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

$A$ satellite is to be placed in an equatorial geostationary orbit around the Earth for communication purposes. The height of such a satellite is $(M_{E} = 6 \times 10^{24} \,kg, R_{E} = 6400 \,km)$.

The Sun revolves around the galaxy with a speed of $250 \, km/s$ and its radius is $3 \times 10^{4} \, \text{light years}$. The mass of the Milky Way is:

The earth (mass $M = 6 \times 10^{24} \ kg$) revolves around the sun with an angular velocity $\omega = 2 \times 10^{-7} \ rad/s$ in a circular orbit of radius $R = 1.5 \times 10^8 \ km$. The force exerted by the sun on the earth in newtons is:

$A$ planet moves around the sun in an elliptical orbit with the sun at one of its foci. The physical quantity associated with the motion of the planet that remains constant with time is

If due to air drag,the orbital radius of a satellite decreases from $R$ to $R - \Delta R$,where $\Delta R << R$,the expression for the change in orbital velocity $\Delta v$ is (mass of Earth is $M$):