$A$ satellite revolving around the earth at a certain height experiences acceleration due to gravity equal to $\frac{16}{49} g_0$,where $g_0$ is the acceleration due to gravity on the earth's surface. If $R$ is the radius of earth,then the square of time period of the satellite's revolution is equal to $K\left[\frac{\pi^2 R^3}{G M}\right]$. The value of $K$ is

Two satellites of mass $50\,kg$ and $100\,kg$ revolve around the earth in circular orbits of radius $9\,R$ and $16\,R$ respectively,where $R$ is the radius of the earth. The ratio of the speeds of the two satellites will be:

$A$ particle of mass $m$ is under the influence of the gravitational field of a body of mass $M (\gg m)$. The particle is moving in a circular orbit of radius $r_0$ with time period $T_0$ around the mass $M$. Then,the particle is subjected to an additional central force,corresponding to the potential energy $V_c(r) = m \alpha / r^3$,where $\alpha$ is a positive constant of suitable dimensions and $r$ is the distance from the center of the orbit. If the particle moves in the same circular orbit of radius $r_0$ in the combined gravitational potential due to $M$ and $V_c(r)$,but with a new time period $T_1$,then $(T_1^2 - T_0^2) / T_1^2$ is given by [ $G$ is the gravitational constant.]

The time period of revolution of a satellite orbiting very close to a planet of radius $R$ is $T$. What is the period of revolution around another planet,whose radius is $3R$ but having the same density?

If the orbital speed of a body revolving in a circular path near the surface of the earth is $8 \ km s^{-1}$,then the orbital speed of a body revolving around the earth in a circular orbit at a height of $19,200 \ km$ from the surface of the earth is (Radius of the earth $R = 6400 \ km$). (in $km s^{-1}$)

The Earth moves around the Sun in an elliptical orbit as shown in the figure. The ratio $OA/OB = x$. The ratio of the speed of the Earth at $B$ to that at $A$ is nearly