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Video: Determining a Length Change Due to Length Contraction

Ed Burdette

A spaceship that has a proper length of 2.00 ×10² m, moves at 0.970𝑐 relative to the Earth. What is the spaceship’s length as measured by an observer on the Earth?

03:05

Video Transcript

A spaceship has a proper length of 2.00 times 10 to the two meters moves at 0.970𝑐 relative to the Earth. What is the spaceship’s length as measured by an observer on the Earth?

Let’s start by highlighting some of the important information given. We’re told that the proper length of the spaceship is 2.00 times 10 to the two meters. We’ll call that length 𝑙 zero. And we’re also told that the spaceship moves at a speed relative to the earth of 0.970𝑐, where 𝑐 is the speed of light. We’ll call that speed simply 𝑣. We want to solve for the length of the spaceship as measured by an observer on Earth. Let’s call that length 𝑙.

We’ll approach this problem by recalling that there is a correlation or a connection between length observed in one inertial reference frame in length observed in another. This correlation says that the length observed in one inertial reference frame, we’ll call it 𝑙 prime, is equal to 𝛾 times 𝑙, the length observed in the other reference frame. Here, 𝛾 is defined as one divided by the square root of one minus 𝑣 squared over 𝑐 squared, where 𝑣 is the speed of one reference frame relative to the other.

If we apply this length correlation relationship to our situation, we find that 𝑙, the length we’re looking to solve for, is equal to 𝛾 times 𝑙 sub zero, the proper length of the spaceship. And we can replace 𝛾 with its defined expression. So the spaceship’s length, as measured by an observer on Earth, is equal to 𝑙 sub zero divided by the square root of one minus 𝑣 squared over 𝑐 squared. We’re given 𝑙 sub zero and 𝑣 in terms of the speed of light 𝑐. So we can enter those values into our equation now.

𝑙 is equal to 2.00 times 10 to the two meters divided by the square root of one minus the quantity 0.970𝑐 squared all over 𝑐 squared. In this denominator, the 𝑐 squared terms cancel out so that our expression inside the square root simplifies to one minus 0.970 squared.

When we enter these numbers into our calculator, we find a value for 𝑙 of 48.6 meters. Compared to its proper length of 200 meters, this is the length of the spaceship as measured by an observer on Earth.