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Chapter 6: Problem 20

A 2150 kg satellite used in a cellular telephone network is in a circular orbit at a height of 780 \(\mathrm{km}\) above the surface of the earth. What is the gravitational force on the satellite? What fraction is this force of the satellite's weight at the surface of the earth?

Short Answer

Expert verified
The gravitational force on the satellite is approximately 18,498.06 N. This force is about 87.73% of the satellite's weight at Earth's surface.

Step by step solution

01

Identify and Define Variables

Given:- Mass of the satellite, \( m = 2150 \, \text{kg} \).- Height of the satellite above Earth's surface, \( h = 780 \, \text{km} = 780,000 \, \text{m} \).- Radius of the Earth, \( R = 6,371,000 \, \text{m} \).- Gravitational constant, \( G = 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \).- Mass of the Earth, \( M_E = 5.972 \times 10^{24} \, \text{kg} \).
02

Calculate the Distance from the Center of the Earth

The distance from the center of the Earth to the satellite is the sum of Earth's radius and the satellite's height above the surface. Thus, the distance \( r \) is given by:\[ r = R + h = 6,371,000 \, \text{m} + 780,000 \, \text{m} = 7,151,000 \, \text{m} \]
03

Use the Gravitational Force Equation

The gravitational force \( F \) acting on the satellite is calculated using Newton's law of universal gravitation:\[ F = \frac{G M_E m}{r^2} \]Substitute the known values to calculate \( F \):\[ F = \frac{6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \times 5.972 \times 10^{24} \, \text{kg} \times 2150 \, \text{kg}}{(7,151,000 \, \text{m})^2} \]Calculate \( F \) using a calculator to find:\[ F \approx 18,498.06 \, \text{N} \]
04

Calculate the Satellite's Weight on Earth's Surface

The weight \( W \) of the satellite on the Earth's surface is given by:\[ W = m g \]where \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity on Earth's surface).\[ W = 2150 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 21,070 \, \text{N} \]
05

Determine the Fraction of Force to Weight

To determine what fraction of the satellite's weight at the Earth's surface is exerted as gravitational force in orbit, calculate:\[ \text{Fraction} = \frac{F}{W} = \frac{18,498.06 \, \text{N}}{21,070 \, \text{N}} \approx 0.8773 \]

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Key Concepts

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

Satellite Motion
A satellite is an object that orbits a planet in space. For a satellite like in our exercise, which is used in a cellular phone network, maintaining a stable orbit is vital. The forces acting on it need to be balanced. One essential aspect of satellite motion is the balance between gravitational pull and the satellite's tendency to move in a straight line at constant speed.
This balance keeps the satellite in a consistent circular orbit. The velocity it needs to maintain this motion depends on the height of the orbit.
  • If the satellite moves too slowly, Earth's gravity can pull it back, potentially causing it to fall into the atmosphere.
  • If it moves too fast, it might escape gravitational pull and drift away into space.
Thus, orbit dynamics are precisely calculated to ensure the satellite remains on its intended path, employing principles of orbital mechanics.
Universal Gravitation
Universal gravitation is the concept introduced by Sir Isaac Newton. It explains how every mass in the universe exerts a pull on every other mass. This pull is called gravity. Newton's Law of Universal Gravitation provides the formula to quantify this force:
\[ F = \frac{G M_1 M_2}{r^2} \]Where:
  • \( F \) is the gravitational force between two masses.
  • \( G \) is the gravitational constant, \( 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \).
  • \( M_1 \) and \( M_2 \) are the masses involved.
  • \( r \) is the distance between the centers of the two masses.
In the exercise, this principle is employed to calculate the gravitational force acting on the satellite. Here, \( M_1 \) is the Earth's mass and \( M_2 \) is the satellite's mass. The equation shows how gravitational pull decreases with increased distance but is significant enough to keep satellites in orbit.
Earth's Radius
Understanding the Earth's radius is crucial when calculating satellite orbits. In gravitational calculations, the effective distance from the Earth's center to the satellite is vital. The radius of the Earth is approximately \( 6,371,000 \, \text{m} \).
When satellites orbit Earth, we must add their altitude to this radius to determine the complete distance from the Earth's center to the satellite. This value along with the universal gravitation equation helps ensure we get accurate force calculations.
  • For satellites orbiting, consider both the Earth's radius and the satellite's height above Earth's surface.
  • Adding the altitude of 780 km (or \( 780,000 \, \text{m} \)) to the radius, provides the overall distance used in force equations.
This concept is fundamental in designing satellite orbits and ensuring orbits stay operationally viable.
Orbital Mechanics
Orbital mechanics is the study of the physics of moving bodies under the influence of gravitational fields. In this domain, crucial laws like Kepler's laws of planetary motion help in comprehending orbital paths and dynamics.
For artificial satellites, orbital mechanics ensures the correct speed and altitude are maintained for specific missions. Key factors involved include:
  • The gravitational pull from Earth, calculated using universal gravitation.
  • Velocity, which must match a finely-tuned speed to ensure the satellite does not veer off its path.
  • The altitude of the satellite impacts both the gravitational force exerted and the velocity necessary to maintain orbit.
When designing and launching a satellite, these principles ensure the satellite achieves its intended purpose, such as providing communication links for cellular networks in our exercise scenario. These calculations enable precise predictions and adjustments, preserving the satellite in its path over time.

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Most popular questions from this chapter

\(\bullet\) The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 \(\mathrm{m}\) . Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 \(\mathrm{s} )\) . (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs 882 \(\mathrm{N}\) at the weight- guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

Stunt pilots and fighter pilots who fly at high speeds in a downward-curving arc may experience a "red out," in which blood is forced upward into the flier's head, potentially swelling or breaking capillaries in the eyes and leading to a reddening of vision and even loss of consciousness. This effect can occur at centripetal accelerations of about 2.5\(g^{\prime}\) s. For a stunt plane flying at a speed of 320 \(\mathrm{km} / \mathrm{h}\) , what is the minimum radius of downward curve a pilot can achieve without experiencing a red out at the top of the arc? (Hint: Remember that gravity provides part of the centripetal acceleration at the top of the arc; it's the acceleration required in excess of gravity that causes this problem.)

A small button placed on a horizontal rotating platform with diameter 0.320 \(\mathrm{m}\) will revolve with the plattorm when it is brought up to a speed of 40.0 \(\mathrm{rev} / \mathrm{min}\) , provided the button is no more than 0.150 \(\mathrm{m}\) from the axis. (a) What is the coefficient of static friction between the button and the platform? (b) How far from the axis can the button be placed, without slipping, if the platform rotates at 60.0 rev/min?

A \(\mathrm{A} 50.0 \mathrm{kg}\) stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane. (a) If the plane's speed at the lowest point of the circle is \(95.0 \mathrm{m} / \mathrm{s},\) what should the minimum radius of the circle be in order for the centripetal acceleration at this point not to exceed 4.00 \(\mathrm{g}\) ? (b) What is the apparent weight of the pilot at the lowest point of the pullout?

Artificial gravity in space stations. One problem for humans living in outer space is that they are apparently weight- less. One way around this problem is to design a cylindrical space station that spins about an axis through its center at a constant rate. (See Figure \(6.32 . )\) This spin creates "artificial gravity" at the outside rim of the station. (a) If the diameter of the space station is \(800.0 \mathrm{m},\) how fast must the rim be moving in order for the "artificial gravity" acceleration to be \(g\) at the outer rim? (b) If the space station is a waiting area for travelers going to Mars, it might be desirable to simulate the acceleration due to gravity on the Martian surface. How fast must the rim move in this case? (c) Make a free-body diagram of an astronaut at the outer rim.
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