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 $= 6.4 \times 10^3\, km$]
$1.25 \times {10^{ - 3}}$
$1.56 \times {10^{ - 3}}$
$1.25 \times {10^{ - 1}}$
$1.56$
A projectile is projected with velocity $k{v_e}$ in vertically upward direction from the ground into the space. ($v_e$ is escape velocity and $k < 1$). If air resistance is considered to be negligible then the maximum height from the centre of earth to whichit can go, will be : ($R =$ radius of earth)
A satellite moving with velocity $v$ in a force free space collects stationary interplanetary dust at a rate of $\frac{{dM}}{{dt}} = \alpha v$ where $M$ is the mass (of satellite + dust) at that instant . The instantaneous acceleration of the satellite is
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
Two particles of equal mass $'m'$ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is
Radius of the earth is $R$. If a body is taken to a height $3R$ from the surface of the earth than change in potential energy will be