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A spaceship and its occupants have a total mass of \(180,000 \mathrm{~kg} .\) The occupants would like to travel to a star that is 35 light-years away at a speed of \(0.70 c .\) To accelerate, the engine of the spaceship changes mass directly to energy. How much mass will be converted to energy to accelerate the spaceship to this speed? Assume the acceleration is rapid, so the speed for the entire trip can be taken to be \(0.70 c,\) and ignore decrease in total mass for the calculation. How long will the trip take according to the astronauts on board?

Short Answer

Expert verified
7200 kg of mass is converted to energy, and the trip takes about 3.58 years for the astronauts.

Step by step solution

01

Understanding the Concepts

First, recognize the key concepts: according to Einstein's equation, mass can be converted into energy, and for a spaceship, energy is needed to reach a certain fraction of the speed of light, given by velocity \(v\). The equation involved is: \(E = \Delta m c^2\), where \(\Delta m\) is the mass converted to energy.
02

Determine Kinetic Energy

Calculate the kinetic energy needed to reach the speed \(0.70c\). The relativistic kinetic energy is given by the formula: \(K = (\gamma - 1)m c^2\). Here, \(\gamma = \frac{1}{\sqrt{1-(v^2/c^2)}}\), \(m\) is the mass of the spaceship, and \(c\) is the speed of light.
03

Calculate Gamma \(\gamma\)

Calculate the Lorentz factor \(\gamma\) for the speed \(v = 0.70c\): \[\gamma = \frac{1}{\sqrt{1-(0.70)^2}} = \frac{1}{\sqrt{1-0.49}} = \frac{1}{\sqrt{0.51}} \approx 1.400\].
04

Compute Kinetic Energy

Substitute values into the kinetic energy formula: \[K = (1.400 - 1) \times 180,000 \times (3 \times 10^8)^2\]. Calculate the excess energy:\[K = 0.4 \times 180,000 \times 9 \times 10^{16} = 6.48 \times 10^{21} \text{ Joules}\].
05

Calculate Mass Converted to Energy

Using \(E = \Delta m c^2\), find the mass \(\Delta m\) converted to energy:\[\Delta m = \frac{K}{c^2} = \frac{6.48 \times 10^{21}}{9 \times 10^{16}} = 7200 \text{ kg} \].
06

Calculate Time of Travel According to Astronauts

The time measured in the spaceship's frame (proper time) is different due to relativistic effects. The formula for time dilation is: \[t' = \frac{t}{\gamma}\]. First, calculate the time in the stationary observer's frame: \[t = \frac{d}{v} = \frac{35 \times 9.46 \times 10^{15}}{0.70 \times 3 \times 10^{8}} \approx 1.58 \times 10^8 \text{ seconds}\]. Now, adjust for the spaceship's frame using \(\gamma\): \[t' = \frac{1.58 \times 10^8}{1.400} \approx 1.13 \times 10^8 \text{ seconds}\].
07

Convert Time to Years

Convert the time from seconds to years: \[1.13 \times 10^8 \text{ seconds} \times \frac{1 ext{ hour}}{3600 ext{ seconds}} \times \frac{1 ext{ day}}{24 ext{ hours}} \times \frac{1 ext{ year}}{365.25 ext{ days}} \approx 3.58 ext{ years}\].

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Key Concepts

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

Mass-Energy Equivalence
In the realm of relativistic mechanics, one of the most profound concepts is the mass-energy equivalence, encapsulated by Einstein’s famous equation, \(E = mc^2\). This equation illustrates the idea that mass can be converted into energy and vice versa, which is a cornerstone in understanding how high-speed travel, like that of a spaceship nearing the speed of light, might be achieved.
This concept states that a small amount of mass can produce a significant amount of energy due to the speed of light \(c\) being a large number \((3 \times 10^8 \text{ m/s})^2\).
  • In practical applications, like that of the spaceship in the exercise, this principle allows the spaceship's engine to convert a portion of its mass directly into the energy needed to accelerate.
  • The conversion is crucial for scenarios where traditional fuel would be insufficient to achieve the necessary speeds for long-distance space travel.
Understanding this equivalence provides us with a fundamental tool for contemplating how future technologies might approach the challenges of high-speed and long-distance travel.
Lorentz Factor
The Lorentz factor \(\gamma\) is a critical component of relativistic physics, especially when dealing with velocities approaching the speed of light. The Lorentz factor accounts for the relativistic effects that come into play at such high speeds.
It is given by the equation \(\gamma = \frac{1}{\sqrt{1-(v^2/c^2)}}\).
  • As velocity \(v\) approaches the speed of light, \(\gamma\) increases significantly, leading to noticeable changes in how time and space are perceived.
  • In the context of the spaceship problem, \(\gamma\) is calculated for the spaceship's speed \(0.70c\), resulting in a value of approximately 1.400.
This factor is essential for calculating the relativistic kinetic energy and understanding time dilation, both of which are vital for planning space travel at relativistic speeds.
Time Dilation
Time dilation is a fascinating consequence of Einstein's theory of relativity. It explains how time can "slow down" for objects moving at a significant fraction of the speed of light, relative to a stationary observer. In the case of our spaceship, this means the travel time experienced by the astronauts on board differs from what someone on Earth would observe.
The formula for time dilation is \(t' = \frac{t}{\gamma}\), where \(t'\) is the time experienced by those in the moving object, \(t\) is the actual time as observed from a stationary point, and \(\gamma\) is the Lorentz factor.
  • For the spaceship traveling to a star 35 light-years away, the perceived time onboard is shorter due to the Lorentz factor, calculated using the speed \(0.70c\).
  • This effect demonstrates the non-intuitive nature of high-speed travel scenarios, emphasizing the importance of precise calculations in physics.
Understanding time dilation helps explain not only theoretical constructs in physics but also the planning and execution of real-world space missions, where time differences are more than mere theoretical curiosities.
Kinetic Energy Calculation
In relativistic mechanics, the calculation of kinetic energy changes when objects move at speeds close to the speed of light. Unlike classical mechanics, where kinetic energy is \( K = \frac{1}{2}mv^2 \), the relativistic form incorporates the Lorentz factor.
The relativistic kinetic energy is calculated as \(K = (\gamma - 1)mc^2\).
  • The equation accounts for changes in energy required as velocity increases substantially, emphasizing how energy demands grow non-linearly at high speeds.
  • As demonstrated in the spaceship exercise, this calculation informs us of how much energy is needed and subsequently, how much mass must be converted to energy to achieve the desired speed.
This concept is pivotal not only in understanding energy requirements for spacecraft but also in broader applications within high-energy physics and cosmology.

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Most popular questions from this chapter

(II) A spaceship traveling at \(0.76 c\) away from Earth fires a module with a speed of \(0.82 c\) at right angles to its own direction of travel (as seen by the spaceship). What is the speed of the module, and its direction of travel (relative to the spaceship's direction), as seen by an observer on Earth?

The Sun radiates energy at a rate of about \(4 \times 10^{26} \mathrm{~W}\). (a) At what rate is the Sun's mass decreasing? ( \(b\) ) How long does it take for the Sun to lose a mass equal to that of Earth? (c) Estimate how long the Sun could last if it radiated constantly at this rate.

(I) When a uranium nucleus at rest breaks apart in the process known as fission in a nuclear reactor, the resulting fragments have a total kinetic energy of about \(200 \mathrm{MeV}\). How much mass was lost in the process?

Suppose a \(14,500-\mathrm{kg}\) spaceship left Earth at a speed of \(0.98 c\) What is the spaceship's kinetic energy? Compare with the total U.S. annual energy consumption (about \(10^{20} \mathrm{~J}\) ).

(III) In the old West, a marshal riding on a train traveling 35.0 \(\mathrm{m} / \mathrm{s}\) sees a duel between two men standing on the Earth 55.0 \(\mathrm{m}\) apart parallel to the train. The marshal's instruments indicate that in his reference frame the two men fired simultaneously. (a) Which of the two men, the first one the train passes (A) or the second one (B) should be arrested for firing the first shot? That is, in the gunfighter's frame of reference, who fired first? \((b)\) How much earlier did he fire? (c) Who was struck first?

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