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    A clock is placed in a satellite that orbits the earth with a period of 90 min. By what time interval will this clock differ from an identical clock on the earth after 1 year?

    My thoughts:

    Use T_e = \gamma T_s

    Where Te is the period of the satellite as viewed in the reference frame of earth (i.e. due to time dilation) and Ts is the period of the satellite in its reference frame.

    First find v using 2 \pi R / T where in this case T=T_s and R is the orbital radius.

    Found Orbital Radius using Kepler's laws, and after a bit of algebra, I got v = 7751 m/s = (2.5E-5)c m/s.

    Use the standard formula for the factor \gamma = 1/(1 - v^2 / c^2)^{0.5}... but this gives me  \gamma \approx 1, unsurprisingly. So, obviously, T_s \approx T_e \rightarrow T_e - T_s \approx 0... :rolleyes:

    Have I done this wrong?
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    (Original post by trm90)
    A clock is placed in a satellite that orbits the earth with a period of 90 min. By what time interval will this clock differ from an identical clock on the earth after 1 year?

    My thoughts:

    Use T_e = \gamma T_s

    Where Te is the period of the satellite as viewed in the reference frame of earth (i.e. due to time dilation) and Ts is the period of the satellite in its reference frame.

    First find v using 2 \pi R / T where in this case T=T_s and R is the orbital radius.

    Found Orbital Radius using Kepler's laws, and after a bit of algebra, I got v = 7751 m/s = (2.5E-5)c m/s.

    Use the standard formula for the factor \gamma = 1/(1 - v^2 / c^2)^{0.5}... but this gives me  \gamma \approx 1, unsurprisingly. So, obviously, T_s \approx T_e \rightarrow T_e - T_s \approx 0... :rolleyes:

    Have I done this wrong?
    Probably not, use a binomial expansion for gamma.
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    (Original post by TableChair)
    Probably not, use a binomial expansion for gamma.
    Ah okay... so I used \gamma \approx 1 + \frac{1}{2}\frac{v^2}{c^2} and I managed to get 1 again, according to my calculator. Argh! The higher order terms probably wont change anything cause they're too negligible...
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    (Original post by trm90)
    Ah okay... so I used \gamma \approx 1 + \frac{1}{2}\frac{v^2}{c^2} and I managed to get 1 again, according to my calculator. Argh! The higher order terms probably wont change anything cause they're too negligible...
    You're putting in numbers too early. Never put numbers into an equation until the end.

    Find an algebraic equation for the difference in time.

    And yes, the higher order terms are negligible, hence why you should be using a binomial expansion.
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    (Original post by TableChair)
    You're putting in numbers too early. Never put numbers into an equation until the end.

    Find an algebraic equation for the difference in time.

    And yes, the higher order terms are negligible, hence why you should be using a binomial expansion.
    Right, so, from the start I guess (excluding velocity part)

    The time difference is:

    \Delta t = t_e - t_s = \gamma t_s - t_s = (\gamma - 1) t_s. Using lower case t here to distinguish between time and period, a mistake I made before I believe.

    But \gamma - 1 \approx \frac{1}{2}\frac{v^2}{c^2}

    And now when I plug in v, I get gamma -1 = 3.125E-10, and \Delta t = 1.64 \times 10^{-4} minutes (after I converted a year into minutes).

    That works, right?
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    (Original post by trm90)
    Right, so, from the start I guess (excluding velocity part)

    The time difference is:

    \Delta t = t_e - t_s = \gamma t_s - t_s = (\gamma - 1) t_s. Using lower case t here to distinguish between time and period, a mistake I made before I believe.

    But \gamma - 1 \approx \frac{1}{2}\frac{v^2}{c^2}

    And now when I plug in v, I get gamma -1 = 3.125E-10, and \Delta t = 1.64 \times 10^{-4} minutes (after I converted a year into minutes).

    That works, right?
    Seems fine, but you might want to use seconds instead of minutes
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    (Original post by TableChair)
    Seems fine, but you might want to use seconds instead of minutes
    Cool - thanks!
 
 
 
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