/*! 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 127 The third stage of a rocket fire... [FREE SOLUTION] | 91Ó°ÊÓ

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The third stage of a rocket fired vertically up over the north pole coasts to a maximum altitude of \(500 \mathrm{km}\) following burnout of its rocket motor. Calculate the downward velocity \(v\) of the rocket when it has fallen \(100 \mathrm{km}\) from its position of maximum altitude. (Use the mean value of \(9.825 \mathrm{m} / \mathrm{s}^{2}\) for \(g\) and \(6371 \mathrm{km}\) for the mean radius of the earth.)

Short Answer

Expert verified
The downward velocity is approximately 1401.78 m/s.

Step by step solution

01

Understanding the Problem

We need to calculate the downward velocity of a rocket when it falls 100 km from its maximum altitude of 500 km. We're given gravitational acceleration (\(g = 9.825 \text{ m/s}^2\)) and the radius of the Earth (\(R = 6371 \text{ km}\)).
02

Setup the Energy Conservation Equation

Using the conservation of mechanical energy, the total energy at maximum height (potential energy only) and the total energy at 100 km fall (potential + kinetic energy) should be equal. The equation for potential energy is \( U = mgh \) and kinetic energy is \( K = \frac{1}{2}mv^2 \).
03

Calculate Energy At Maximum Altitude

At the maximum altitude (500 km) potential energy \( U_1 = mg(H_1 + R)\) where \(H_1\) = 500 km converted to meters, \(R = 6371\times 10^3\) meters. Kinetic energy at this point is zero since it's at rest.
04

Calculate Energy After 100 km Descent

At 400 km (500 km - 100 km), the potential energy is \( U_2 = mg(H_2 + R)\) where \(H_2\) is 400 km converted to meters. Here we add kinetic energy since it's moving: \( K_2 = \frac{1}{2}mv^2 \) for velocity \( v \).
05

Setting Up Energy Conservation Equation

Since energy is conserved, set up the equation: \( U_1 = U_2 + K_2 \). This becomes: \[ mg(H_1 + R) = mg(H_2 + R) + \frac{1}{2}mv^2 \] where \( m \) cancels out.
06

Solve for Velocity, \(v\)

Rearrange the equation to solve for \( v \): \[ v = \sqrt{2g(H_1 - H_2)} \] Substitute \( g = 9.825 \text{ m/s}^2\), \( H_1 = 500000 \text{ m}\), \( H_2 = 400000 \text{ m}\) into the equation. So, \( v = \sqrt{2 \times 9.825 \times (500000 - 400000)} = \sqrt{1965000} \approx 1401.78 \text{ m/s} \).

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Key Concepts

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

Energy Conservation
In physics, energy conservation is a fundamental principle that states that the total energy in a closed system remains constant over time. This principle is incredibly useful when analyzing projectile motion, such as the movement of a rocket.

When a rocket reaches its maximum altitude and begins to descend, the energy it possesses transforms but doesn’t disappear. Initially, at the maximum height, the rocket's energy is predominantly potential due to its elevation above the Earth. As it falls, this potential energy is converted to kinetic energy, increasing the rocket's speed as it descends.

## Key Components of Energy Conservation
  • **Mechanical Energy**: The sum of kinetic and potential energy in an object.
  • **Closed System**: No external forces doing work, ensuring energy remains conserved.
Understanding energy conservation allows scientists and engineers to predict the motion and speed of rockets and other projectiles at various points in their trajectories.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It is given by the formula \( K = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. When a rocket is descending from its maximum altitude, it begins to move faster, increasing its kinetic energy.

As the rocket falls from 500 km to 400 km, it loses potential energy, which is converted into kinetic energy, causing an increase in velocity. At this point, the mathematical relationship becomes crucial. By applying the energy conservation principle, we derive the kinetic energy at the lower altitude to calculate the rocket's speed.

## Important Aspects of Kinetic Energy
  • Kinetic energy depends on the square of velocity, so as the speed doubles, kinetic energy quadruples.
  • It is always a positive value, indicative of motion.
This principle can help explain changes in speed during ballistic trajectories where potential energy decreases and is transformed into kinetic energy, obtaining speeds necessary for calculating motion parameters.
Potential Energy
Potential energy is the stored energy of an object due to its position or condition. For a rocket, potential energy is calculated based on its height above the Earth, expressed as \( U = mgh \), where \( m \) is mass, \( g \) is gravitational acceleration, and \( h \) is the height.

In the case of our projectile problem, when the rocket is at its maximum height of 500 km, it has significant potential energy. As it descends by 100 km, a portion of this potential energy converts into kinetic energy. This energy transformation is what results in the increased velocity of the rocket as it nears the ground again.

## Core Elements of Potential Energy
  • Potential energy increases with height. The higher the object, the greater its potential energy.
  • It plays a critical role in energy transformations, especially in gravitational fields.
By understanding potential energy and how it is converted to kinetic energy, we can better comprehend and predict the behaviors of moving objects like rockets as they travel through different altitudes.

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Most popular questions from this chapter

A projectile is launched from the north pole with an initial vertical velocity \(v_{0} .\) What value of \(v_{0}\) will result in a maximum altitude of \(R / 3 ?\) Neglect aerodynamic drag and use \(g=9.825 \mathrm{m} / \mathrm{s}^{2}\) as the surface-level acceleration due to gravity.

A jet-propelled airplane with a mass of \(10 \mathrm{Mg}\) is flying horizontally at a constant speed of \(1000 \mathrm{km} / \mathrm{h}\) under the action of the engine thrust \(T\) and the equal and opposite air resistance \(R\). The pilot ignites two rocket-assist units, each of which develops a forward thrust \(T_{0}\) of \(8 \mathrm{kN}\) for \(9 \mathrm{s}\). If the velocity of the airplane in its horizontal flight is \(1050 \mathrm{km} / \mathrm{h}\) at the end of the \(9 \mathrm{s}\), calculate the timeaverage increase \(\Delta R\) in air resistance. The mass of the rocket fuel used is negligible compared with that of the airplane.

At a steady speed of \(200 \mathrm{mi} / \mathrm{hr}\) along a level track, the racecar is subjected to an aerodynamic force of \(900 \mathrm{lb}\) and an overall rolling resistance of 200 lb. If the drivetrain efficiency is \(e=0.90,\) what power \(P\) must the motor produce?

An electromagnetic catapult system is being designed to replace a steam-driven system on an aircraft carrier. The requirements include accelerating a 12000 -kg aircraft from rest to a speed of \(70 \mathrm{m} / \mathrm{s}\) over a distance of \(90 \mathrm{m} .\) What constant force \(F\) must the catapult exert on the aircraft?

A projectile is launched from \(B\) with a speed of \(2000 \mathrm{m} / \mathrm{s}\) at an angle \(\alpha\) of \(30^{\circ}\) with the horizontal as shown. Determine the maximum altitude \(h_{\max }\).

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