Length Contraction Formula: How to Calculate Observed Length at High Speeds

[Matt21]

Homework Statement

This is a question asked in a entrance examination[/B]
A spacecraft moves at a speed of 0.900c with respect to the ground. If its length is L, as measured by an observer on the spacecraft , what is the length measured by a ground observer?

Homework Equations

observed length(delta L) = proper length(delta Lo) * sqrt(1-(velocity/speed of light)^2)

The attempt at a solution
I am unable to proceed with this problem as I am not given the proper length. I don't understand if I am supposed to be using a different formula or if the answer is supposed to be an actual value or not. Any help in regards to the proper formula would be appreciated.

 

[Nugatory]

What is the relationship between $L$ and the proper length?

 

[sophiecentaur]

If its length is L, as measured by an observer on the spacecraft ,

I am not given the proper length.

The required answer is not numerical, I think.

 

[Dale]

I am not given the proper length

Yes, you are given the proper length. It is L.

Be very careful in your substitution, this could get confusing.

 

[Matt21]

When plugging in the values that I have into the equation I get Lo = L/((0.900c)^2/c^2) = L/0.316. Unless I did something wrong, I'm wondering if this is the right answer because it certainly doesn't seem like it.

 

[sophiecentaur]

I can't see how you got where you are by proper re-arrangement of that formula. Try to do that manipulation again- sticking carefully to the rules of algebra.
For instance, what happened to the "1-" bit? And the square root?

 

[MarkFL]

What you want here is:

$L=\dfrac{L_0}{\gamma(v)}$

where:

$\gamma(v)\equiv\dfrac{1}{\sqrt{1-\left(\dfrac{v}{c}\right)^2}}$

And so we have:

$L=L_0\sqrt{1-\left(\dfrac{v}{c}\right)^2}\tag{1}$

This is what you gave initially, but you wound up dividing, and not by $\gamma(v)$. Try using the formula you initially gave (1).

 

[sophiecentaur]

What you want here is:

$L=\dfrac{L_0}{\gamma(v)}$

where:

$\gamma(v)\equiv\dfrac{1}{\sqrt{1-\left(\dfrac{v}{c}\right)^2}}$

And so we have:

$L=L_0\sqrt{1-\left(\dfrac{v}{c}\right)^2}\tag{1}$

This is what you gave initially, but you wound up dividing, and not by $\gamma(v)$. Try using the formula you initially gave (1).

That's fine but I was hoping that he would dig himself out of that one. (I am a hateful person sometimes )

 

[MarkFL]

That's fine but I was hoping that he would dig himself out of that one. (I am a hateful person sometimes )

My apologies; during the time it took me to compose my post you had posted. If I had seen your post first, I certainly wouldn't have replied.

 

[sophiecentaur]

No harm done. It's very hard not to give the answer when it's staring you in the face. But, hopefully, we have solved the OP's problem now.