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.

Explanation

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.