# Answer to Question #108672 in Mechanics | Relativity for ALI

Question #108672
A CLOCK KEEP THE CORRECT TIME. WITH WHAT SPEED SHOULD IT MOVE RELATIVE TO OBSERVER SO THAT IT MAY SEEM TO LOSE 4 MIN IN 24 HOURS?
1
2020-04-10T08:42:23-0400

Solution: The Lorentz transformation for time intervals in different inertial frames of reference is

(1) "\\Delta t'=\\Delta t\\cdot \\sqrt{1-\\beta^2}" , where "\\beta=\\frac{V}{c}" , "V" - speed of primed system relative reference frame. Denote "\\alpha=\\frac{\\Delta t'}{\\Delta t}" and from (1) find "\\beta", we have

(2) "\\beta=\\sqrt{1-\\alpha^2}"

The clock will display the correct time in its own reference system "\\Delta t=24hour". In a moving frame of reference it will seem that it is wrong "\\Delta t'=24h-4min"

Lets astimate "\\alpha=\\frac{24h-4min}{24h}=1-\\frac{4}{24\\cdot 60}=1-2.78\\cdot10^{-3}" and denote "\\delta=2.78\\cdot10^{-3}"

(3) "\\alpha^2=(1-\\delta)^2=1-2\\cdot \\delta+\\delta^2\\simeq1-5.56\\cdot 10^{-3}" the square "\\delta" can be ignored its value is less than the third significant digit. Substituting (3) in (2) we get

"\\beta=\\sqrt{1-(1-2\\cdot\\delta)}=\\sqrt{5.56\\cdot 10^{-3}}=7.45\\cdot 10^{-2}"

The speed of clock should be "V=7.45\\cdot 10^{-2}\\cdot c=22.4\\cdot 10^6 m\/s"

Answer: Clock should move relative to observer with speed "2.24\\cdot 10^7m\/s"

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