/*! 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 66 One way to attack a satellite in... [FREE SOLUTION] | 91Ó°ÊÓ

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One way to attack a satellite in Earth orbit is to launch a swarm of pellets in the same orbit as the satellite but in the opposite direction. Suppose a satellite in a circular orbit \(500 \mathrm{~km}\) above Earth's surface collides with a pellet having mass \(4.0 \mathrm{~g}\). (a) What is the kinetic energy of the pellet in the reference frame of the satellite just before the collision? (b) What is the ratio of this kinetic energy to the kinetic energy of a \(4.0 \mathrm{~g}\) bullet from a modern army rifle with a muzzle speed of \(950 \mathrm{~m} / \mathrm{s} ?\)

Short Answer

Expert verified
(a) 465 kJ, (b) Ratio: 258.3

Step by step solution

01

Determine Satellite's Orbital Speed

Calculate the orbital speed of the satellite using the formula for velocity in a circular orbit: \( v = \sqrt{\frac{GM}{R}} \). Here, \( G = 6.674 \times 10^{-11} \mathrm{~m^3~kg^{-1}~s^{-2}} \) is the gravitational constant, and \( M = 5.972 \times 10^{24} \mathrm{~kg}\) is Earth's mass. The radius \( R = 6371 \text{ km} + 500 \text{ km} \), converted to meters gives \( 6871000 \text{ m} \). Thus, \( v \approx 7627 \text{ m/s} \).
02

Calculate Relative Velocity

The relative velocity of the pellet to the satellite is twice the orbital speed because they travel in opposite directions: \( v_{rel} = 2 \times 7627 \text{ m/s} = 15254 \text{ m/s}. \)
03

Calculate Kinetic Energy of Pellet

Use the kinetic energy formula \( KE = \frac{1}{2}mv^2 \). The mass \( m \) of the pellet is \( 4.0 \times 10^{-3} \text{ kg} \). Substitute \( m = 4.0 \times 10^{-3} \text{ kg} \) and \( v_{rel} = 15254 \text{ m/s} \) into the formula to find \( KE = \frac{1}{2} \times 4.0 \times 10^{-3} \times (15254)^2 = 465 \text{ kJ}. \)
04

Calculate Kinetic Energy of a Bullet

For the bullet, use the same kinetic energy formula. The mass of the bullet is \( 4.0 \times 10^{-3} \text{ kg} \) and its speed is \( 950 \text{ m/s} \). Substitute these values: \( KE_{bullet} = \frac{1}{2} \times 4.0 \times 10^{-3} \times 950^2 = 1.8 \text{ kJ}. \)
05

Compute the Ratio of Kinetic Energies

The ratio of the kinetic energy of the pellet to that of the bullet is \( \frac{KE_{pellet}}{KE_{bullet}} = \frac{465}{1.8} \approx 258.3. \)

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Key Concepts

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

Orbital Mechanics
Orbital mechanics is a fascinating field that deals with the motion of objects in space, especially under the influence of gravity. It essentially teaches us how planets, satellites, and other celestial bodies move. When we talk about satellites orbiting Earth, they move along a path dictated by the balance between gravitational forces pulling them towards Earth and their inertia trying to continue in a straight line.

For a satellite in a circular orbit, its velocity is crucial. This velocity can be calculated using the formula given by orbital mechanics: \[v = \sqrt{\frac{GM}{R}} \]where
  • \( G \) is the gravitational constant,
  • \( M \) is the mass of Earth,
  • \( R \) is the distance from the center of Earth to the satellite.
With this formula, we can determine how fast the satellite travels around Earth.

Understanding orbital mechanics is essential not only for satellite technology but also for planning space missions and predicting the behavior of celestial bodies.
Gravitational Constant
The gravitational constant, often denoted as \( G \), is a fundamental value in physics that facilitates the calculation of gravitational forces. Its approximate value is \( 6.674 \times 10^{-11} \, \mathrm{m^3\,kg^{-1}\,s^{-2}} \). This tiny number appears in Newton's law of universal gravitation, which is expressed as:\[F = \frac{G \cdot m_1 \cdot m_2}{r^2}\]
  • \( F \) is the gravitational force between two masses,
  • \( m_1 \) and \( m_2 \) are the two masses,
  • \( r \) is the distance between the centers of the masses.
The gravitational constant makes it possible to calculate how strong the force of gravity is between two objects. In the context of space and satellites, \( G \) is vital as it helps determine the gravitational pull that keeps satellites in orbit around the Earth. Without this constant, understanding and predicting the motion of celestial bodies would be much more complex.

Knowing \( G \) allows scientists and engineers to not only keep satellites in the correct orbits but also to plan interplanetary missions.
Relative Velocity
Relative velocity refers to the velocity of an object as observed from another moving object. This concept is crucial when dealing with bodies moving in different directions, such as a satellite and a pellet traveling in opposite orbits.

For objects in space, determining relative velocity helps us understand how quickly two objects will meet or collide. For the satellite and pellet scenario we discussed, the relative velocity is calculated by adding their speeds due to opposite directions:\[v_{rel} = v_{satellite} + v_{pellet}\]For a satellite orbiting Earth, its speed can be found using the circular orbital speed formula. If the pellet is moving in the opposite direction along the same path, the relative velocity becomes twice the satellite's orbital speed.

Understanding relative velocity is crucial for predicting the outcomes of potential collisions in space, making it a key concept in maintaining safe and efficient space travel and operations.

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Most popular questions from this chapter

An asteroid, whose mass is \(2.0 \times 10^{-4}\) times the mass of Earth, revolves in a circular orbit around the Sun at a distance that is twice Earth's distance from the Sun. (a) Calculate the period of revolution of the asteroid in years. (b) What is the ratio of the kinetic energy of the asteroid to the kinetic energy of Earth?

The Martian satellite Phobos travels in an approximately circular orbit of radius \(9.4 \times 10^{6} \mathrm{~m}\) with a period of \(7 \mathrm{~h} 39 \mathrm{~min}\). Calculate the mass of Mars from this information.

A satellite, moving in an elliptical orbit, is \(360 \mathrm{~km}\) above Earth's surface at its farthest point and \(180 \mathrm{~km}\) above at its closest point. Calculate (a) the semimajor axis and (b) the eccentricity of the orbit.

Some people believe that the Moon controls their activities. If the Moon moves from being directly on the opposite side of Earth from you to being directly overhead, by what percent does (a) the Moon's gravitational pull on you increase and (b) your weight (as measured on a scale) decrease? Assume that the Earth-Moon (center-to-center) distance is \(3.82 \times 10^{8} \mathrm{~m}\) and Earth's radius is \(6.37 \times 10^{6} \mathrm{~m} .\)

A satellite is in elliptical orbit with a period of \(8.00 \times 10^{4} \mathrm{~s}\) about a planet of mass \(7.00 \times 10^{24} \mathrm{~kg} .\) At aphelion, at radius \(4.5 \times 10^{7} \mathrm{~m},\) the satellite's angular speed is \(7.158 \times 10^{-5} \mathrm{rad} / \mathrm{s} .\) What is its angular speed at perihelion?

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