Assuming that the moon is a sphere of the same mean density as that of the earth and one quarter of its radius the length of a seconds on the moon (its length on the earth's surface is `99.2cm)` is .

The ratio of the mass of the moon to that of the earth is 1:81. If the radius of the moon is assumed to be one-fourth of that of the earth , then what is the acceleration due to gravity on the surface of the moon in terms of that on the earth's surface ? Hence, find the escape velocity from the surface of the moon in terms of that from the earth's surface.

The value of 'g' on moon is five times less than that on earth. If the length of the seconds pendulum on the earth is 100cm , its length on the moon is

The acceleration due to gravity on the moon is (1)/(6) th the acceleration due to gravity on the surface of the earth. If the length of a second's pendulum is 1 m on the surface of the earth, then its length on the surface of the moon will be

The mass of the moon is 1/(80) th of the mass of the earth and its radius is 1/4 th of that of the earth. Compare the accelerations due to gravity on the earth's and the moon's surface

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

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?

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