Relativistic and proper length

[AClass]

Homework Statement

A spaceship travels past a planet at a speed of 0.80 c as measured from the planet’s frame of reference. An observer on the planet measures the length of a moving spaceship to be 40 m.
a) How long is the spaceship, according to the astronaut?
b) At what speed would the spaceship have to travel for its relativistic length to be half its “proper” length?

Homework Equations

L= Lo x $${\sqrt{1 - \frac{ v^2 }{ c^2 }}}$$

The Attempt at a Solution

a)

Lo=40m
v=0.80c

Using the equation above I came to a answer of L = 24m

b)

L= 0.5 Lo

0.5Lo = Lo x $${\sqrt{1 - \frac{ v^2 }{ c^2 }}}$$

v^2 = (0.75)(c^2)
v= Square root [(0.75)(c)]
c= 3.0x10^8
v= 15000 m/s

The speed of the spaceship must be 15000 m/s.

Could someone verify this? The 15,000 m/s seems small.

 

[collinsmark]

a)

Lo=40m
v=0.80c

Using the equation above I came to a answer of L = 24m

You're not applying the lengths correctly. The astronaut is the one that will measure the longer length for his own spaceship. Putting that into better words, people on the planet will measure L to be less than L₀.

Anytime an object moves relative to a particular observer's frame of reference, that object is always shorter (squished up) according to that particular observer. In other words, it's not squished up in his own frame of reference.

The astronaut will see the entire planet "squished" and even the distances between planets as being shorter (if the planets are along the astronaut's line of motion), i.e. meter sticks on the planet, along the astronaut's line of motion, are shorter than 1 m. But the astronaut doesn't observe his own ship as squished. On his own ship, a meter stick is 1 m long.

People on the planet on the other hand measure a meter stick on their own planet to be exactly 1 m long. But they measure meter sticks in the passing spaceship (meter sticks oriented along the spaceship's line of motion) to be less than 1 m.

Length is always less in the moving frame; where the stationary frame is the frame where the measurements are being performed. There is no length contraction in the stationary frame. (And again, the "stationary frame" is whichever frame of reference is doing the measuring.)*

*This assumes that the frame of reference doing the measuring is an inertial frame -- a frame that is not accelerating or rotating.

b)

L= 0.5 Lo

0.5Lo = Lo x $${\sqrt{1 - \frac{ v^2 }{ c^2 }}}$$

v^2 = (0.75)(c^2)
v= Square root [(0.75)(c)]

You've already taken the square root of c², so c is not inside of the square root.

 

[AClass]

I'm not getting another answer for a)

From what I was taught,
L is the relativistic contracted distance measured by an observer moving. [Astronaut]
Lo is the proper distance measured by a observer stationary. [Planet Observer]

For b) I realize where I went wrong.

I used a different technique this time.
If v= X x C where X is a multiple of the speed of light.
Then

0.5Lo= Lo x $${\sqrt{1 - \frac{ X^2 x C^2 }{ c^2 }}}$$

0.25 = 1-X^2
0.75= X^2

X= +/- Root 0.75

V= SQR[0.75] x 3.0x^8

Thanks the heads up.

 

[collinsmark]

I'm not getting another answer for a)

From what I was taught,
L is the relativistic contracted distance measured by an observer moving. [Astronaut]
Lo is the proper distance measured by a observer stationary. [Planet Observer]

(red emphasis mine)

Fine, but in this problem, you are being asked how long the spaceship is according to the astronaut. The astronaut is not moving relative to his/her own spaceship. The length of the spaceship according to the astronaut is L₀.

The planet on the other hand is moving relative to the spaceship when the 40 m figure was measured. That's L, the contracted length.

 

[rwisz]

Your calculations for part a) are correct, the length would be 24m according to the astronaut.

For part b), your calculations are also correct. It may seem like a small speed, but keep in mind that this is a significant fraction of the speed of light. Even a small percentage of the speed of light can result in significant relativistic effects.