Specific relativity Length Contraction and Time dialation question

[Galgenstrick]

1. The problem statement, all variables and given/known data

Planet X is 18 Light Years from earth, A spaceship moving towards the earth at a speed of 0.82C (Measured from Planet X) fires a series of 1.2 cm long and 2.1 gram mass (as measured by the pilot) projectiles with a speed of 0.72C (Measured by the pilot) towards the earth. At the moment the projectiles were fired the pilot measured his distance from the earth to be 1.3 light years.

a.) From the point of view of the observer on earth, how far away is the spaceship from Earth when it starts firing?

There are parts b-f, That I will add later if I get stuck on these as well.


2. Relevant equations

L'=L / \gamma


3. The attempt at a solution

I am having two difficulties. First, the speed of the spaceship is 0.82C as measured by Planet X, is this the velocity I need to answer the question? or do I need to find the velocity as it would be measured by the earth?

Second, Is there "negative velocity" in this equation? What I mean is, since the spaceship is traveling towards the earth, would the velocity be negative? Or is velocity always positive whether or not an object is moving towards or away from you?

[Doc Al]

First, the speed of the spaceship is 0.82C as measured by Planet X, is this the velocity I need to answer the question? or do I need to find the velocity as it would be measured by the earth?
Presumably, Planet X and the earth are at rest with respect to each other. So they both measure the same speed for the spaceship.
Second, Is there "negative velocity" in this equation? What I mean is, since the spaceship is traveling towards the earth, would the velocity be negative? Or is velocity always positive whether or not an object is moving towards or away from you?
Does it matter? In the equation you need for this question, v enters into the expression for gamma as v^2.

[Galgenstrick]

Presumably, Planet X and the earth are at rest with respect to each other. So they both measure the same speed for the spaceship.

Does it matter? In the equation you need for this question, v enters into the expression for gamma as v^2.

You just saved the day, thank you so much!

[Galgenstrick]

There is one other detail I missed. I was using 1.3 Light Years as the proper length, But I don't think this is correct. how can I calculate the proper length?

[Doc Al]

If the 1.3 LY distance is the measured contracted length (in the ship's frame), what must be the proper length (in the earth's frame)?

[Galgenstrick]

If the 1.3 LY distance is the measured contracted length (in the ship's frame), what must be the proper length (in the earth's frame)?

I came up with this, 1.3/sqrt(1-v^2/c^2)... Since 1.3=L/gama, does that look correct?


Edit: this doesn't seem to work, because it will give me 1.3 LY observed by the earth, which doesn't make sense.

[Doc Al]

I came up with this, 1.3/sqrt(1-v^2/c^2)... Since 1.3=L/gama, does that look correct?
Looks OK.

Edit: this doesn't seem to work, because it will give me 1.3 LY observed by the earth, which doesn't make sense.
:confused: L = 1.3*gamma.

[Galgenstrick]

Looks OK.


:confused: L = 1.3*gamma.

My text book gives me L'=L/gamma, or L=L(proper) / gamma

1.3/sqrt(1-v^2/c^2) is the same as 1.3*gamma

[Galgenstrick]

Here is what I have so far, and why I think it is not correct:

Step 1:
Find the proper distance between ship and earth,

L=L(proper) / gamma, so 1.3 = L(proper) * sqrt(1-(0.82c)^2/c^2)

Solving for L(proper) = 2.2713 LY.

Step 2:
Find the contracted distance as observed by the earth.

L=L(proper) / gamma

L= 2.2713 * sqrt(1-(0.82c)^2/c^2)

L= 1.3 LY

So you see I just went in a circle. I found the proper length, then used the proper length and got the contracted length as observed by the pilot as my answer. Somewhere I am missing something.

[Doc Al]

Here is what I have so far, and why I think it is not correct:

Step 1:
Find the proper distance between ship and earth,

L=L(proper) / gamma, so 1.3 = L(proper) * sqrt(1-(0.82c)^2/c^2)

Solving for L(proper) = 2.2713 LY.
In this context the proper distance is that measured in the earth frame. You're done.

Step 2:
Find the contracted distance as observed by the earth.

L=L(proper) / gamma

L= 2.2713 * sqrt(1-(0.82c)^2/c^2)

L= 1.3 LY

So you see I just went in a circle. I found the proper length, then used the proper length and got the contracted length as observed by the pilot as my answer. Somewhere I am missing something.
Yes, you went in a circle. Imagine that there's a ruler stretching from earth to planet X. That ruler measures 'proper' distance between those two, which is the distance according to earth observers. At some point along that ruler, the spaceship starts firing projectiles. According to the spaceship, the distance is 1.3 LY from earth. The spaceship, which is moving, sees the contracted distance. To find the proper distance, just multiply by gamma.

[Galgenstrick]

In this context the proper distance is that measured in the earth frame. You're done.


Yes, you went in a circle. Imagine that there's a ruler stretching from earth to planet X. That ruler measures 'proper' distance between those two, which is the distance according to earth observers. At some point along that ruler, the spaceship starts firing projectiles. According to the spaceship, the distance is 1.3 LY from earth. The spaceship, which is moving, sees the contracted distance. To find the proper distance, just multiply by gamma.


AHH! I see now. Thank you again!

[Galgenstrick]

Looks like I have one more question for the last part (part f) of this problem.

f.) Given that one of the projectiles misses the Earth. What is the mass of the projectile as observed from earth.

I am probably going to need the equation E=gamma*m*c^2. I know the mass of the projectile as measured from the pilot, however; I don't know the rest mass of the projectile to use this equation. I am not exactly sure where to start. I also know the speed of the projectile as observed (that was part b).

[Doc Al]

f.) Given that one of the projectiles misses the Earth. What is the mass of the projectile as observed from earth.
I presume that they want the relativistic mass of the projectile.
I am probably going to need the equation E=gamma*m*c^2.
What's the definition of relativistic mass?
I know the mass of the projectile as measured from the pilot, however; I don't know the rest mass of the projectile to use this equation.
The mass given is the rest mass of the projectile.
I also know the speed of the projectile as observed (that was part b).
Good--you'll need that speed.

[Galgenstrick]

I couldn't find anything in my textbook about relativistic mass, but here is what I found online: m = gamma * m0,

Also, Why is the mass observed by the pilot the rest mass if it is moving 0.82c?

[Doc Al]

I couldn't find anything in my textbook about relativistic mass, but here is what I found online: m = gamma * m0,
That's what you need.
Also, Why is the mass observed by the pilot the rest mass if it is moving 0.82c?
When they give the mass of the projectile as 2.1 grams as measured by the pilot, presumably they mean before he fires them. So that's the rest mass of the projectiles.

[Galgenstrick]

When they give the mass of the projectile as 2.1 grams as measured by the pilot, presumably they mean before he fires them. So that's the rest mass of the projectiles.

Right, but since the pilot is moving at 0.82c relative to the earth, so is the projectile before it is fired. So wouldn't I need the mass of the projectile as it would be if it were motionless on Earth?

[Doc Al]

Right, but since the pilot is moving at 0.82c relative to the earth, so is the projectile before it is fired. So wouldn't I need the mass of the projectile as it would be if it were motionless on Earth?
The rest mass of the projectile will be the same in any frame in which it is at rest. When the pilot measured the mass of the projectiles, presumably they were at rest with respect to him. So 2.1 g is their rest mass.

To find their relativistic mass, you need to know how fast they are moving with respect to the observer.