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Chapter 14: Problem 15

A rocket is accelerated to a speed of \(v=2 \sqrt{g R_{\mathrm{E}}}\) near the Earth's surface and then coasts upward. (a) Show that it will escape from the Earth. ( \(b\) ) Show that very far from the Earth its speed is \(v=\sqrt{2 g R_{\mathrm{E}}}\).

Short Answer

Expert verified
The rocket will escape from the Earth as its speed is greater than Earth's escape velocity. The speed of the rocket far away from the Earth will be \(v=\sqrt{2 g R_{\mathrm{E}}}\).

Step by step solution

01

Understanding the Escape Velocity Concept

The escape velocity is the minimum velocity an object must attain to escape the gravitational pull of a planet or other body. The formula of escape velocity is \(v_e = \sqrt{2gR_{\mathrm{E}}}\), where \(g\) is the acceleration due to gravity and \(R_{\mathrm{E}}\) is the Earth's radius.
02

Compare Given Velocity with Escape Velocity

The rocket's velocity is given as \(v=2 \sqrt{g R_{\mathrm{E}}}\). Compare it with the escape velocity formula. It can be observed that the given velocity is greater than the escape velocity. Therefore, the rocket will escape from the Earth's surface.
03

Applying the Law of Conservation of Energy

To find the velocity of the rocket far away from the Earth, use the law of conservation of energy. According to this law, the total energy of a system remains constant if no external force acts on it. In the case of the rocket, the total energy includes kinetic and potential energy.
04

Set Up the Energy Conservation Equation

The total energy at the surface of Earth (when the rocket is just launched) is the sum of kinetic and potential energy, i.e., \(E_1 = \frac{1}{2} m v^2 - \frac{GmM}{R_{\mathrm{E}}}\). Far away from the Earth (when the influence of Earth's gravity is negligible), the total energy is \(E_2 = \frac{1}{2} m v_f^2\), where \(v_f\) is the final velocity of the rocket. Since the total energy is conserved, one can equate \(E_1\) and \(E_2\) and determine \(v_f\).
05

Calculation of Rocket's Speed Far Away from Earth

Based on the energy conservation equation obtained in Step 4, one can solve for \(v_f\). This results in \(v_f=\sqrt{2 g R_{\mathrm{E}}}\), which finally gives the speed of the rocket far away from the Earth.

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Most popular questions from this chapter

If \(g\) is to be determined by dropping an object through a distance of (exactly) \(10 \mathrm{~m}\), how accurately must the time be measured to obtain a result good to \(0.1 \%\) ? Calculate a percent error and an absolute error, in milliseconds.

The orbit of the Earth about the Sun is almost circular. The closest and farthest distances are \(1.47 \times 10^{8} \mathrm{~km}\) and \(1.52 \times\) \(10^{8} \mathrm{~km}\), respectively. Determine the maximum variations in ( \(a\) ) potential energy, (b) kinetic energy, ( \(c\) ) total energy, and (d) orbital speed that result from the changing Earth-Sun distance in the course of 1 year. (Hint: Use conservation of energy and angular momentum.)

It is conjectured that a "burned-out" star could collapse to a "gravitational radius," defined as the radius for which the work needed to remove an object of mass \(m\) from the star's surface to infinity equals the rest energy \(m c^{2}\) of the object. Show that the gravitational radius of the Sun is \(G M_{\mathrm{S}} / c^{2}\) and determine its value in terms of the Sun's present radius. (For a review of this phenomenon see "Black Holes: New Horizons in Gravitational Theory," by Philip C. Peters, American Scientist, September- October \(1974, \mathrm{p} .575 .)\)

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