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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 \(1.20 \times 10^{4}\) s. What is the speed at which the satellite travels?

Short Answer

Expert verified
The satellite travels at approximately 7.58 km/s.

Step by step solution

01

Identify Given Variables and Constants

We are given the mass of the Earth, \(M_E = 5.98 \times 10^{24}\) kg, and the orbital period of the satellite, \(T = 1.20 \times 10^4\) s. We need to find the speed \(v\) of the satellite.
02

Apply the Orbital Period Formula

The orbital period formula relates the period \(T\), the radius of the orbit \(r\), and the gravitational constant \(G\):\[T = 2\pi \sqrt{\frac{r^3}{GM_E}}\]We will rearrange this to solve for \(r\).
03

Solve for Orbital Radius r

Rearrange the formula:\[r^3 = \left(\frac{T}{2\pi}\right)^2 GM_E\]Substitute the given values and the gravitational constant, \(G = 6.674 \times 10^{-11} \text{m}^3/\text{kg s}^2\), into the formula to calculate \(r\).
04

Calculate the Orbital Radius

Substitute the constants and calculated values:\[r^3 = \left(\frac{1.20 \times 10^4}{2\pi}\right)^2 \times (6.674 \times 10^{-11}) \times (5.98 \times 10^{24})\]Calculate \(r\) by taking the cube root of both sides.
05

Determine the Satellite's Speed

The speed of the satellite in a circular orbit is given by:\[v = \frac{2\pi r}{T}\]Substitute the calculated \(r\) and given \(T\) to find \(v\).
06

Calculate the Speed

Insert the values:\[v = \frac{2\pi \times (calculated \ r)}{1.20 \times 10^4}\]Perform the calculation to find the speed \(v\).

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Key Concepts

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

Satellite Speed
The speed of a satellite in orbit is crucial for maintaining its path around a planet. Speed, in this context, is the rate at which a satellite travels along its circular path. For a satellite in a circular orbit, speed can be determined using the equation:
  • \(v = \frac{2\pi r}{T}\)
This formula tells us that speed \(v\) relies on two main factors:
  • The radius \(r\) of the orbit, which is the distance from the center of the Earth to the satellite.
  • The orbital period \(T\), which is the time taken for one complete revolution around the Earth.
The greater the radius, the faster the satellite needs to travel to maintain its path without being pulled back by Earth's gravity. Similarly, a shorter orbital period demands a higher speed. By calculating the speed, scientists ensure that the satellite doesn't fall back to Earth or drift into space.
Orbital Period
The orbital period is the time it takes for a satellite to make one complete orbit around the Earth. This period depends heavily on the gravitational pull of the Earth and the distance from the Earth to the satellite.

For a circular orbit, the orbital period can be calculated using:
  • \(T = 2\pi \sqrt{\frac{r^3}{GM_E}}\)
Here:
  • \(r\) is the radius of the orbit,
  • \(G\) is the gravitational constant, and
  • \(M_E\) is the mass of the Earth.
The gravitational constant \(G\) is a fixed value that reflects the strength of gravity's pull. By knowing Earth's mass and the orbit's radius, we can determine how long it will take for a satellite to circle the planet. This calculation is crucial for timing satellite communications and predicting satellite positions.
Newton's Law of Universal Gravitation
Newton's law of universal gravitation is foundational for understanding how objects in space interact. According to this law, every point mass attracts every other point mass by a force acting along the line intersecting both points. The formula is:
  • \(F = G \frac{m_1 m_2}{r^2}\)
Where:
  • \(F\) is the force between the masses,
  • \(G\) is the gravitational constant,
  • \(m_1\) and \(m_2\) are the masses of two objects, and
  • \(r\) is the distance between the centers of the two masses.
For satellites orbiting Earth, \(m_1\) is often the mass of the Earth, and \(m_2\) is the mass of the satellite. This law helps us calculate gravitational forces and ensure that satellites remain in stable orbits.
Circular Orbits
Circular orbits are a special type of orbit where a satellite moves around the Earth in a perfect circle. This pathway is beneficial for maintaining a constant distance and speed, which is ideal for certain satellite functions such as GPS and communication satellites. The motion in a circular orbit requires a precise balance between gravitational forces and the satellite's speed.

To achieve a circular orbit, a satellite must have a specific speed that corresponds to its orbit's radius. This synchronization ensures that the gravitational pull is perfectly balanced by the inertia of the satellite moving forward. Any deviation from this balance can lead to an elliptical orbit or, if disrupted considerably, the satellite might descend towards Earth or escape into space. Understanding circular orbits is essential for designing satellite missions and ensuring their longstanding success.

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Most popular questions from this chapter

A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius \(r .\) A passenger feels the seat of the car pushing upward on her with a force equal to twice her weight as she goes through the dip. If \(r=20.0 \mathrm{m},\) how fast is the roller coaster traveling at the bottom of the dip?

Computer-controlled display screens provide drivers in the Indianapolis 500 with a variety of information about how their cars are performing. For instance, as a car is going through a turn, a speed of \(221 \mathrm{mi} / \mathrm{h}\) \((98.8 \mathrm{m} / \mathrm{s})\) and centripetal acceleration of \(3.00 \mathrm{g}\) (three times the acceleration due to gravity) are displayed. Determine the radius of the turn (in meters).

A satellite circles the earth in an orbit whose radius is twice the earth's radius. The earth's mass is \(5.98 \times 10^{24} \mathrm{kg}\), and its radius is \(6.38 \times 10^{6} \mathrm{m}\). What is the period of the satellite?

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 \(80.0 \mathrm{kg}\) and is \(6.10 \mathrm{m}\) from the pivot. He is skating at a speed of \(6.80 \mathrm{m} / \mathrm{s}\). Determine the magnitude of the centripetal force that acts on him.

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 \(\theta=70.0^{\circ} ?\)

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