Twin Paradox: Solving for Time Interval in Satellite Orbit | Special Relativity

[Benzoate]

Homework Statement

A clock is placed in a satellite that orbits Earth for a period of 5400 seconds. By what time interval will this clock differ from an identical clock on Earth after 1 year(assume that special relativity applies.)

Homework Equations

v(escape velocity)=11186 m/s
possibly delta(t) = delta(t')/gamma
gamma =1/sqrt(1-v^2/c^2)

The Attempt at a Solution

First I converted 1 year => 31536000 s

after 1 year , the observers on the satelitte measured the new time of the sattilitte to be

t(new)= 31536000 seconds + 5400 seconds = 31541400 seconds

after 1 year , the observers measured the Earth notices that the velocity of the satelitte to approach the speed of light.

t'(new) = 31536000 seconds + 5400 seconds/gamma=31541400.000004

the time interval between the times measured on Earth and on the satelitte after 1 year = 31541400 seconds -31541400.000004=.000004 seconds

but the back of my books says 9.6 ms. Where did I go wrong ?

 

[Doc Al]

What's the speed of the satellite?

 

[Benzoate]

What's the speed of the satellite?

It does explicity state the speed of the satellite in the problem . But I assumed the escape velocity for all objects leaving the Earth is about 11186 m/s , according to the list of physical constants in the appendix of my textbooks.

 

[Doc Al]

It does explicity state the speed of the satellite in the problem .

Does? Or does not? If not, you have to figure it out.

But I assumed the escape velocity for all objects leaving the Earth is about 11186 m/s , according to the list of physical constants in the appendix of my textbooks.

What's escape velocity have to do with it? It's orbiting the earth, not escaping.

 

[Benzoate]

Does? Or does not? If not, you have to figure it out.

What's escape velocity have to do with it? It's orbiting the earth, not escaping.

It doesn't. wouldn't the velocity be v=2*Pi*(R(earth))/1 year?

 

[Doc Al]

wouldn't the velocity be v=2*Pi*(R(earth))/1 year?

Only if its altitude is zero. Read this: Earth Orbits

 

[Benzoate]

Only if its altitude is zero. Read this: Earth Orbits

so then finding the velocity of the orbit would be pointless if we are not given the distance between the Earth and sattelite

 

[learningphysics]

so then finding the velocity of the orbit would be pointless if we are not given the distance between the Earth and sattelite

Set the gravitational force equal to centripetal force... ie:

GMm/r^2 = mv^2/r
GM/r^2 = v^2/r

using this equation and

v = 2pir/5400

You can solve for v... you have 2 equations 2 unknowns (v and r).

 

[Benzoate]

Set the gravitational force equal to centripetal force... ie:

GMm/r^2 = mv^2/r
GM/r^2 = v^2/r

using this equation and

v = 2pir/5400

You can solve for v... you have 2 equations 2 unknowns (v and r).

Isn't r just the sum of the Radius of the Earth + the altitude of the satelitte?
r=R(earth) + h

 

[learningphysics]

Isn't r just the sum of the Radius of the Earth + the altitude of the satelitte?
r=R(earth) + h

Yes. But you don't know the altitude.

 

[Benzoate]

Yes. But you don't know the altitude.

I understand how calculate the velocity but for some reason I came up with the same time interval
I'll take you through my calculations.
GM(earth)/r^2 = v^2/r => v=sqrt(G*M(earth)/r) = 2*Pi*r/T

G*M(earth)/r = 4*pi^2 *r^2 /(5400s)^2 => r^3 = ((5400s)^2/(4*pi^2))*(6.67e-11)*(5.97e24 kg) => r= 6650321.521 m

now I can find the velocity: v =2*pi*(6650321.521 m)/(5400 s) = 7738 m/s

1 year = 31536000 seconds as measured by the person on the satellite

the time measured by the observer on Earth is gamma*proper time = 31536000 seconds*(1/(sqrt(1-(7738 m)^2/c^2) =3153600seconds

so there is no time differences according to my calculations. What am I doing wrong ?

 

[Doc Al]

Since the speed is so small compared to light speed, the time difference won't be much. You won't be able to find it by plugging the numbers directly into a (typical) calculator, since 1 minus a tiny number will show up as 1. Instead, use a Taylor expansion of gamma for small v/c to calculate the time difference.

 

[Benzoate]

But are my equations and calculations are correct

You think a TI -83 or TI-89 would approximate gamma better than a mere scientific calculator?

 

[Doc Al]

You work looks OK to me. I don't know anything about those calculators, but I doubt they'll do any different. Do the Taylor expansion--that's the easy way.