Question

A spacecraft flies away from the earth with a speed of$\mathbf{4}\mathbf{.}\mathbf{80}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}$relative to the earth and then returns at the same speed. The spacecraft carries an atomic clock that has been care- fully synchronized with an identical clock that remains at rest on earth. The spacecraft returns to its starting point 365 days (1 year) later, as measured by the clock that remained on earth. What is the difference in the elapsed times on the two clocks, measured in hours? Which clock, the one in the spacecraft or the one on earth, shows the shorter elapsed time?

Short answer

A) The time measured by the clock on Earth is 8761.12h b) The clock is on the spacecraft.

Step-by-step solution

Concept of the time dilation equation.

The time dilation equation is given as$\mathbf{\Delta t}\mathbf{=}{\mathbf{\gamma \Delta t}}_{\mathbf{0}}\mathbf{/}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{u}}^{\mathbf{2}}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}}$ where is the proper time in the rest frame,$\mathbf{∆}\mathbf{t}$is the time interval measured in the second frame of reference u is the speed of the second frame with respect to the rest frame, c is the speed of light

Calculate the time measured by the clock on Earth

The time measured by the clock on Earth can be solved by

$$\begin{gathered} ∆{\mathrm{t}}_0=\frac{1}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}∆\mathrm{t} \\ =\frac{8700}{\sqrt{1-{\displaystyle \frac{{\left(4.80\times {10}^6\right)}^2}{{\left(3\times {10}^8\right)}^2}}}} \\ =8761.12\mathrm{h} \end{gathered}$$

Therefore, the time measured by the clock on Earth is 8761.12h

Calculate the time difference between the two clocks.

The time difference between the two clocks can be solved by

$\mathrm{\Delta t}-{\mathrm{\Delta t}}_0=8761.12\mathrm{h}-8760\mathrm{h}$

The clock on the spacecraft showed smaller elapsed time since this is the clock in the rest frame producing proper time.