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A projectile is fired vertically from Earth's surface with an initial speed of \(10 \mathrm{~km} / \mathrm{s}\). Neglecting air drag, how far above the surface of Earth will it go?

Short Answer

Expert verified
The projectile will reach approximately 13,000 kilometers above Earth's surface.

Step by step solution

01

Understanding the Problem

We need to find the maximum height a projectile reaches when fired vertically from the Earth's surface with an initial speed of \(10 \, \text{km/s}\). We will use the conservation of energy principle to find this maximum height.
02

Applying Conservation of Energy

The total mechanical energy at the Earth's surface consists of kinetic energy and potential energy. At maximum height, the kinetic energy is zero, and we have only potential energy. The energy conservation equation is \( \frac{1}{2}mv_i^2 = GMm \left( \frac{1}{R} - \frac{1}{R + h} \right) \), where \( m \) is the mass of the projectile, \( v_i = 10,000 \, \text{m/s} \) is the initial velocity, \( G = 6.674 \times 10^{-11} \, \text{N(m/kg)}^2 \) is the gravitational constant, \( M = 5.972 \times 10^{24} \, \text{kg} \) is the Earth's mass, \( R = 6.371 \times 10^6 \, \text{m} \) is the Earth's radius, and \( h \) is the maximum height above Earth's surface.
03

Simplifying the Equation

Since \( m \) appears on both sides of the equation, we can cancel it out. Rearrange the equation to solve for \( h \): \( \frac{v_i^2}{2G M} + \frac{1}{R} = \frac{1}{R + h} \). Our task is to solve this equation for \( h \).
04

Solving for Maximum Height \( h \)

Rearrange the equation to isolate \( \frac{1}{R + h} \): \[ \frac{1}{R + h} = \frac{1}{R} + \frac{v_i^2}{2G M} \]. Plug in numerical values: \( v_i = 10,000 \, \text{m/s}, \ G = 6.674 \times 10^{-11} \, \text{N(m/kg)}^2, \ M = 5.972 \times 10^{24} \, \text{kg}, \ R = 6.371 \times 10^6 \, \text{m} \). Compute: \( \frac{1}{R + h} = \frac{1}{6.371 \times 10^6} + \frac{(10,000)^2}{2 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24}} \).
05

Calculating \( h \)

Calculate \( \frac{1}{R + h} \) with the given values. This gives us a small number which we can then take the reciprocal of to find \( R + h \). Finally, compute: \( h = (R + h) - R \) to find the maximum height above Earth's surface.
06

Final Calculation and Answer

Upon performing the calculations: \( \frac{1}{R + h} \approx 1.567 \times 10^{-7} \). Taking the reciprocal gives \( R + h = 6.384 \times 10^7 \). Thus, \( h = 6.384 \times 10^7 - 6.371 \times 10^6 = 1.3 \times 10^7 \) meters.

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is essential in understanding projectile motion, particularly when analyzing the maximum height a projectile can reach when fired from Earth's surface. This principle states that the total energy of an isolated system remains constant over time. For a projectile fired into the sky, this means that the sum of its kinetic and potential energies at launch is equal to the sum of its energies at its peak height.

When the projectile is launched, it possesses a high amount of kinetic energy and relatively low gravitational potential energy. As it ascends, its kinetic energy decreases because it's moving against gravitational pull, while its potential energy increases. At the maximum height, all the initial kinetic energy has been converted into potential energy.

By setting the initial total energy (kinetic plus potential) equal to the total potential energy at the peak, we can derive expressions to calculate maximum heights. This foundational concept avoids needing to know the specific forces acting at each point during the projectile's flight and simplifies computations dramatically.
Gravitational Potential Energy
Gravitational potential energy (GPE) plays a significant role whenever we're dealing with heights and gravity. It's the energy an object possesses due to its position in a gravitational field, which, in our exercise, is Earth's gravity. The formula for GPE is: \[ PE = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) \]
  • GMm: Represents the gravitational interaction between the Earth and the projectile.
  • \(\frac{1}{R}\): Describes the object's initial position, or the potential energy at the surface of Earth.
  • \(\frac{1}{R+h}\): Indicates the object's position when it reaches maximum height.
As the projectile rises, it moves higher in Earth's gravitational field, resulting in increased gravitational potential energy. This means its initial kinetic energy transforms into gravitational potential energy until maximum height is achieved. By understanding GPE's role, you recognize that the potential energy accounts for the changes in height during projectile motion.
Kinetic Energy
Kinetic energy (KE) concerns the energy an object possesses due to its motion. For our projectile, this energy comes from its initial velocity as it's fired up into the sky. The expression for kinetic energy is \[ KE = \frac{1}{2}mv^{2} \]where:
  • m: Is the mass of the projectile.
  • v: Is the velocity of the projectile.
Kinetic energy is highest at the launch because the projectile is moving fastest at this point. As the projectile travels upward, the forces of gravity slow it down, causing the kinetic energy to diminish until it reaches zero at the maximum height. By starting with this kinetic energy, we can calculate how far the projectile will ascend by using the conservation of energy. Recognizing this energy transition offers insights into how projectile motion proceeds against gravity.
Earth's Gravity
Earth's gravity is the force that processes the entire framework of projectile motion. This constant force pulls objects toward the center of the Earth, and it's why objects in free-fall accelerate at approximately 9.81 m/s². However, for objects traveling significant distances from the surface, such as our projectile, the simplification of a constant gravitational force is replaced by applying the gravitational law with mass and radius.

The force of gravity not only influences how far the projectile travels but also how energy converts from kinetic to potential. It acts as a retarding force that continuously decelerates the projectile on its upward path, converting the kinetic energy into gravitational potential energy until peak height is achieved. Understanding Earth's gravity thereby ties together all the concepts of energy, emphasizing why the projectile eventually stops ascending and begins to fall back.

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