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 $= 6.4 \times 10^3\, km$]
$1.25 \times {10^{ - 3}}$
$1.56 \times {10^{ - 3}}$
$1.25 \times {10^{ - 1}}$
$1.56$
The additional kinetic energy to be provided to a satellite of mass $m$ revolving around a planet of mass $M$, to transfer it from a circular orbit of radius $R_1$ to another of radius $R_2\,(R_2 > R_1)$ is
In a satellite if the time of revolution is $T$, then $K.E.$ is proportional to
The Earth is assumed to be a sphere of radius $R$. A platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $fv$, where $v$ is its escape velocity from the surface of the Earth. the value of $f$ is
A geostationary satellite is orbiting the earth at a height of $6\,R$ above the surface of earth ($R$ is the radius of earth). The time period of another satellite at a height of $2.5\,R$ from the surface of the earth is :-
In a satellite if the time of revolution is $T$, then $P E$ is proportional to ..........