The time period of a pendulum is directly proportional to ______

Explanation

Let M and R `-=` mass and radius of the Earth, G `-=` gravitional constant, `h -=` altitude of an orbit, `r = R + h -=` radius of the circular orbit, `v_(c )` and `T -=` critical orbitical speed and period of the satellite.
`v_(c)=sqrt((GM)/(r))" "...(1)`
In one period (T), the satellite travels a distance equal to the circumference of its circular orbit.
`therefore T =("circumference")/("critical orbital speed")=(2pir)/(v_(c))`
Substituting for `v_(c)` from Eq. (1),
`T = (2pir)/(sqrt(GM//r))=2pisqrt((r^(3))/(GM))" "...(2)`
`therefore T^(2)=((4pi^(2))/(GM))^(3)" "...(3)`
Since `4pi^(2)//GM ` is constant for a givn planet,
`T^(2) prop r^(3)" "...(4)`
THus, the square of the period or revolution of a satellite is proportional to the cube of its orbital radius.
Equation (2) shows that `T prop (1)/(sqrt(M))`.