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Chapter 2: Problem 5

A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes \(8 \mathrm{~min}\) and \(20 \mathrm{~s}\) to cover this distance?

Short Answer

Expert verified
The distance is 500 new units.

Step by step solution

01

Understanding the Problem

In this problem, we are tasked with determining the distance between the Sun and the Earth in a new unit of length where the speed of light is given the value of 1. It is stated that light takes 8 minutes and 20 seconds to travel from the Sun to Earth.
02

Calculating the Time in Seconds

Convert the total time for light to travel from the Sun to Earth into seconds. 8 minutes is equal to 480 seconds. Add the remaining 20 seconds to find the total time: \[ 8 ext{ minutes} = 480 ext{ seconds} \]Total time = 480 + 20 = 500 seconds.
03

Using the Defined Speed of Light

Given that the speed of light in this new unit is 1, the distance (in these new units) is equal to the time it takes for light to travel the distance. Thus, the distance between the Sun and the Earth is 500 of these new units.

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Key Concepts

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

New Units of Measurement
When measuring something as vast as the universe, having appropriate units can make complex concepts more manageable. In this problem, a new unit of length has been chosen. This unit is designed to simplify our calculations by making the speed of light equal to unity, or 1. This implies that instead of using standard units like kilometers or miles, light's speed itself becomes the yardstick for measurement.
  • Why make the speed of light equal to 1? This simplifies physics equations, turning light's speed into a simple, convenient value.
  • In this new unit, if light travels some distance in one unit of time, that distance is simply the time taken, since the speed is 1 unit per second.
This method of measurement becomes powerful in theoretical physics, where dealing with the constants of physical laws is common.
Distance Calculation
The task is to find the distance between the Sun and Earth in this novel unit of measurement. The provided information is that light covers this distance in 8 minutes and 20 seconds. To perform these calculations, we need to convert this time into a simpler form.
  • Convert 8 minutes into seconds: 8 minutes = 480 seconds.
  • Add the remaining 20 seconds: Total = 480 + 20 = 500 seconds.
Now, since light takes 500 seconds to travel from the Sun to Earth, and using our new measurement unit where the speed of light equals 1, this directly translates into 500 units of this new measure as the distance.
Light Travel Time
Understanding how long light takes to travel across vast distances is key in astrophysics. In this context, 8 minutes and 20 seconds is the travel time for light from our nearest star, the Sun, to reach Earth.
  • This duration, 500 seconds, captures how far and fast light can move in space.
  • Measuring delays in light travel time helps calibrate astronomical distances across the cosmos.
Because light is the fastest thing in the universe, its travel time is a fundamental standard for astronomical measurements. By knowing how long light takes to travel certain distances, scientists can track everything from the orbits of planets to the expansion of galaxies.

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Most popular questions from this chapter

Fill in the blanks by suitable conversion of units (a) \(1 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}=\ldots . \mathrm{g} \mathrm{cm}^{2} \mathrm{~s}^{-2}\) (b) \(1 \mathrm{~m}=\ldots . \mathrm{ly}\) (c) \(3.0 \mathrm{~m} \mathrm{~s}^{2}=\ldots . \mathrm{km} \mathrm{h}^{2}\) (d) \(G=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2}(\mathrm{~kg})^{-2}=\ldots .\left(\mathrm{cml}^{3} \mathrm{~s}^{-2} \mathrm{~g}^{-1} .\right.\)

When the planet Jupiter is at a distance of \(824.7\) million kilometers from the Earth. its angular diameter is measured to be \(35.72^{\prime \prime}\) of arc. Calculate the diameter of Jupiter.

The unit of length convenient on the atomic scale is known as an angstrom and is denoted by \(\AA\) : \(1 \AA^{\circ}=10^{-10} \mathrm{~m}\). The size of a hydrogen atom is about \(0.5 \AA\). What is the total atomic volume in \(\mathrm{m}^{3}\) of a mole of hydrogen atoms?

Estimate the average mass density of a sodium atom assuming its size to be about \(2.5 \AA\) 脜. (Use the known values of Avogadro's number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase : \(970 \mathrm{~kg} \mathrm{~m}^{-3}\). Are the two densities of the same order of magnitude? If so, why?

Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.
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