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

Step-by-step solution

The time taken by the car to travel from tower A to tower B in the Earth's frame is given by t = d v . Substituting d = 1000 km and v = 100 km h ⁻¹ , we get t = 1000 100 = 10 h = 36000 s . The speed of the car in m s ⁻¹ is v = 100 5 18 = 250 9 m s ⁻¹ . According to the time dilation formula, the proper time t₀ measured by the wristwatch in the car is related to the time t in the Earth's frame by t₀ = t 1 - v^2 c^2 . Since v c , using the binomial expansion, we can approximate this as t₀ t (1 - v^2 2c^2 ) . The time