A rocket ship flies past the earth at 91.0% of the speed of...

A rocket ship flies past the earth at 91.0% of the speed of light. Inside, an astronaut who is undergoing a physical examination is having his height measured while he is lying down parallel to the direction in which the ship is moving. (a) If his height is measured to be 2.00 m by his doctor inside the ship, what height would a person watching this from the earth measure? (b) If the earth-based person had measured 2.00 m, what would the doctor in the spaceship have measured for the astronaut's height? Is this a reasonable height? (c) Suppose the astronaut in part (a) gets up after the examination and stands with his body perpendicular to the direction of motion. What would the doctor in the rocket and the observer on earth measure for his height now?

Transcript

00:01 This is chapter 37 problem number 14.

00:03 We have a rocket.

00:03 It goes 95 % of the speed of light, so it's going to be 0 .95c, right? the speed.

00:11 Now, in part a, we're given the proper length to be 2 meters in the frame of the ship.

00:21 So l0 equals 2 meters.

00:24 So in part a, we're asked to calculate a person on earth, how tall, would the pilot of that ship would appear to be for an observer on earth.

00:39 So then we have the length expansion formula, those like l -nat over gamma factor.

00:47 So all we need to do is actually calculate the l from here, provided that l -0 is the proper length, well, the height of that person, right? so, and not whatever gamma factor is going to be square root of 1 minus we squared over c squared.

01:06 Let's put 2 meters here, square root of 1 minus .91c squared over c squared.

01:14 Now when we do the algebra, we find it.

01:18 We find the height to be 0 .83 meters to an observer on earth.

01:25 Still a little on the short side, right? so part b, this time we are given the height of this pilot based on an observer on earth.

01:39 So l equals two meters now, and what would be the proper height of his pilot in the frame of the ship that moves with this relative speed.

01:50 Now, we're going to use the same formula, right? l0 equals l not over ga.

01:56 This time, we're trying to calculate what lnu is.

01:59 So lnut is going to be gamma times l.

02:01 Gamma is going to be 1 over square of 1 minus 3 squared over c squared...