Find the length of a simple pendulum on the surface of the moon that has a time period same as that of a simple pendulum of the earth's surface. Mass of earth is 80 times that of moon and the radius of earth is 4 times that of moon.

Find the length of a simple pendulum on the surface of the moon that h

1.	Find the length of a simple pendulum on the surface of the moon that has a time period same as that of a simple pendulum of the earth's surface. Mass of earth is 80 times that of moon and the radius of earth is 4 times that of moon.
Answer» Solution :If MASSES of the EARTH and the moon are `M_(1)andM_(2)` respectively, `M_(1)/M_(2)=80`. Again, their radii are `R_(1)andR_(2)` (SAY). So, `R_(1)/R_(2)=4` Acceleration due to gravity on earth, `g_(1)=(GM_(1))/R_(1)^(2)`, acceleration due to gravity on moon, `g_(2)=(GM_(2))/R_(2)^(2)` `therefore" "(g_(1))/(g_(2))=M_(1)/M_(2)xxR_(2)^(2)/R_(1)^(2)=M_(1)/M_(2)xx(R_(2)/R_(1))^(2)=80xx(1/4)^(2)=5` Also, in the case of simple pendulum, time period on earth surface `T=2pisqrt(L_(1)/g_(1))` and time period on the moon.s surface `T=2pisqrt(L_(2)/g_(2))` `therefore" "sqrt(L_(1)/g_(1))=sqrt(L_(2)/g_(2))or,L_(1)/L_(2)=(g_(1))/(g_(2))=5` `therefore" "L_(2)=L_(1)/5` Hence, length of pendulum on the moon.s surface should be `1/5`th of its length on the earth.s surface.