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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