A spaceship is moving away from Earth with a constant...

A spaceship is moving away from Earth with a constant velocity of $0.80 c$ with respect to Earth. The spaceship and an Earth station synchronize their clocks, setting both to zero, at an instant when the ship is near Earth. By prearrangement, when the clock on Earth reaches a reading of $1.0 \times 10^{4} \mathrm{~s}$, the Earth station sends out a light signal to the spaceship. (a) In the frame of reference of the Earth station, how far must the signal travel to reach the spaceship? (b) According to an Earth observer, what is the reading of the clock on Earth when the signal is received?

Key Concepts

Clock Synchronization and Relativity of Simultaneity

Clock synchronization in one inertial frame, such as setting clocks to zero at the moment two objects meet, does not guarantee that the timing of events will appear simultaneous in another frame moving relative to the first. This concept, known as the relativity of simultaneity, is essential for understanding how events like the emission and reception of signals are coordinated across different reference frames.

Inertial Reference Frames

An inertial reference frame is one in which an observer is either at rest or moving at a constant velocity, with no acceleration. In the theory of relativity, measurements of time and space depend on the inertial frame in which events are observed, and this problem involves comparing observations from the Earth frame and the spaceship's frame.

Constancy of the Speed of Light

One of the fundamental postulates of special relativity is that the speed of light in a vacuum is constant and the same for all observers, regardless of their state of motion. This principle ensures that light signals propagate at speed c in all inertial frames and is crucial when determining how light travels relative to moving objects.

Relativistic Kinematics

Relativistic kinematics involves analyzing the motion of objects at speeds comparable to the speed of light. It extends classical kinematics to account for time dilation, length contraction, and the invariance of the speed of light, enabling one to calculate distances and time intervals correctly in different inertial frames.

Transcript

00:01 All right, so for question a, well, we know that the spaceship will already travel a distance, let's say d1, okay, which is simply equal to b times t1, okay? v is given as 0 .80c, which is 0 .80 times 3 times 10 to 8 meter per second, which will give us the speed of a spaceship is about 2 .4.

00:31 Times 10 to the power of 8 meter per second.

00:35 And we know the t1 was given as 1 .0 times 10 to the power of 4 seconds.

00:44 So therefore before the signal was sent out, the spaceship would travel a distance d1, which is equal to 2 .4 times 10 to the power of 8 meter per second and then times 1 .0 times 10 to power of 4.

01:06 Seconds and this will give us 2 .4 times 10 to the power of 12 meter.

01:14 Okay, so in order to determine how far should the signal travel, well, we need to first determine the time first.

01:26 Well, we know that if we assume that the distance that the signal travel is equal to d.

01:30 Therefore, it should be equal to the speed of signal, which is a speed of light, times the time you need to take.

01:37 Let's say.

01:37 Say it, okay? so therefore we know that t can be equal to the distance divided by the speed of signal, which is c.

01:46 Okay? so for this time here, the spaceship will also travel this much time before the signal catching up, okay? therefore, the total distance, which is simply equal to v times t plus d1, okay? and when the t is equal to d over c, therefore we have v times d over c plus d1 and then we'll have d is equal to v over c times d plus d1 and we know v which is 0.

02:33 A 0c and then divide by c times d we don't know okay and we know d1 is 2 .4 times 10 to the power of 12 meter as you can tell here c and c can be canceled out.

02:56 And if we move the 0 .80d to the left side, we'll have d minus 0 .80d, which is 0 .20d.

03:05 And this is equal to 2 .4 times 10 to the power of 12 meter...

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