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An Earth satellite moves in a circular orbit \(750 \mathrm{~km}\) above Earth's surface with a period of \(98.0 \mathrm{~min}\). What are the (a) speed and (b) magnitude of the centripetal acceleration of the satellite?

Short Answer

Expert verified
The speed of the satellite is approximately 7610 m/s, and its centripetal acceleration is 8.13 m/s².

Step by step solution

01

Understanding the Problem

The satellite is orbiting Earth at a height of 750 km above the surface with a period of 98 minutes. We need to find both the speed of the satellite and the magnitude of its centripetal acceleration.
02

Calculate Total Radius of Orbit

To find the total radius of the orbit, add Earth's radius (approximately 6371 km) to the satellite's height above Earth. The total radius is:\[ r = 6371 ext{ km} + 750 ext{ km} = 7121 ext{ km} = 7.121 imes 10^6 ext{ m} \]
03

Convert Orbital Period to Seconds

Convert the period from minutes to seconds to use in the formula for speed:\[ T = 98 ext{ minutes} \times 60 ext{ seconds/minute} = 5880 ext{ seconds} \]
04

Calculate Orbital Speed

Use the formula for orbital speed, which is the circumference of the orbit divided by the period:\[ v = \frac{2\pi r}{T} \]\[ v = \frac{2\pi \times 7.121 \times 10^6 ext{ m}}{5880 ext{ s}} \approx 7610 ext{ m/s} \]
05

Calculate Centripetal Acceleration

Use the formula for centripetal acceleration, \( a_c = \frac{v^2}{r} \):\[ a_c = \frac{(7610 ext{ m/s})^2}{7.121 \times 10^6 ext{ m}} \approx 8.13 ext{ m/s}^2 \]

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Key Concepts

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

Circular Orbits
In orbital mechanics, a circular orbit refers to the path a celestial object takes around another body where the distance from the central body remains constant. In the case of Earth satellites, like the one in the exercise, the satellite moves in a circular orbit meaning its path is a perfect circle around Earth. The circular nature ensures that both the speed and direction of the satellite are constant. This consistent path is crucial for communication satellites and weather monitoring, as it ensures consistent data transfer and forecast accuracy. However, to maintain this kind of orbit, several forces need to be balanced, primarily the gravitational pull of Earth and the inertia of the satellite's motion. Together, these result in a stable orbit where the satellite doesn't drift away or get pulled into Earth's atmosphere.
Centripetal Acceleration
Centripetal acceleration is the force required to keep an object moving in a circular path. It's directed towards the center of the circle, ensuring the satellite doesn't shoot off in a straight line due to inertia. The centripetal acceleration of an orbiting satellite can be calculated using the formula:
  • \[ a_c = \frac{v^2}{r} \]
  • where \(v\) is the orbital speed and \(r\) is the radius of the orbit.
In our exercise, the calculated centripetal acceleration is approximately 8.13 m/s². This acceleration is crucial for satellites because it ensures that they stay in their designated orbit without deviating. If this acceleration were incorrectly calculated or unbalanced, the satellite might drift into a much closer or more distant orbit, disrupting its mission.
Orbital Speed
Orbital speed is the speed at which a satellite or object needs to travel to maintain a stable orbit around a celestial body. This speed ensures that the gravitational pull is balanced by the inertia of the moving body. In our exercise, the orbital speed is calculated as approximately 7610 m/s.
A satellite needs to maintain this speed for several reasons:
  • To remain in a stable orbit at a fixed altitude.
  • To ensure that it doesn't fall back to Earth or escape into a more distant orbit.
The calculation involves determining the circumference of the orbit, which is dependent on the total radius (Earth's radius plus the satellite's height above Earth), and dividing it by the period it takes for the satellite to complete one orbit. This ensures that the satellite completes its orbit at the right pace.
Satellite Motion
Satellite motion involves the complex interplay of speed, gravitational forces, and inertia to keep a satellite in a consistent path around Earth. Understanding satellite motion is crucial, as it forms the foundation for the operation of various satellite types, from GPS to communication satellites.
Key aspects of satellite motion include:
  • Maintaining precise orbital speed to ensure stability.
  • Handling changes in altitude, which can alter orbital speed requirements.
The motion of a satellite is governed by Newton's laws of motion and the law of universal gravitation, which ensure continuous orbiting around Earth without needing constant propulsion. The balance between gravitational pull and the inertia from its speed keeps it from falling back to Earth while preventing it from escaping into space. For successful satellite missions, understanding and controlling these dynamics is vital.

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Most popular questions from this chapter

A cat rides a merry-go-round turning with uniform circular motion. At time \(t_{1}=2.00 \mathrm{~s}\), the cat's velocity is \(\vec{v}_{1}=\) \((3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\), measured on a horizontal \(x y\) coordinate system. At \(t_{2}=5.00 \mathrm{~s}\), the cat's velocity is \(\vec{v}_{2}=(-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+\) \((-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). What are (a) the magnitude of the cat's centripetal acceleration and (b) the cat's average acceleration during the time interval \(t_{2}-t_{1}\), which is less than one period?

In the 1991 World Track and Field Championships in Tokyo, Mike Powell jumped \(8.95 \mathrm{~m}\), breaking by a full \(5 \mathrm{~cm}\) the 23-year long-jump record set by Bob Beamon. Assume that Powell's speed on takeoff was \(9.5 \mathrm{~m} / \mathrm{s}\) (about equal to that of a sprinter) and that \(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\) in Tokyo. How much less was Powell's range than the maximum possible range for a particle launched at the same speed?

Two seconds after being projected from ground level, a projectile is displaced \(40 \mathrm{~m}\) horizontally and \(58 \mathrm{~m}\) vertically above its launch point. What are the (a) horizontal and (b) vertical components of the initial velocity of the projectile? (c) At the instant the projectile achieves its maximum height above ground level, how far is it displaced horizontally from the launch point?

You throw a ball toward a wall at speed \(25.0 \mathrm{~m} / \mathrm{s}\) and at angle \(\theta_{0}=\) \(40.0^{\circ}\) above the horizontal (Fig. 426). The wall is distance \(d=22.0 \mathrm{~m}\) from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall? (d) When it hits, has it passed the highest point on its trajectory?

During a tennis match, a player serves the ball at \(23.6 \mathrm{~m} / \mathrm{s}\), with the center of the ball leaving the racquet horizontally \(2.42 \mathrm{~m}\) above the court surface. The net is \(12 \mathrm{~m}\) away and \(0.90 \mathrm{~m}\) high. When the ball reaches the net, (a) does the ball clear it and (b) what is the distance between the center of the ball and the top of the net? Suppose that, instead, the ball is served as before but now it leaves the racquet at \(5.00^{\circ}\) below the horizontal. When the ball reaches the net, (c) does the ball clear it and (d) what now is the distance between the center of the ball and the top of the net?

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