What is the time period of a seconds pendulum | vedclass

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(B) The acceleration due to gravity $g$ is given by $g = \frac{GM}{R^2}$.For the new planet,$M' = 2M$ and $R' = 2R$.Thus,the new acceleration due to g

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$2\sqrt{2} \ s$

Vedclass · Answer

(B) The acceleration due to gravity $g$ is given by $g = \frac{GM}{R^2}$.For the new planet,$M' = 2M$ and $R' = 2R$.Thus,the new acceleration due to gravity $g'$ is:$g' = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2} = \frac{1}{2} \left( \frac{GM}{R^2} ight) = \frac{g}{2}$.The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.Therefore,$\frac{T'}{T} = \sqrt{\frac{g}{g'}} = \sqrt{\frac{g}{g/2}} = \sqrt{2}$.Given that for a seconds pendulum on Earth,$T = 2 \ s$.So,$T' = T 	imes \sqrt{2} = 2\sqrt{2} \ s$.

What is the time period of a seconds pendulum on a planet whose mass and radius are twice that of the Earth?

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