Question bank: Theory of Relativity: Time Dilation

According to the Special Theory of Relativity, proposed by Albert Einstein in 1905, time is not absolute and can be dilated or contracted depending on the relative velocity between observers. An astronaut traveling at a speed close to the speed of light experiences a time dilation effect, where time passes more slowly for him compared to an observer at rest. Consider an experiment in which an astronaut leaves Earth in a spaceship that reaches 80% of the speed of light. If on Earth, a clock shows 60 minutes from the spaceship's departure to a certain event, what would be the time recorded by the astronaut's clock when he perceives the same event in space? Consider the speed of light to be 3 x 10^8 m/s and use the Lorentz factor (γ) to calculate the time dilation.

In a physics experiment, a group of students set out to empirically verify the time dilation predicted by Einstein's Special Theory of Relativity. They use an extremely precise clock capable of measuring time with nanosecond precision. They set up an experiment to measure the time it takes for a beam of light to travel a distance of 300 million meters at rest and aboard a hypothetical spaceship traveling at a constant speed of 0.8c, where c is the speed of light in a vacuum (c = 3 x 10^8 m/s). Considering the Lorentz factor (γ) as γ = 1/sqrt(1 - (v^2/c^2)), where v is the spaceship's velocity: (a) Calculate the time measured for the beam of light to travel the distance at rest and compare it with the time measured aboard the spaceship. (b) Explain how time dilation manifests in this experiment and the importance of this phenomenon for understanding the universe.

A satellite equipped with a high-precision clock is orbiting the Earth at a constant speed of 7,500 km/h. In this context, an observer on Earth will measure the duration of time recorded by the clock on the satellite compared to their own clock. As special relativity theory is applicable to this phenomenon, calculate the difference in the duration of time measured by the observer on Earth in relation to the satellite after 24 hours of observation.

An astronaut travels in a spaceship at a constant speed of 0.8c, where c is the speed of light in a vacuum (c = 3.00 x 10^8 m/s), towards a space station. According to an observer at the station, the astronaut takes 60 Earth minutes to complete a specific task. Using the special theory of relativity and the Lorentz factor (γ = 1 / (1 - (v^2 / c^2))^0.5), calculate the dilated time that the astronaut observes to complete the task, considering that the space station is at rest relative to the Earth observer.

An astronaut travels in a spaceship at a constant speed of 0.8c, where c is the speed of light in a vacuum. He intends to conduct an experiment involving the emission of a light signal to a satellite in orbit. According to the special theory of relativity, the time for an observer at rest on Earth between the emission and reception of the light signal would be different from the time recorded by the astronaut in the spaceship. Considering the time dilation given by the Lorentz factor, where γ = 1 / sqrt(1 - (v^2 / c^2)), and that c = 3 x 10^8 m/s, calculate: 1) The γ factor for the astronaut's speed. 2) The time dilation, that is, the relationship between the time measured by the astronaut and the time that an observer on Earth would measure for the light signal to travel the round trip distance of 4.8 x 10^8 meters. Consider the speed of light as c = 3 x 10^8 m/s and respond in terms of c and the exact value of the γ factor, without rounding.