Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Can you help me with this one? I\'m not given thevelocity, so I don\'t know even

ID: 1728577 • Letter: C

Question

Can you help me with this one? I'm not given thevelocity, so I don't know even how to approach it. Any helpis greatly appreciated.
A clock is placed in a satellite that orbits Earth with an orbitalperiod of 105 min. By what timeinterval will this clock differ from an identical clock on Earthafter 1 y? (Assume that specialrelativity applies and neglect generalrelativity.)  [answer in seconds]
Can you help me with this one? I'm not given thevelocity, so I don't know even how to approach it. Any helpis greatly appreciated.

Explanation / Answer

We know that F = ms*a =G*ms*mE/r2 a = G*mE/r2 We also know the centripetal acceleration is: a = v2/rand a = v2/r =G*mE/r2v2 =G*mE/r And we know that v = xr or v = *r*sin(). If we assumea circular orbit, then sin() = 1, and v2 = G*mE/(v/) v3 = G*mE* We can get the velocity: v = (G*mE*)(1/3) G = 6.673*10-11 m3/kg-s2 mE = 5.9742*1024 kg = 2*/((105 min)*(60 s/min)) = 9.973*10-4rad/s The velocity is: v = ((6.673*10-11m3/kg-s2)*(5.9742*1024kg)*(9.973*10-4 rad/s))(1/3) =7.353*103 m/s By the Lorentz transformation: diff = t0/(1 -v2/c2)0.5 - t0 = (1y)/(1 - (7.353*103m/s)2/(2.99792458*108m/s)2)05 - 1 y = (1 y)*3.008*10-10s diff = (1 y)*(3.008*10-10) = (3.15576*107s)(3.008*10-10) = 0.0095 s