# Homework Help: Special Relativity and spacecraft

1. May 6, 2010

### sean-820

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

A spacecraft is traveling around earth at 1.8x10^8m/s relative to the earth. If the spacecraft determines two events on earth to be 32H, what time interval would they find if the spacecraft is traveling at 2.82x10^8m/s?

2. Relevant equations

delta Tm=delta Ts/[1-(v^2)/(c^2)]

Where Tm= time viewed by a moving object
Ts= time viewed y a stationary object
v= velocity
c= speed of light=3x10^8m/s

3. The attempt at a solution

This is the equation i know, but i don't know how to incorporate the 32h from just the spacecrafts frame of reference

2. May 6, 2010

### Andrew Mason

The spacecraft can't be travelling "around" the earth at this speed if you mean that it is circling the earth. Does 32H mean 32 hours? Are the events at the same location on the earth? Can you give us the exact wording of the question?

AM

3. May 8, 2010

### sean-820

The spacecraft can't be travelling "around" the earth at this speed if you mean that it is circling the earth.Yes its path is around earth at that speed. Does 32H mean 32 hours? yesAre the events at the same location on the earth?yes Can you give us the exact wording of the question?

A spacecraft has a speed of 1.8x10^8m/s with respect to earth. The spacecraft determines two events on earth to be 32 hours apart. What time interval would they find if the spacecraft is traveling at 2.82x10^8m/s (instead of 1.8x10^8)?

4. May 8, 2010

### Andrew Mason

Ok. The question simply states that it is moving at a constant speed relative to the earth ie. in a straight line. It has more than enough escape velocity so it cannot be going around the earth.

How do you transform time from one inertial frame to another?

AM

Last edited: May 8, 2010
5. May 9, 2010

### sean-820

I solved it. The answer is 75 hours. I think it was more the wording of the question that messed me up. Escape velocity wouldn't matter as it isn't necessarily orbiting earth.

To solve, i just used the formula given twice. First time to sub in the initial velocity and time to get the stationary observers time, then did the equation again using this stationary time to find time observed by the spacecraft at its new speed.

Thanks for trying to help