Post a New Question

Physics

posted by on .

A GPS satellite orbits at an altitude of 2.0 x 10^7 m and a speed of 3.9 x 10^3 m/s, Earth’s radius is 6.4 x 10^6 m, and a point at its equator has tangential speed of 460 m/s, By approximately what fraction must the time be adjusted to account for both regular/speed-dependent and gravitational time dilation?

  • Physics - ,

    t(Earth) = t(satellite) / (square root) 1 - (v/c)^2

    = t(satellite) /
    (square root) 1 - [ (3.9 x 10^3) / (3x 10^8)^2 ]

    = (1 - 1.7 x 10^-10)^ - 1/2 [ t(satellite) ]

    t(Earth) = (1 + 8.5 x 10^-11) [ t(satellite) ]


    GMEarth = (6.67 x 10^-11 N x m^2/kg^2)
    (5.98 x 10^24 kg)

    = 4.0 x 10^14 m^3/s^2


    But how did you get (5.98 x 10^24 kg)??

    To plug in

    t(Earth) = t(satellite) [1 - (1/c^2) [ ( GM/r, Earth )

    - ( GM/r, satellite )

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question