The masses and radii of the earth and the moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are distance $d$ apart. The minimum speed with which particle of mass $m$ should be projected from a point midway between the two centres so as to escape to infinity is
$v = \sqrt {\frac{{4g({M_1} + {M_2})}}{d}} $
$v = \sqrt {\frac{{4G({M_1} + {M_2})}}{d}} $
$v = \sqrt {4G({M_1 M_2})} $
$v = \sqrt {4Gd({M_1} + {M_2})} $
Three identical bodies of equal mass $M$ each are moving along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each body is
The orbit of geostationary satellite is circular, the time period of satellite depends on $(i)$ mass of the satellite $(ii)$ mass of the earth $(iii)$ radius of the orbit $(iv)$ height of the satellite from the surface of the earth
A body of mass $m$ is situated at a distance equal to $2R$ ($R-$ radius of earth) from earth's surface. The minimum energy required to be given to the body so that it may escape out of earth's gravitational field will be
When a body is taken from pole to the equator its weight
A satellite in force free space sweeps stationary interplanetary dust at a rate of $\frac{d M}{d t}=\alpha v$ where $M$ is mass and $v$ is the speed of satellite and $\alpha$ is a constant. The acceleration of satellite is