Q152
General SciencePhysics
A pendulum clock that keeps correct time on the earth, is taken to moon. It will then run
- a.at correct rate.
- b.6 times faster.
- c.$\sqrt{6}$ times faster.
- d.$\sqrt{6}$ times slower.
Answer: (D) $\sqrt{6}$ times slower.
The period of a pendulum is $T=2\pi\sqrt{\frac{L}{g}}$. On the Moon $g$ is about one-sixth of its value on the Earth, so $T_{moon}=\sqrt{6}\,T_{earth}\approx2.45\,T_{earth}$: each swing takes $\sqrt{6}$ times longer and the clock runs $\sqrt{6}$ times slower. Option (C) reverses the effect and option (B) forgets the square root. A spring-driven or quartz watch would be unaffected, since it does not depend on $g$; in an orbiting satellite (effective $g=0$) a pendulum clock stops altogether.