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A planet has a mass twice that of the earth and a diameter twice that of the earth. A pendulum that beats seconds (period 2 s) on the earth is taken to this planet. What will its period of oscillation be there?
- sqrt(2) seconds
- 2*sqrt(2) seconds
- 1/sqrt(2) second
- 1/(2*sqrt(2)) second
- 4 seconds
Correct answer: 2*sqrt(2) seconds
Solution
The pendulum period depends on local g. Compute the planet's surface gravity relative to earth's from the mass and radius scaling, then scale the seconds-pendulum period accordingly.
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