/*! 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 42 A rocket ship approaching Earth ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A rocket ship approaching Earth at \(0.90 c\) fires a missile toward Earth with a speed of \(0.50 c,\) relative to the rocket ship. As viewed from Earth, how fast is the missile approaching Earth?

Short Answer

Expert verified
Answer: The total speed of the missile as viewed from Earth is approximately 0.9655c.

Step by step solution

01

Write down the given values

We are given the speed of the rocket ship (\(u\)) as \(0.90c\) and the speed of the missile relative to the rocket ship (\(v'\)) as \(0.50c\). Our goal is to find the speed of the missile as seen from Earth (\(v\)).
02

Use the relativistic velocity addition formula

Now, plug in the given values into the relativistic velocity addition formula: $$v = \frac{u + v'}{1 + \frac{u \cdot v'}{c^2}}$$
03

Calculate the speed of the missile as seen from Earth

By substituting the given values into the formula, we have: $$v = \frac{(0.90c) + (0.50c)}{1 + \frac{(0.90c) \cdot (0.50c)}{c^2}}$$ To simplify this expression, we can divide both the numerator and denominator by \(c\): $$v = \frac{0.90 + 0.50}{1 + \frac{0.90 \cdot 0.50}{c}} = \frac{1.40}{1 + \frac{0.45}{c}}$$ Then, use the fact that \(c\) is a constant (the speed of light) and note that the denominator becomes unitless: $$v = \frac{1.40}{1 + 0.45} = \frac{1.40}{1.45}c$$ After dividing the numerator by the denominator, we obtain: $$v = \frac{1.40}{1.45}c = 0.9655c$$ Thus, as viewed from Earth, the missile is approaching Earth at a speed of approximately \(0.9655c\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Special Relativity
Imagine you are aboard a spaceship, zooming through the cosmos close to the speed of light. As counterintuitive as it seems, the universe changes rules when you travel that fast. This is where Albert Einstein's theory of special relativity comes into play. It provides a new framework for looking at the physics of objects moving at a significant fraction of the speed of light, known as relativistic speeds.

One of the most groundbreaking ideas of special relativity is that the laws of physics are the same for all observers, regardless of their constant velocity. This means that whether you're sitting still on Earth or flying through space, the same physical laws apply. However, special relativity introduces modifications to more classical concepts like time, distance, mass, and energy for objects in high-speed motion. Most notably, it tells us that as an object's speed approaches the speed of light, significant changes occur to these quantities that classical physics can't account for.

An essential application of special relativity is in the equation of relativistic velocity addition. This concept is crucial when determining how fast an object appears to be moving relative to another object when both are moving at speeds close to the speed of light.
Speed of Light
The speed of light, denoted by the symbol 'c,' is more than just a constant—it is the cosmic speed limit, clocking in at approximately 299,792,458 meters per second (in a vacuum). This 'c' plays a pivotal role in the theories of relativity and in understanding the fabric of space-time.

One of Einstein's postulates in special relativity is that the speed of light in a vacuum is the same for all observers, regardless of their motion or the motion of the light source. In other words, no matter how fast you're moving, you'll measure the same speed of light as someone who's standing still. This invariant speed has profound implications, for it leads to effects such as time dilation and length contraction for objects in rapid motion. It acts as a fundamental constant in equations such as E=mc^2, which shows the equivalence of mass and energy, and in the relativistic velocity addition formula.
Velocity Calculation in Physics
Velocity calculation in classical physics is quite simple: if you want to compute the speed at which two objects approach each other, you just add up their velocities. If a car traveling north at 60 km/h passes another car going south at 70 km/h, their relative speed is 130 km/h. But, as the speeds involved get close to the speed of light, this simple addition no longer gives the correct result. Relativistic effects come into play.

The relativistic velocity addition formula becomes necessary when dealing with velocities that are a significant fraction of the speed of light. As seen in the exercise, when a rocket ship travels at 0.90c and fires a missile at 0.50c relative to itself, the velocities cannot be simply added up. Instead, the formula:
\[\begin{equation}v = \frac{u + v'}{1 + \frac{u \cdot v'}{c^2}}\end{equation}\] must be used where 'u' and 'v' are the velocities of the objects and 'c' is the constant speed of light. This equation ensures that the resulting speed does not exceed the speed of light and adheres to the principles of special relativity.

To provide a clearer image for the solution from the exercise, we observed that even when the rocket ship and the missile are both moving at high speeds, their resulting velocity, as observed from Earth, does not merely equal their arithmetic sum but instead is calculated by this more complex formula, resulting in a value of approximately 0.9655c.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use light cones and world lines to help solve the following problem. Eddie and Martin are throwing water balloons very fast at a target. At \(t=-13 \mu s,\) the target is at \(x=0,\) Eddie is at \(x=-2 \mathrm{~km},\) and Martin is at \(x=5 \mathrm{~km},\) and all three remain in these positions for all time. The target is hit at \(t=0 .\) Who made the successful shot? Prove this using the light cone for the target. When the target is hit, it sends out a radio signal. When does Martin know the target has been hit? When does Eddie know the target has been hit? Use the world lines to show this. Before starting to draw your diagrams, consider: If your \(x\) position is measured in \(\mathrm{km}\) and you are plotting \(t\) versus \(x / c,\) what units must \(t\) be in, to the first significant figure?

In some proton accelerators, proton beams are directed toward each other for head-on collisions. Suppose that in such an accelerator, protons move with a speed relative to the lab of \(0.9972 c\). a) Calculate the speed of approach of one proton with respect to another one with which it is about to collide head on. Express your answer as a multiple of \(c\), using six significant digits. b) What is the kinetic energy of each proton beam (in units of \(\mathrm{MeV}\) ) in the laboratory reference frame? c) What is the kinetic energy of one of the colliding protons (in units of \(\mathrm{MeV}\) ) in the rest frame of the other proton?

A HeNe laser onboard a spaceship moving toward a remote space station emits a beam of red light toward the space station. The wavelength of the beam, as measured by a wavelength meter on board the spaceship, is \(632.8 \mathrm{nm}\). If the astronauts on the space station see the beam as a blue beam of light with a measured wavelength of \(514.5 \mathrm{nm},\) what is the relative speed of the spaceship with respect to the space station? What is the shift parameter \(z\) in this case?

An astronaut in a spaceship flying toward Earth's Equator at half the speed of light observes Earth to be an oblong solid, wider and taller than it appears deep, rotating around its long axis. A second astronaut flying toward Earth's North Pole at half the speed of light observes Earth to be a similar shape but rotating about its short axis. Why does this not present a contradiction?

A gravitational lens should produce a halo effect and not arcs. Given that the light travels not only to the right and left of the intervening massive object but also to the top and bottom, why do we typically see only arcs?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.