/*! 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 115 A missile is fired from the grou... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 115

A missile is fired from the ground with an initial velocity \(v_{0}\) forming an angle \(\phi_{0}\) with the vertical. If the missile is to reach a maximum altitude equal to \(\alpha R,\) where \(R\) is the radius of the earth, \((a)\) show that the required angle \(\phi_{0}\) is defined by the relation $$\sin \phi_{0}=(1+\alpha) \sqrt{1-\frac{\alpha}{1+\alpha}\left(\frac{v_{\mathrm{esc}}}{v_{0}}\right)^{2}}$$ where \(v_{\mathrm{esc}}\) is the escape velocity, ( \(b\) ) determine the range of allowable values of \(v_{0}\).

Short Answer

Expert verified
(a) Shown by conservation of energy. (b) \( v_0 \geq v_{\mathrm{esc}} \sqrt{\frac{\alpha}{1+\alpha}} \).

Step by step solution

01

Understanding Maximum Altitude Condition

The condition that the missile reaches a maximum altitude equal to \(\alpha R\) means that the energy conservation principle should be applied, where kinetic plus potential energy at launch equals potential energy at maximum altitude.
02

Express Energy at Launch

At launch from the ground, the missile's total energy consists of kinetic energy \( \frac{1}{2}mv_{0}^{2} \) and potential energy \( -\frac{GMm}{R} \), where \( M \) is the Earth's mass and \( G \) is the gravitational constant.
03

Express Energy at Maximum Altitude

At the maximum altitude \( \alpha R \), the missile's energy is purely potential \( -\frac{GMm}{(1+\alpha)R} \). The missile has zero kinetic energy at this point if it stops at the top momentarily.
04

Apply Conservation of Energy

By conservation of energy, equate the total energy at the launch point to that at maximum altitude: \[ \frac{1}{2}mv_{0}^{2} - \frac{GMm}{R} = -\frac{GMm}{(1+\alpha)R} \]. This simplifies to \[ \frac{1}{2}v_{0}^{2} = \frac{GM}{R} \left( \frac{1}{1+\alpha} - 1\right) \].
05

Relate Escape Velocity

The escape velocity is defined by \( v_{\mathrm{esc}} = \sqrt{\frac{2GM}{R}} \), so we can express \( \frac{GM}{R} = \frac{1}{2}v_{\mathrm{esc}}^{2} \). Substitute this into the previous equation: \[ \frac{1}{2}v_{0}^{2} = \frac{1}{2}v_{\mathrm{esc}}^{2} \frac{-\alpha}{1+\alpha} \].
06

Solve for Kinetic Energy Portion Related to Angle

The kinetic energy depends on the angle \( \phi_0 \) such that the vertical component of the initial velocity allows for reaching altitude \( \alpha R \). We need the vertical component \( v_{0}\cos \phi_{0} \) to satisfy this energy condition.
07

Solve for \( \sin \phi_0 \)

Substitute and rearrange the equation for energy: \( v_{0}^{2} \left(\sin^{2} \phi_0 \right) = v_{0}^{2} - v_{\mathrm{esc}}^{2} \frac{\alpha}{1+\alpha} \). Thus, \( \sin^{2} \phi_0 = 1 - \frac{\alpha}{1+\alpha} \left(\frac{v_{ ext{esc}}}{v_0}\right)^{2} \). Solving gives the desired expression: \( \sin \phi_0 = (1 + \alpha) \sqrt{1 - \frac{\alpha}{1+\alpha} \left(\frac{v_\mathrm{esc}}{v_0}\right)^2} \).
08

Determine Allowable Values of \(v_0\)

The velocity \(v_0\) must be such that the term under the square root in the expression for \(\sin \phi_0\) is non-negative: \( 1 - \frac{\alpha}{1+\alpha} \left(\frac{v_\mathrm{esc}}{v_0}\right)^2 \geq 0 \). Solving this inequality provides the allowable range: \( v_0^2 \geq \frac{\alpha}{1+\alpha} v_\mathrm{esc}^2 \). Thus, \( v_0 \geq v_\mathrm{esc} \sqrt{\frac{\alpha}{1+\alpha}} \).

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Key Concepts

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

Energy Conservation
The concept of energy conservation is key to understanding projectile motion, especially for something like a missile being launched into the air. Energy conservation tells us that the total energy in an isolated system remains constant over time. In this case, it's all about balancing the missile's kinetic and potential energy.
  • Initially, the missile has kinetic energy due to its velocity and potential energy because of its position relative to Earth.
  • As it ascends, its potential energy increases while its kinetic energy decreases, until it reaches its maximum altitude.
At maximum altitude, all the initial kinetic energy is converted into potential energy. Using this energy shift enables us to derive equations relating to motion and forces acting on the projectile, ensuring its successful reach of the desired height.
Initial Velocity
Initial velocity is where our projectile motion starts. It鈥檚 essential because it determines how far and how high the projectile will go. In this context, it鈥檚 the speed of the missile when it first leaves the ground. The initial velocity often has two components:
  • Horizontal component, affected by the angle at launch.
  • Vertical component, also affected by the angle, but crucial for altitude.
The vertical component of the initial velocity, expressed as \(v_0 \cos \phi_0\), is critical for reaching the desired altitude. It dictates if enough energy is available to fight against gravity and reach the maximum height of \(\alpha R\). Both the magnitude and direction matter, and tweaking either can drastically change the trajectory.
Maximum Altitude
Achieving maximum altitude is crucial as it represents the furthest vertical distance the missile can travel from Earth. In this problem, the target altitude is given as \(\alpha R\), or some fraction of Earth's radius.
  • To reach this height, the missile must convert all its initial kinetic energy into potential energy.
  • The point at maximum altitude is where the kinetic energy is theoretically zero.
This point demonstrates the limits of projectile motion under given conditions. Understanding how much energy is needed to reach a certain height can help predict whether a projectile will hit its intended mark or fall short.
Escape Velocity
Escape velocity is a unique and exciting concept. It鈥檚 the speed needed for an object to break free from Earth's gravitational pull without any further propulsion. For Earth, escape velocity is about 11.2 km/s.
  • In this exercise, escape velocity is used as a reference point to set upper bounds on the missile's launch requirements.
  • If initial velocity is too close to escape velocity, the missile could exit Earth's orbit instead of reaching the desired altitude.
Understanding escape velocity helps define the required speed and energy for the missile to enter a specific orbit or reach a certain height without overshooting. It's a critical factor in ensuring that the missile meets its destination, be it the maximum altitude or staying within Earth's gravitational sphere.

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Most popular questions from this chapter

A roller coaster starts from rest at \(A,\) rolls down the track to \(B\), describes a circular loop of 40 -ft diameter, and moves up and down past point \(E .\) Knowing that \(h=60 \mathrm{ft}\) and assuming no energy loss due to friction, determine \((a)\) the force exerted by his seat on a 160 -lb rider at \(B\) and \(D,(b)\) the minimum value of the radius of curvature at \(E\) if the roller coaster is not to leave the track at that point.

A 4-lb collar can slide without friction along a horizontal rod and is released from rest at A. The undeformed lengths of springs BA and CA are 10 in. and 9 in., respectively, and the constant of each spring is 2800 lb/in. Determine the velocity of the collar when it has moved 1 in. to the right.

In anticipation of a long \(6^{\circ}\) upgrade, a bus driver accelerates at a constant rate from \(80 \mathrm{km} / \mathrm{h}\) to \(100 \mathrm{km} / \mathrm{h}\) in 8 s while still on a level section of the highway. Knowing that the speed of the bus is \(100 \mathrm{km} / \mathrm{h}\) as it begins to climb the grade at time \(t=0\) and that the driver does not change the setting of the throttle or shift gears, determine (a) the speed of the bus when \(t=10 \mathrm{s},(b)\) the time when the speed is \(60 \mathrm{km} / \mathrm{h}\).

A 2-kg collar is attached to a spring and slides without friction in a vertical plane along the curved rod ABC. The spring is undeformed when the collar is at C and its constant is 600 N/m. If the collar is released at A with no initial velocity, determine its velocity (a) as it passes through B, (b) as it reaches C.

A satellite is projected into space with a velocity \(\mathbf{v}_{0}\) at a distance \(r_{0}\) from the center of the earth by the last stage of its launching rocket. The velocity \(\mathbf{v}_{0}\) was designed to send the satellite into a circular orbit of radius \(r_{0}\). However, owing to a malfunction of control, the satellite is not projected horizontally but at an angle \(\alpha\) with the horizontal and, as a result, is propelled into an elliptic orbit. Determine the maximum and minimum values of the distance from the center of the earth to the satellite.
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