A satellite is orbitting around the earth with areal speed $v_a$. At what height from the surface of the earth, it is rotating, if the radius of earth is $R$
$\frac{{4v_a^2}}{{g{R^2}}} - R$
$\frac{{2v_a^2}}{{g{R^2}}} - R$
$\frac{{v_a^2}}{{g{R^2}}} - R$
$\frac{{v_a^2}}{{2g{R^2}}} - R$
In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be $(g = 10\,m{s^{ - 2}}$ and radius of earth is $6400 \,kms)$
If the gravitational acceleration at surface of Earth is $g$ , then increase in potential energy in lifting an object of mass $m$ to a height equal to half of radius of earth from surface will be
If the change in the value of ' $g$ ' at a height ' $h$ ' above the surface of the earth is same as at a depth $x$ below it, then ( $x$ and $h$ being much smaller than the radius of the earth)
The rotation of the earth having $R$ radius about its axis speeds up to a value such that a man at latitude angle $60^o$ feels weightlessness. The duration of the day in such a case is.
A satellite $S$ is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Then