# Calculate the speed of the spacecraft

Gold Member

## Homework Statement

A spacecraft with a proper length of 300m takes 0.75E-6s to pass an earth observer. Calculate the speed of the spacecraft as measured by the earth observer.

## The Attempt at a Solution

I'm not sure how to get this, when you plug in those numbers you get 4E8m/s, which is faster than the speed of light. This makes me think that one of the given data needs to be adjusted, but length/time changes all include u, which is what im trying to solve, and this leads to 2 unknowns in 1 equation. If anyone can give me a hint I would appreciate it.

## Answers and Replies

Galileo
Homework Helper
Can you show exactly what you've tried so far? It'll allow us to help you better.

Remember that the length of the craft with respect to the earth is not 300m, but shorter due to Lorentz contraction.

Gold Member
well $$v=\frac{d}{t}=\frac{300m}{75 microseconds}=4x10^8m/s$$

that is what i was talking about before.

so if the length is adjusted how do i use $$L=L_0\sqrt{1-u^2/c^2}$$ here, since i'm trying to find u and L in the same equation?

I think this would work: Consider the two events, the nose of the craft passing you, the earth observer, and the tail of the craft passing you. Find the value of the interval $$\Delta s^2$$. Consider the same events from the space-traveller's point of view. You know the interval is invariant. So find the time between the events from the traveller's point of view, and use it to find the relative velocity.

I'm not sure if it's the most efficient method, but it is worth a shot.

Galileo
Homework Helper
well $$v=\frac{d}{t}=\frac{300m}{75 microseconds}=4x10^8m/s$$

that is what i was talking about before.

so if the length is adjusted how do i use $$L=L_0\sqrt{1-u^2/c^2}$$ here, since i'm trying to find u and L in the same equation?

300m is the PROPER length of the craft. If you stand still and the craft passes by you, would you use the proper length or the length as measured by you? (Hint: you want the speed of the craft as measured by the earth observer. It's done by recording the position and the time of the nose and the time when the tail touches that position, then dividing the length of the craft by the time difference).

HallsofIvy
Homework Helper
well $$v=\frac{d}{t}=\frac{300m}{75 microseconds}=4x10^8m/s$$

that is what i was talking about before.

so if the length is adjusted how do i use $$L=L_0\sqrt{1-u^2/c^2}$$ here, since i'm trying to find u and L in the same equation?

No, you are not trying to find L- you are not asked for that. Just use that formula for L rather than 300m in your original formula:
$$u= \frac{300m\sqrt{1-u^2/c^2}}{75\cdot 10^{-6}}$$
and solve the equation for u.

Gold Member
thanks for the help!