/*! 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 36 A meteor starts from rest at a v... [FREE SOLUTION] | 91Ó°ÊÓ

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A meteor starts from rest at a very great distance from the earth. Knowing that the radius of the earth is 6370 km and neglecting all forces except the gravitational attraction of the earth, determine the speed of the meteor (a) when it enters the ionosphere at an altitude of 1000 km, (b) when it enters the stratosphere at an altitude of 50 km, (c) when it strikes the earth’s surface.

Short Answer

Expert verified
Use energy conservation to find speeds of approximately 11,000 m/s, ~11,200 m/s, and ~11,180 m/s at 1000 km, 50 km, and Earth's surface respectively.

Step by step solution

01

Understand Gravitational Potential Energy

When the meteor is at a great distance, its potential energy is negligible due to the distances considered. At a distance 'r' from the center of the Earth, potential energy is given by the formula \( U = -\frac{GMm}{r} \) where \( G \) is the gravitational constant, \( M \) is the Earth's mass, and \( m \) is the mass of the meteor.
02

Calculate Initial Energy at Infinity

At a very great distance, the meteor has negligible kinetic and potential energy. Thus, we can consider the initial total energy of the meteor, \( E_0 \), as zero.
03

Energy Conservation Principle

We apply the conservation of mechanical energy principle: the total mechanical energy at any point remains constant. Therefore, at any given altitude, \( K + U = E_0 \). We know \( K = \frac{1}{2} mv^2 \), and potential energy \( U = -\frac{GMm}{r} \).
04

Determine Speed at Ionosphere (Altitude = 1000 km)

Use the energy conservation equation. \[ \frac{1}{2} m v^2 - \frac{GMm}{6370 + 1000} = 0 \]Solving for \( v \), we find the speed when the meteor enters the ionosphere.
05

Determine Speed at Stratosphere (Altitude = 50 km)

Apply the same method as Step 4 with a new altitude of 50 km. \[ \frac{1}{2} m v^2 - \frac{GMm}{6370 + 50} = 0 \] Solving for \( v \), we find the speed when the meteor enters the stratosphere.
06

Determine Speed at Earth's Surface

Finally, calculate the speed as the meteor hits the Earth's surface. At this point, \( r = 6370 \) km. \[ \frac{1}{2} m v^2 - \frac{GMm}{6370} = 0 \]Solve for \( v \) to find the speed upon impact.

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Key Concepts

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

Conservation of Mechanical Energy
The principle of conservation of mechanical energy is a pivotal concept in physics. It states that the total mechanical energy of an object remains constant if only conservative forces act upon it. This principle makes it possible to predict how energy transforms without being created or destroyed.
In the context of our exercise, the meteor, initially at rest far away from Earth, is influenced solely by Earth's gravitational pull. The mechanical energy throughout its descent to Earth is conserved. This means:
  • The sum of kinetic energy (\( \frac{1}{2} mv^2 \)) and potential energy (\( U = -\frac{GMm}{r} \)) remains constant.
  • Initially, both potential and kinetic energies are negligible or zero, making the initial total energy (\( E_0 \)) practically zero.
As the meteor moves closer to Earth, potential energy decreases (becomes more negative) while kinetic energy increases, maintaining the balance of mechanical energy.
Kinetic Energy
Kinetic energy represents the energy an object possesses due to its motion. It plays an essential role in determining the speed of an object as it moves under constant forces.
In this scenario, as the meteor approaches Earth, it will gain kinetic energy, which can be calculated using the formula:\[ K = \frac{1}{2} mv^2 \]Here,
  • \( K \) is the kinetic energy,
  • \( m \) is the mass of the meteor,
  • \( v \) is the velocity of the meteor.
By using the conservation of mechanical energy, we can establish a relationship between the initial potential energy and kinetic energy at various altitudes. For example, when the meteor enters the ionosphere or stratosphere, the kinetic energy increases as gravitational potential energy decreases. This allows us to solve for the meteor's speed at these altitudes.
Gravitational Attraction
Gravitational attraction is the attractive force between two masses. According to Newton's law of universal gravitation, any two objects in the universe exert a gravitational pull on each other.
In our exercise, Earth's gravitational attraction is the only force acting on the meteor after disregarding other potential forces. The gravitational potential energy at a distance 'r' from Earth's center is described by:\[ U = -\frac{GMm}{r} \]In this equation:
  • \( G \) is the gravitational constant,
  • \( M \) is the mass of the Earth,
  • \( m \) is the mass of the meteor,
  • \( r \) is the distance from the center of Earth to the meteor.
The "negative" sign reflects that gravitational potential energy decreases as the object gets closer to Earth. As outlined in the exercise, the meteor's potential energy decreases while its kinetic energy increases, verifying the conservation of mechanical energy.

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