/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 A satellite is in a circular orb... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 35

A satellite is in a circular orbit about the earth \(\left(M_{\mathrm{E}}=5.98 \times\right.\) \(10^{24} \mathrm{kg} .\) The period of the satellite is \(1.20 \times 10^{4} \mathrm{s}\) . What is the speed at which the satellite travels?

Short Answer

Expert verified
The satellite travels at approximately 4399.65 m/s.

Step by step solution

01

Write Down Known Values

We begin by noting the given values: Mass of the Earth \(M_{\mathrm{E}} = 5.98 \times 10^{24} \text{ kg}\) and the period of the satellite \(T = 1.20 \times 10^{4} \text{ s}\). We want to find the speed \(v\) of the satellite.
02

Use the Formula for Orbital Speed

The orbital speed \(v\) of a satellite can be found using the formula:\[v = \frac{2\pi r}{T}\]where \(r\) is the orbital radius and \(T\) is the period of the satellite. We need to find \(r\) in the next step.
03

Calculate Orbital Radius

We can find the orbital radius \(r\) using the relationship derived from the gravitational force and centripetal force:\[T = 2\pi \sqrt{\frac{r^3}{GM_{\mathrm{E}}}}\]Rearranging for \(r\), we have:\[r = \left(\frac{T^2 G M_{\mathrm{E}}}{4\pi^2}\right)^{1/3}\]Let's plug the values for \(T\), \(G = 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}\), and \(M_{\mathrm{E}}\).
04

Calculate Orbital Radius Value

Substituting the known values, we find:\[r = \left(\frac{(1.20 \times 10^{4})^2 \times 6.674 \times 10^{-11} \times 5.98 \times 10^{24}}{4\pi^2}\right)^{1/3}\approx 8381250 \text{ m}\]
05

Calculate Orbital Speed

Now that we have \(r\), substitute it back into the formula for orbital speed to find \(v\):\[v = \frac{2 \pi \times 8381250}{1.20 \times 10^{4}} \approx 4399.65 \text{ m/s}\]
06

Conclusion: Final Answer

After completing the calculations, we find that the speed at which the satellite travels is approximately \(4399.65 \text{ m/s}\).

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Key Concepts

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

Gravitational Force
Gravitational force is the attractive force that pulls two masses toward each other. Essentially, every object in the universe that has mass exerts this force. However, we predominantly notice gravitational attraction between large masses, such as planets or stars, due to their considerable impact.
  • The formula for gravitational force between two objects is given by Newton's Law of Universal Gravitation: \[F = G \frac{m_1 m_2}{r^2}\]
  • Here, \(F\) is the gravitational force, \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}\)), \(m_1\) and \(m_2\) are the masses of the objects, and \(r\) is the distance between the centers of the two masses.
When we talk about objects like satellites orbiting Earth, the gravitational force is what keeps them moving in their orbits. This force acts as a centripetal force to ensure that the satellite consistently revolves around the Earth, rather than drifting away into space.
Orbital Speed
Orbital speed refers to the constant speed at which an object travels along its orbital path. For a satellite circling the Earth, this speed ensures that it remains in a stable orbit without descending back to Earth due to gravity or escaping into outer space.
  • The formula used to determine the orbital speed \(v\) of a satellite is:\[v = \frac{2\pi r}{T}\]
  • Where \(r\) is the radius of the orbit, and \(T\) is the orbital period 鈥� the time it takes for the satellite to complete one full orbit.
This formula shows that as the orbit's radius increases, the orbital speed for a given period decreases. Understanding orbital speed is crucial for placing satellites in the correct orbits for tasks like communication or weather observation, ensuring they cover the necessary areas adequately.
Centripetal Force
Centripetal force is the inward force required to keep an object moving in a circular path. Without this force, the object would move off in a straight line due to inertia. In space, gravitational force often acts as the centripetal force for orbiting objects.
  • For a satellite orbiting a planet like Earth, the gravitational pull of the planet provides this centripetal force necessary for circular motion.
  • The relationship can be described using: \[F_c = \frac{mv^2}{r}\]
  • Where \(F_c\) is the centripetal force, \(m\) is the satellite's mass, \(v\) is its orbital speed, and \(r\) is the orbital radius.
The balance between gravitational pull and the satellite's velocity is crucial; if the speed is too high, it might escape orbit, and if it's too low, it could spiral towards Earth. Thus, the interactions leading to centripetal force ensure satellites maintain consistent, stable orbits.
Satellite Motion
Satellite motion involves understanding how satellites orbit larger celestial bodies like Earth. This motion depends heavily on gravitational interactions, orbital mechanics, and the principles of centripetal force. A satellite follows a specific path determined by its speed and the gravitational pull of the planet it orbits. The steps needed to maintain this motion involve balancing forces precisely.
  • Continuous gravitational attraction acts on the satellite, directing it towards the center of the planet and contributing to its orbital stability.
  • To achieve a stable circular orbit, the satellite needs a specific speed matching its altitude, which is calculated to prevent it from crashing into the planet or drifting away.
  • Natural forces like atmospheric drag are minimal in high orbits, allowing the satellite to maintain its trajectory with minimal adjustment.
Understanding satellite motion is key for technologies such as GPS, telecommunications, and Earth observation, impacting daily life and scientific exploration.

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Most popular questions from this chapter

The hammer throw is a track-and-field event in which a 7.3-kg ball (the 鈥渉ammer鈥�) is whirled around in a circle several times and released. It then moves upward on the familiar curving path of projectile motion and eventually returns to earth some distance away. The world record for this distance is 86.75 m, achieved in 1986 by Yuriy Sedykh. Ignore air resistance and the fact that the ball is released above the ground rather than at ground level. Furthermore, assume that the ball is whirled on a circle that has a radius of 1.8 m and that its velocity at the instant of release is directed 41 above the horizontal. Find the magnitude of the centripetal force acting on the ball just prior to the moment of release.

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 kg and is 6.10 m from the pivot. He is skating at a speed of 6.80 m/s. Determine the magnitude of the centripetal force that acts on him.

A car travels at a constant speed around a circular track whose radius is 2.6 km. The car goes once around the track in 360 s. What is the magnitude of the centripetal acceleration of the car?

A jet flying at 123 m/s banks to make a horizontal circular turn. The radius of the turn is 3810 m, and the mass of the jet is \(2.00 \times 10^{5} \mathrm{kg}\). Calculate the magnitude of the necessary lifting force.

Multiple-Concept Example 7 reviews the concepts that play a role in this problem. Car A uses tires for which the coefficient of static friction is 1.1 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 25 m/s. Car B uses tires for which the coefficient of static friction is 0.85 on the same curve. What is the maximum speed at which car B can negotiate the curve?
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