The angular speed of earth, so that the object on equator may appear weightless, is $(g = 10\,m/{s^2}$, radius of earth $6400\, km$)

In order to find time, the astronaut orbiting in an earth satellite should use

Gravitational acceleration on the surface of a planet is $\frac{\sqrt 6}{11}g$ , where $g$ is the gravitational acceleration on the surface of the earth. The average mass  density of the planet is $\frac{2}{3}\, times$ that of the earth. If the escape speed on the surface of the earth is taken to be $11\, kms^{-1}$, the escape speed on the surface of  the  planet in $kms^{-1}$ will be

$T$ is the time period of simple pendulum on the earth's surface. Its time period becomes $x T$ when taken to a height $R$ (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be:

Obtain an expression of acceleration produced by gravity of earth.