Problem 35 - Fundamentals of Physics

Question

A spaceship, moving away from Earth at a speed of \(0.900 c\), reports back by transmitting at a frequency (measured in the spaceship frame) of \(100 \mathrm{MHz}\). To what frequency must Earth receivers be tuned to receive the report?

Step-by-Step Solution

Answer

The frequency Earth receivers must be tuned to is approximately 22.92 MHz.

Step 1: Understand the Problem

We need to find the frequency received on Earth, considering the spaceship is moving away from Earth. The transmitted frequency is reduced due to the Doppler effect because the source is receding from the observer.

Step 2: Apply the Doppler Effect Formula for Light

We'll use the formula for the relativistic Doppler effect for receding sources. The formula is \( f' = f \sqrt{\frac{1 - \beta}{1 + \beta}} \), where \( f \) is the emitted frequency, \( f' \) is the observed frequency, and \( \beta = \frac{v}{c} \).

Step 3: Substitute Given Values

Substitute \( f = 100 \) MHz and \( \beta = 0.900 \) into the formula. Thus the formula becomes \( f' = 100 \sqrt{\frac{1 - 0.900}{1 + 0.900}} \).

Step 4: Simplify the Expression Under the Square Root

The expression under the square root simplifies to \( \frac{1 - 0.900}{1 + 0.900} = \frac{0.100}{1.900} \). Compute this ratio.

Step 5: Calculate the Frequency

Calculate \( f' = 100 \times \sqrt{\frac{0.100}{1.900}} = 100 \times \sqrt{0.05263} \). Find the square root of 0.05263 and then multiply by 100.

Step 6: Result Calculation

\( \sqrt{0.05263} \approx 0.2292 \). Therefore, the frequency received on Earth is \( 100 \times 0.2292 \approx 22.92 \) MHz.

Key Concepts

Frequency

In the context of this problem, frequency refers to the number of oscillations or cycles per second of a wave, commonly measured in Hertz (Hz). For this exercise, the spaceship sends a signal at a frequency of 100 MHz (megahertz), which means the wave oscillates 100 million times per second. Understanding frequency is crucial as it determines the energy and type of the wave being transmitted or received. Some key points about frequency include:

• Higher frequency waves have more energy and shorter wavelengths.
• Lower frequency waves have less energy and longer wavelengths.
• The frequency shifts when the source or observer is moving relative to each other, due to the Doppler effect.

This shift in frequency is essential in understanding how signals sent from a moving spaceship are received on Earth, especially when the spaceship is receding, as it directly influences the frequency we must tune to.

Spaceship

In this exercise, the spaceship is a moving source that transmits a signal back to Earth. Since the spaceship is receding from Earth at a significant fraction of the speed of light, the relativistic Doppler effect comes into play. Here are some important considerations regarding the spaceship in this context:

• The speed of the spaceship is given as 0.900 times the speed of light (denoted as \( 0.900 c \)).
• As the spaceship moves away, the frequency of the transmitted signals decreases for an Earth observer, which is a key trait of the Doppler effect when the source is in motion.
• Understanding the role of the spaceship helps in applying the correct formula to determine the new frequency of the signal received on Earth.

This forms the basis for calculating how the observed frequency changes due to the high speed at which the spaceship is retreating from Earth.

Speed of Light

The speed of light, denoted by \( c \), is a fundamental constant in physics, valued at approximately \( 299,792,458 \) meters per second (m/s). In relativistic physics, it's often a pivotal factor, especially when dealing with moving sources like the spaceship in this exercise.

Understanding the speed of light in this context:

• It provides the reference speed (\( c \)), used to express the spaceship's speed as a fraction (0.900 of \( c \)).
• In calculations involving the Doppler effect, the ratio \( \beta = \frac{v}{c} \) is used, where \( v \) is the spaceship's speed.
• The constancy of the speed of light is a postulate of Einstein's theory of relativity, influencing how time, space, and frequency relate when objects approach the speed of light.

This foundational speed is crucial for determining how signals are perceived differently when their sources travel at high velocities.

Doppler Effect Formula

The Doppler effect formula is central to solving the given problem, allowing us to calculate the frequency shift observed for signals sent from a moving source, like the spaceship. The relativistic Doppler effect formula for a receding source is given by:

\[ f' = f \sqrt{\frac{1 - \beta}{1 + \beta}} \]

Here's how each component of this formula works:

• \( f' \) is the observed frequency, the one that an Earth observer will receive.
• \( f \) is the emitted frequency, in this case, 100 MHz sent by the spaceship.
• \( \beta \) is a dimensionless quantity defined as the ratio of the source's speed to the speed of light (\( \beta = 0.900 \)).

By substituting these values into the formula, we adjust the emitted frequency to find the observed frequency that accounts for the spaceship's high velocity away from Earth. This calculation highlights how relative motion alters the observed properties of waves, such as frequency, through the Doppler effect.