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
$\frac{-\alpha v^{2}}{M}$
$-\alpha v^{2}$
$\frac{-2 \alpha v^{2}}{M}$
$\frac{-\alpha v^{2}}{2 M}$
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 $= 6.4 \times 10^3\, km$]
The value of escape velocity on a certain planet is $2\, km/s$ . Then the value of orbital speed for a satellite orbiting close to its surface is
Two identical spheres are placed in contact with each other. The force of gravitation between the spheres will be proportional to ($R =$ radius of each sphere)
The variation of acceleration due to gravity $g$ with distance $d$ from centre of the earth is best represented by ($R =$ Earth's radius)
The radius of a planet is $R$. A satellite revolves around it in a circle of radius $r$ with angular velocity $\omega _0.$ The acceleration due to the gravity on planet’s surface is