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 $6400 \,kms)$
$0\,\;rad\,{\sec ^{ - 1}}$
$\frac{1}{{800}}rad\,se{c^{ - 1}}$
$\frac{1}{{80}}rad\,se{c^{ - 1}}$
$\frac{1}{8}rad\,se{c^{ - 1}}$
What should be the angular speed of earth, so that body lying on equator may appear weightlessness $ (g = 10\,m/{s^2},\,\,R = 6400\,km)$
A satellite can be in a geostationary orbit around a planet at a distance $r$ from the centre of the planet. If the angular velocity of the planet about its axis doubles, a satellite can now be in a geostationary orbit around the planet if its distance from the centre of the planet is
The potential energy of a satellite of mass $m$ and revolving at a height $R_e$ above the surface of earth where $R_e =$ radius of earth, is
If $v_e$ is escape velocity and $v_0$ is orbital velocity of satellite for orbit close to the earth's surface. Then these are related by
A spherical part of radius $R/2$ is excavated from the asteroid of mass $M$ as shown in the figure. The gravitational acceleration at a point on the surface of the asteroid just above the excavation is