# Relativity, time dilation

1. May 21, 2009

### fredrick08

1. The problem statement, all variables and given/known data
in 1962 a person orbited the earth 22 times, the press stated that each orbit he aged 2millionths of a second less, then if he remained on earth.

a.assuming he was 160km above the earth, in a circular orbit, determine the time difference between on earth and the orbiting astronaut for 22orbits.

2. Relevant equations
V=root(GM/R)
tau=gamma*t

3. The attempt at a solution
v=root(6.67x10^-11*5.98x10^24/6.37x10^6+160x10^3)=7815.5m/s

distance(circumference)=2*pi*r=2*pi*(6.37x10^6+160x10^3)=4.1x10^7m

t=4.1x10^7/7815.5=5249.7s

gamma =approx (1+.5*(7815.5/3x10^8))=1.000000001

tau=gamma*t=5249.719894s

tau-t=3.563x10^-6s????? is this rite, because, i cant see anywhere i have gone wrong... used exact answers in all my calculations. i guess that is kind of close to 2x10^-6s

2. May 22, 2009

### Andrew Mason

What was the relative speed of the astronaut relative to Houston? I think you have to factor in the rotation of the earth here.

AM

3. May 22, 2009

### chiro

I think you might have to fix up your gamma, the calculation for gamma should be
gamma = 1 / SQRT(1 - v^2/c^2)) which is going to be slightly different from your answer.

The formula for t = t0*gamma and you should get hopefully the dilated time if everything
is relative.

4. May 22, 2009

### fredrick08

ok thanks

5. May 22, 2009

### D H

Staff Emeritus
You are looking at only part of the time dilation. The astronaut ages less because of special relativistic and gravitational time dilation.

Fortunately, you don't need to know either special or general relativity to answer this question because you have already been given the answer to this question. All you need to compute is how much time passed for a person on the surface of the Earth and use the given relativistic time dilation to compute how much time passed on-orbit.

6. May 22, 2009

### Andrew Mason

We can see that this is non-relativisitic speed so we can approximate gamma by 1 + v^2/2c^2. I think that is where you went wrong. You were approximating by 1+v/2c.

The orbital speed is 7816 m/sec as you found. So the orbital period is 5.25 x10^3 seconds as you have found. 22 periods amounts to 1.14 x 10^5 seconds or about 32 hours. I don't see the 22 orbits in your calculation.

But the earth is also rotating, and during that period a person on the earth is moving at a speed of 460 m/sec. So the relative speed (depending on the direction of the orbit) is a range from (7816 + 460) m/sec to (7816 - 460) m/sec. (8276 m/sec to 7356 m/sec)

From this I get the maximum time dilation to be:

$$t'-t = 4.2 x 10^{-5}$$ sec

The minimum time dilation is about 3.4 x 10^-5 seconds.

AM

7. May 22, 2009

### D H

Staff Emeritus
You are forgetting gravitational time dilation. A person 160 km above the surface of the Earth experiences less gravitational time dilation than someone on the surface of the Earth.

There is no reason to calculate the time dilation at all because the problem already gives that answer: 2 microseconds per orbit.

8. May 23, 2009

### Andrew Mason

The first rule of being a scientist should be: don't to believe what you read in the press about science.

I think the question is asking the student to determine whether the press report was right. It appears to be right.

AM