In a satellite if the time of revolution is $T$, then $K.E.$ is proportional to
$\frac{1}{T}$
$\frac{1}{T^2}$
$\frac{1}{T^3}$
$T^{-2/3}$
Two planets move around the sun. The periodic times and the mean radii of the orbits are ${T_1},\,{T_2}$ and ${r_1},\,{r_2}$ respectively. The ratio ${T_1}/{T_2}$ is equal to
Suppose the gravitational force varies inversely as the $n^{th}$ power of distance. Then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to
The angular speed of earth in $rad/s$, so that bodies on equator may appear weightless is : [Use $g = 10\, m/s^2$ and the radius of earth $= 6.4 \times 10^3\, km$]
The mean radius of earth is $R$, and its angular speed on its axis is $\omega$. What will be the radius of orbit of a geostationary satellite?
If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth, the height of the satellite above the surface of the earth is