In 1985 the space shuttle Challenger flew a cesium clock and compared its time with a fixed clock left on Earth. The shuttle orbited at approximately 330 km above Earth with a speed of 7712 m/s (∼17,250 mph). (a) Calculate the expected time lost per second for the moving clock and compare with the measured result of -295.02 \pm 0.29 ps/s, which includes a predicted effect due to general relativity of 35.0 \pm 0.06 ps/s. (b) How much time would the clock lose due to special relativity alone during the entire shuttle flight that lasted for 7 days?
Strategy This should be a straightforward application of the time dilation effect, but we have the complicating fact that the space shuttle is moving in a noninertial system (orbiting around Earth). We don’t want to consider this now, so we make the simplifying assumption that the space shuttle travels in a straight line with respect to Earth and the two events in the calculations are the shuttle passing the starting point (launch) and the ending point (landing). We are not including the effects of general relativity.
We know the orbital speed of the shuttle with respect to Earth, which allows us to determine β and the relativistic factor γ. We let T be the time measured by the clock fixed on Earth. Then we can use the time dilation effect given by Equation (2.19) to determine the proper time T_{0}^{\prime} measured by the clock in the space shuttle. The time difference is \Delta T=T-T_{0}^{\prime} \text {. We have } T_{0}^{\prime}=T \sqrt{1-\beta^{2}} \text { and } \Delta T=T-T_{0}^{\prime}=T\left(1-\sqrt{1-\beta^{2}}\right). For part (b) we need to find the total time lost for the moving clock for 7 days.
T^{\prime}=\frac{T_{0}}{\sqrt{1-v^{2} / c^{2}}}=\gamma T_{0} (2.19)
