Question

A person travelling by a car moving at 100 km h⁻¹ finds that his wristwatch agrees with the clock on a tower A. By what amount will his wristwatch lag or lead the clock on another tower B, 1000 km (in the earth's frame) from the tower A when the car reaches there?

Answer 1

Given:
Velocity of the car, v = 100 km/h = $\frac{1000}{36}$ m/s⁻¹

Distance between tower A and tower B, s = 1000 km

If ∆t be the time interval to reach tower B from tower A, then

$∆t=\frac{v}{s}=\frac{1000}{100}=10\mathrm{h}$ = 36000 s
$$\begin{gathered} ∆t\text{'}=\frac{∆t}{\sqrt{1-v^2/c^2}} \\ =\frac{36000}{\sqrt{1-{\left({\displaystyle \frac{1000}{36\times 3\times {10}^8}}\right)}^2}} \\ ∆t\text{'}=36000{\left[1-{\left(\frac{1000}{36\times 3\times {10}^8}\right)}^2\right]}^{-\frac{1}{2}}∆t \end{gathered}$$

Now,
∆t − ∆t' = 0.154 ns

∴ Time will lag by 0.154 ns.