satellite's inertia determines its orbital speed and period, which are essential for calculating how long it takes to complete one orbit.
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Orbital Period

The orbital period is the time taken for a satellite to complete one full orbit around a celestial body. It depends on the altitude of the orbit and the mass of the central body. For a satellite in a circular orbit, the period can be calculated using Kepler's third law, which relates the orbital radius to the period, providing insights into the satellite's motion.
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Related Practice
Textbook Question

In 2004 astronomers reported the discovery of a large Jupiter-sized planet orbiting very close to the star HD 179949 (hence the term 'hot Jupiter'). The orbit was just 1/9 the distance of Mercury from our sun, and it takes the planet only 3.09 days to make one orbit (assumed to be circular). How fast (in km/s) is this planet moving?

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Textbook Question

Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was 2.87 × 106 km from the earth and traveling at 1.20 × 104 km/h relative to the earth. At this time, what were (a) the spacecraft's kinetic energy relative to the earth and (b) the potential energy of the earth–spacecraft system?

Textbook Question

In its orbit each day, the International Space Station makes 15.65 revolutions around the earth. Assuming a circular orbit, how high is this satellite above the surface of the earth?

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Textbook Question

Two satellites are in circular orbits around a planet that has radius 9.00 × 106 m. One satellite has mass 68.0 kg, orbital radius 7.00 × 107 m, and orbital speed 4800 m/s. The second satellite has mass 84.0 kg and orbital radius 3.00 × 107 m. What is the orbital speed of this second satellite?

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Textbook Question

For a satellite to be in a circular orbit 890 km above the surface of the earth, what orbital speed must it be given?

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Textbook Question

A planet orbiting a distant star has radius 3.24 × 106 m. The escape speed for an object launched from this planet’s surface is 7.65 × 103 m/s. What is the acceleration due to gravity at the surface of the planet?

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