/*! 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) 464,698 J; (b) Ratio is approximately 257.4.

Step by step solution

01

Determine Orbital Speed of Satellite

First, we calculate the orbital speed of the satellite using the formula for circular orbit speed: \[ v = \sqrt{\frac{GM}{R}} \]where \( G \) is the gravitational constant \((6.674 \times 10^{-11} \text{ N m}^2 \text{/kg}^2)\), \( M \) is Earth's mass \((5.972 \times 10^{24} \text{ kg})\), and \( R \) is the distance from Earth's center to the satellite. Since the satellite is 500 km above the surface, we add this to Earth's radius (approximately 6371 km), resulting in \( R = 6871 \times 10^3 \text{ m} \). Plug these numbers into the formula to find \( v \).
02

Calculate Speed

Using the formula: \[ v = \sqrt{\frac{(6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2)(5.972 \times 10^{24} \text{ kg})}{6871 \times 10^3 \text{ m}}} \]we obtain the orbital speed \( v \approx 7616 \text{ m/s} \). This is both the speed of the satellite but initially opposite to the speed of the colliding pellet.
03

Find Velocity of Pellet Relative to Satellite

Because the pellet is moving in the opposite direction, the relative speed \( v_{rel} \) between the pellet and the satellite is the sum of their speeds: \[ v_{rel} = 7616 + 7616 = 15232 \text{ m/s} \].
04

Calculate Kinetic Energy of Pellet

The formula for kinetic energy is:\[ KE = \frac{1}{2}mv^2 \]Substitute \( m = 4.0 \times 10^{-3} \text{ kg} \) (convert grams to kilograms) and \( v_{rel} = 15232 \text{ m/s} \). Thus:\[ KE = \frac{1}{2} (4.0 \times 10^{-3} \text{ kg})(15232)^2 \approx 464,698 \text{ J} \].
05

Calculate Kinetic Energy of Bullet

Similarly, for the bullet, again use:\[ KE_{bullet} = \frac{1}{2}mv^2 \]Substitute \( m = 4.0 \times 10^{-3} \text{ kg} \) and \( v = 950 \text{ m/s} \), so:\[ KE_{bullet} = \frac{1}{2} (4.0 \times 10^{-3} \text{ kg})(950)^2 \approx 1,805 \text{ J} \].
06

Compute Ratio of Kinetic Energies

Finally, find the ratio of the pellet's kinetic energy to the bullet's:\[ \text{Ratio} = \frac{464,698 \text{ J}}{1,805 \text{ J}} \approx 257.4 \].

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Key Concepts

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

Kinetic Energy
Kinetic energy is an essential concept in physics, representing the energy possessed by an object due to its motion. Its calculation is grounded in the formula:
  • \( KE = \frac{1}{2}mv^2 \)
where \( m \) stands for mass and \( v \) for velocity. The kinetic energy depends not only on how heavy an object is but also on how fast it is moving.
For instance, in the case of the pellet in the original exercise, which has a mass of 4 grams, converting it to kilograms by dividing by 1000 results in 0.004 kg. Then by having a substantial relative velocity of 15232 m/s, the kinetic energy becomes significant, calculated as approximately 464,698 J.
Experience with such high kinetic energies is rare on Earth, expressed strikingly by the noticeable difference when compared to everyday things like bullets, which are considerably less energetic.
Relative Velocity
Relative velocity is the velocity of one object as observed from another moving object. This is crucial in understanding how fast two objects approach each other.
In the context of the exercise, the satellite and the pellet move in opposite directions, each at a speed of approximately 7616 m/s. The relative velocity of the pellet concerning the satellite, therefore, becomes the sum of their individual speeds, resulting in a staggering 15232 m/s.
  • Relative velocity is not always additive; it depends on direction.
  • Understanding the direction of motion helps prevent calculation errors.
This concept helps predict the potential impact forces during practical scenarios, like satellite collisions, vastly differing from coordinated movements in the same direction.
Circular Orbit
Satellites often move in circular orbits, defined by a consistent distance from the Earth and a steady speed. Such orbits require the gravitational force acting as the centripetal force that keeps the object in orbit. The speed of an object in a circular orbit is calculated using:
  • \( v = \sqrt{\frac{GM}{R}} \)
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the distance from Earth's center to the satellite.
In the original problem, factoring in Earth's radius plus the satellite's 500 km altitude gives an orbital radius \( R \) of 6871 km. The result is a velocity of 7616 m/s, confirming the satellite is in steady orbit while balancing gravitational pull.
Such orbits are exemplary models for understanding basic astrophysics principles and are pivotal in both communications and scientific research.
Satellite Collision
Satellite collisions are rare yet potentially catastrophic events due to their high velocities. In scenarios like in the exercise, collisions can occur by introducing objects moving in opposite orbits to intersect paths.
Colliding at such tremendous relative speeds means there's substantial energy involved, vastly greater than terrestrial accidents.
  • The kinetic energy upon collision translates into enormous disruptive force.
  • Understanding the forces involved helps improve satellite design and inform space debris management.
Explorations into such collisions improve knowledge of space safety standards and disaster mitigation. Contemplating these factors remains critical as humanity continues to expand its activities in outer space.

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Most popular questions from this chapter

Consider a pulsar, a collapsed star of extremely high density, with a mass \(M\) equal to that of the Sun \(\left(1.98 \times 10^{30} \mathrm{~kg}\right)\), a radius \(R\) of only \(12 \mathrm{~km}\), and a rotational period \(T\) of \(0.041 \mathrm{~s}\). By what percentage does the free-fall acceleration \(g\) differ from the gravitational acceleration \(a_{g}\) at the equator of this spherical star?

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In a double-star system, two stars of mass \(3.0 \times 10^{30} \mathrm{~kg}\) each rotate about the system's center of mass at radius \(1.0 \times 10^{11} \mathrm{~m}\). (a) What is their common angular speed? (b) If a meteoroid passes through the system's center of mass perpendicular to their orbital plane, what minimum speed must it have at the center of mass if it is to escape to "infinity" from the two-star system?

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