The acceleration due to gravity on the surface of moon is `1.7 ms ^-2`. What is the time period of a simple pendulum on the surface of moon if its time period of the surface of earth is `3.5 s` ? (g on the surface of the earth is `9.8 ms^-2`

The acceleration due to gravity on the surface of earth varies

The acceleration due to gravity at a place is pi^(2)m//s^(2) . Then, the time period of a simple pendulum of length 1 m is

Calculate the acceleration due to gravity on the surface of moon if mass of the moon is 1/80 times that of the Earth and diameter of the moon is 1/4 times that of the Earth.

If g is acceleration due to gravity on the surface of the Earth and R is radius of the Earth, then the gravitational potential on the surface of the Earth is

The gravitational force on the surface of th Moon is _______ times than that on the surface of the Earth.

The length of a second's pendulum on the surface of earth is 1 m. What will be the length of a second's pendulum on the moon?

If the length of a simple pendulum is equal to the radius of the earth, its time period will be

If the acceleration due to gravity on the surface of the Earth is 9.8m//s^2 ,What will be the acceleration due to ravity on the surface of the planet whose mass and radius both are two times the corresponding quantites for the Earth

Solve the following examples / numerical problems: If the acceleration due to gravity on the surface of the earth is 9.8 m/s^2, what will be the acceleration due to gravity on the surface of a planet whose mass and radius both are two times the corresponding quantities for the earth ?

Acceleration due to gravity at earth's surface is 10ms^(-2) . The value of acceleration due to gravity at the surface of a planet of mass (1/5)^(th) and radius 1/2 of the earth is