# Question Video: Determining a Length Change Due to Length Contraction

A spaceship has a proper length of 2.00 ร 10ยฒ m, moves at 0.970๐ relative to the Earth. What is the spaceshipโs length as measured by an observer on the Earth?

03:04

### Video Transcript

A spaceship has a proper length of 2.00 times 10 to the two meters moves at 0.970๐ relative to the Earth. What is the spaceshipโs length as measured by an observer on the Earth?

Letโs start by highlighting some of the important information given. Weโre told that the proper length of the spaceship is 2.00 times 10 to the two meters. Weโll call that length ๐ zero. And weโre also told that the spaceship moves at a speed relative to the earth of 0.970๐, where ๐ is the speed of light. Weโll call that speed simply ๐ฃ. We want to solve for the length of the spaceship as measured by an observer on Earth. Letโs call that length ๐.

Weโ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, weโll call it ๐ prime, is equal to ๐พ times ๐, the length observed in the other reference frame. Here, ๐พ is defined as one divided by the square root of one minus ๐ฃ squared over ๐ squared, where ๐ฃ is the speed of one reference frame relative to the other.

If we apply this length correlation relationship to our situation, we find that ๐, the length weโre looking to solve for, is equal to ๐พ times ๐ sub zero, the proper length of the spaceship. And we can replace ๐พ with its defined expression. So the spaceshipโs length, as measured by an observer on Earth, is equal to ๐ sub zero divided by the square root of one minus ๐ฃ squared over ๐ squared. Weโre given ๐ sub zero and ๐ฃ in terms of the speed of light ๐. So we can enter those values into our equation now.

๐ is equal to 2.00 times 10 to the two meters divided by the square root of one minus the quantity 0.970๐ squared all over ๐ squared. In this denominator, the ๐ 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 ๐ 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.