## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 26 - Problems - Page 1010: 10

#### Answer

For an observer on the Earth, the time between ticks is an extra $0.36~ns$

#### Work Step by Step

Let $\Delta t_0$ be 1 second in the clock's frame. Let $\Delta t$ be the time measured in the Earth frame. We can find the time measured in the Earth frame: $\Delta t = \frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$ $\Delta t \approx (\Delta t_0)(1+\frac{v^2}{2c^2})$ $\Delta t \approx (1~s)\left(1+\frac{(8000~m/s)^2}{(2)(3.0\times 10^8~m/s)^2}~\right)$ $\Delta t = 1.00000000036~s$ For an observer on the Earth, the time between ticks is an extra $0.36~ns$.

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