Problem 98 - Fundamentals of Physics

Question

An astronaut exercising on a treadmill maintains a pulse rate of 150 per minute. If he exercises for \(1.00 \mathrm{~h}\) as measured by a clock on his spaceship, with a stride length of \(1.00 \mathrm{~m} / \mathrm{s},\) while the ship travels with a speed of \(0.900 \mathrm{c}\) relative to a ground station, what are (a) the pulse rate and (b) the distance walked as measured by someone at the ground station?

Step-by-Step Solution

Answer

(a) 344.16 pulses/min, (b) 8258.313 m

Step 1: Understand the Theory of Time Dilation

Due to the high-speed travel of the spaceship at 0.900c (where c is the speed of light), time dilation, a concept from Einstein's theory of relativity, will occur. Time aboard the spaceship will pass differently compared to time as measured by someone at the ground station.

Step 2: Apply Time Dilation Formula

The time dilation formula is given by \( t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \), where \( t' \) is the time measured by the ground observer, \( t \) is the time measured by the observer on the spaceship, \( v \) is the velocity of the spaceship, and \( c \) is the speed of light.

Step 3: Calculate Time for Ground Observer

Given \( v = 0.900c \) and \( t = 1.00 \text{ h} = 3600 \text{ seconds} \), substitute into the time dilation formula: \[ t' = \frac{3600}{\sqrt{1 - (0.900)^2}} \approx \frac{3600}{0.43588989} \approx 8258.313 \text{ seconds} \].

Step 4: Calculate Pulse Rate as Measured by Ground Observer

The original pulse rate of the astronaut is 150 pulses per minute. Convert to pulses per second (2.5 pulses per second), then multiply by the time dilation factor (inverse of \( \sqrt{1 - (0.900)^2} \)) to find the pulse rate as measured by the ground observer. Therefore, new pulse rate: \[ 2.5 \times \frac{1}{0.43588989} \approx 5.736 \text{ pulses per second} \] or approximately 344.16 pulses per minute.

Step 5: Determine Distance Walked by Astronaut

The astronaut has a stride rate of 1.00 m/s for 1.00 h as per the spaceship clock, meaning he walks 3600 m.

Step 6: Calculate Distance Walked as Measured by Ground Observer

Due to time dilation, the observer on Earth sees the astronaut walking for \( t' = 8258.313 \) seconds at 1.00 m/s. Hence, distance = \( 1.00 \times 8258.313 = 8258.313 \text{ m} \).

Key Concepts

Einstein's Theory of Relativity

At the heart of Einstein's theory of relativity is the idea that time is relative and can vary depending on the speed at which an observer is moving. According to the special theory of relativity, time dilation occurs when an object moves at a significant fraction of the speed of light. This means that time appears to pass at a different rate for observers moving with respect to each other.

In our exercise, the spaceship travels at 0.900c, where c is the speed of light. This high speed causes time aboard the spaceship to pass differently compared to time on Earth. Essentially, the faster you move through space, the slower you experience time. This forms the basis of why the astronaut's exercise on the spaceship is perceived differently by someone at the ground station.

Pulse Rate Calculation

When calculating the pulse rate, it is important to factor in the effects of time dilation. The astronaut maintains a pulse rate of 150 beats per minute on the spaceship. To find out how this heartbeat is perceived from the ground, we must adjust for the time dilation caused by the spaceship's high velocity.

The time dilation factor is the inverse of the square root of the velocity squared divided by the speed of light squared, \(\frac{1}{\sqrt{1 - v^2/c^2}}\). By applying this, the perceived pulse rate increases because the time as observed by the ground station stretches. After converting beats per minute to beats per second, we calculate the ground observer's perspective, finding a much higher pulse rate due to this effect.

Velocity

Velocity is a crucial element in understanding time dilation. It refers to the speed of an object in a specific direction and impacts how time is perceived across different frames of reference. High velocity close to the speed of light causes more pronounced time dilation effects.

In this scenario, the spaceship's velocity is 0.900c, which significantly affects time perception. As the spaceship approaches the speed of light, the greater the time difference between the spaceship and the ground observer. This high velocity stretches time from the perspective of the observer on Earth, affecting all time-related measurements, such as the pulse rate and the distance walked.

Distance Measurement

Measuring distance from different frames of reference also requires accounting for relativity. On the spaceship, distance is straightforward: the astronaut walks 1.00 meter per second for an hour, totaling 3600 meters. However, when seen from the ground station, time dilation alters this simple calculation.

The time from the ground observer's viewpoint expands to 8258.313 seconds due to time dilation. Therefore, the ground observer measures the astronaut walking 8258.313 meters. This discrepancy perfectly illustrates the effects of high-speed travel on distance perception in relativity, highlighting how intertwined time and space are in Einstein's theory.