question_answer

The mass and diameter of a planet are twice those of the earth. What will be the time-period of that pendulum on this planet, which is a second's pendulum on the earth?                      

Answer:

                Initially, \[g=\frac{GM}{{{R}^{2}}}\] and \[T=2\pi \sqrt{\frac{l}{g}}=2s\] (For a second's pendulum)          When M and R are doubled, \[g'=\frac{G(2M)}{{{(2R)}^{2}}}=\frac{g}{2}\] and        \[T'=2\pi \sqrt{\frac{l}{g/2}}=\sqrt{2}T=2\sqrt{2}s.\]