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A satellite is in a circular orbit about the earth \(\left(M_{\mathrm{E}}=5.98 \times 10^{24} \mathrm{~kg}\right) .\) 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 3.07 km/s.

Step by step solution

01

Determine the gravitational constant and Earth's mass

We will need the gravitational constant (G) which is approximately \(6.674 \times 10^{-11} \ \text{Nm}^2/ ext{kg}^2\), and the mass of Earth \(M_{E} = 5.98 \times 10^{24} \ \text{kg}\). These values are necessary to apply Kepler's third law in the next step.
02

Apply Kepler's Third Law to determine the orbital radius

Kepler’s Third Law can be stated as: \(T^2 = \frac{4\pi^2}{GM_{E}}r^3\), where \(T\) is the period of the orbit, and \(r\) is the orbital radius. \(T = 1.20 \times 10^4 \ \text{s}\). Rearrange the formula to solve for \(r\): \[r = \left( \frac{T^2 \cdot G \cdot M_{E}}{4\pi^2} \right)^{1/3}\]. Substitute the known values to find \(r\).
03

Calculate the orbital radius

Using the rearranged formula from Step 2, calculate \[r = \sqrt[3]{\frac{(1.20 \times 10^4 \ \text{s})^2 \cdot 6.674 \times 10^{-11} \cdot 5.98 \times 10^{24}}{4\pi^2}}\]The calculation yields the orbital radius. Compute this value.
04

Calculate the orbital speed

Once you know the orbital radius, use the formula for the orbital speed \(v\), which is given by \[v = \frac{2\pi r}{T}\].Substitute the calculated \(r\) and given \(T = 1.20 \times 10^4 \ \text{s}\) into the formula to find the speed \(v\).
05

Compute the orbital speed

Using the values from the previous steps for \(r\) and \(T\): \[v = \frac{2\pi ( \text{calculated value of } r) }{1.20 \times 10^4}\].Perform the calculation to find the speed of the satellite.

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Key Concepts

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

Kepler's Third Law
Kepler's Third Law is an essential concept in orbital mechanics. It provides a relationship between the period of orbit and the radius of the orbit for objects in space. Specifically, it states that the square of the orbital period (T^2) is directly proportional to the cube of the semi-major axis of the orbit (r^3). For a satellite orbiting Earth, this can be expressed mathematically as:
  • \( T^2 = \frac{4\pi^2}{GM_{E}}r^3 \)
where \(G\) is the gravitational constant and \(M_{E}\) is the mass of the Earth.
This law is useful because it allows us to predict how long an object will take to orbit Earth based on its orbital radius. By rearranging the equation, we can solve for the orbital radius when we know the period.
Using these calculations, we can determine other properties of the orbit.
Gravitational Constant
The gravitational constant (G) is a fundamental constant in physics, crucial for calculations involving gravitational forces. It is approximately \(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\).This constant is part of Newton's law of universal gravitation, which helps us calculate the gravitational attraction between two masses. Specifically, it allows us to understand how mass and distance influence the gravitational force.
  • G helps determine the force between large masses, like Earth and any satellite orbiting it.
  • It's a key player in Kepler’s Third Law, linking the gravitational pull from Earth to the satellite’s orbit.
Understanding G helps us better grasp the forces affecting satellite motion and how these forces affect the speed and trajectory of satellites in space.
Orbital Speed
Orbital speed is the speed at which an object travels around another object in space, such as a satellite orbiting Earth. The orbital speed can be derived from the radius of the orbit and the period of the satellite.
In the case of a circular orbit, this is calculated using the following formula:
  • \( v = \frac{2\pi r}{T} \)
where \(r\) is the radius of the orbit and \(T\) is the orbital period.
To find the speed, first, calculate the orbital radius using the rearranged equation from Kepler's Third Law, and then plug it into the formula above.
Orbital speed is crucial because it determines how quickly a satellite can complete its orbit, influencing everything from the timing of data collection to the stability of its path.
Satellite Motion
Satellite motion refers to the movement of a satellite as it orbits around a planet like Earth. Gravitational forces primarily govern this motion.
Satellites maintain their orbits due to the balance between gravity pulling them toward the planet and their momentum carrying them forward.
  • The gravitational force provides the necessary centripetal force to keep the satellite in a stable orbit.
  • Momentum keeps the satellite moving along its orbital path without crashing into Earth.
By understanding satellite motion, engineers can design orbits that meet specific needs, such as geostationary or polar orbits.
Accurate calculations of orbital speed and radius ensure that these artificial satellites function as intended, whether for communication, navigation, or earth observation purposes.

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Most popular questions from this chapter

A centrifuge is a device in which a small container of material is rotated at a high speed on a circular path. Such a device is used in medical laboratories, for instance, to cause the more dense red blood cells to settle through the less dense blood serum and collect at the bottom of the container. Suppose the centripetal acceleration of the sample is \(6.25 \times 10^{3}\) times as large as the acceleration due to gravity. How many revolutions per minute is the sample making, if it is located at a radius of \(5.00 \mathrm{~cm}\) from the axis of rotation?

There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted so that it can be rotated about its axis by turning a hand crank. The outer wall of the cylinder is perforated with small holes. You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 12 \(\mathrm{cm}\). When the cylinder is rotating at 2.0 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?

In an automatic clothes dryer, a hollow cylinder moves the clothes on a vertical circle (radius \(r=0.32 \mathrm{~m}\) ), as the drawing shows. The appliance is designed so that the clothes tumble gently as they dry. This means that when a piece of clothing reaches an angle of \(\theta\) above the horizontal, it loses contact with the wall of the cylinder and falls onto the clothes below. How many revolutions per second should the cylinder make in order that the clothes lose contact with the wall when \(\theta=70.0^{\circ} ?\)

Concept Questions Interactive LearningWare 5.2 at illustrates good problemsolving techniques for this type of problem. Two cars are traveling at the same speed of \(27 \mathrm{~m} / \mathrm{s}\) on a curve that has a radius of \(120 \mathrm{~m}\). Car \(\mathrm{A}\) has a mass of \(1100 \mathrm{~kg},\) and car \(\mathrm{B}\) has a mass of \(1600 \mathrm{~kg}\). Without doing any calculations, decide (a) which car, if either, has the greater centripetal acceleration and (b) which car, if either, experiences the greater centripetal force. Justify your answers.

Reviews the concepts that are involved in this problem. A child is twirling a \(0.0120-\mathrm{kg}\) ball on a string in a horizontal circle whose radius is \(0.100\) \(\mathrm{m}\). The ball travels once around the circle in \(0.500 \mathrm{~s}\). (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, does the centripetal force double? If not, by what factor does the centripetal force increase?

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