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

\[∆ t = \frac{v}{s} = \frac{1000}{100} = 10 h= 36000 s\]

\[∆ t' = \frac{∆ t}{\sqrt{1 - v^2 / c^2}}\]

\[ = \frac{36000}{\sqrt{1 - \left( \frac{1000}{36 \times 3 \times {10}^8} \right)^2}}\]

\[ ∆ t' = 36000 \left[ 1 - \left( \frac{1000}{36 \times 3 \times {10}^8} \right)^2 \right]^{- \frac{1}{2}} ∆ t\]

Now,

∆t − ∆t' = 0.154 ns

∴ Time will lag by 0.154 ns.