Every planet revolves around the sun in an elliptical orbit :

$A.$ The force acting on a planet is inversely proportional to square of distance from sun.

$B.$ Force acting on planet is inversely proportional to product of the masses of the planet and the sun

$C.$ The centripetal force acting on the planet is directed away from the sun.

$D.$ The square of time period of revolution of planet around sun is directly proportional to cube of semi-major axis of elliptical orbit.

Choose the correct answer from the options given below :

Earth's orbit is an ellipse with eccentricity $0.0167$. Thus the earth's distance from the sun and speed as it moves around the sun varies from day-to-day. This means that the length of the solar day is not constant through the year. Assume that the earth's spin axis is normal to its orbital plane and find out the length of the shortest and the longest day. A day should be taken from noon to noon. Does this explain variation of length of the day during the year ?

The distance between Sun and Earth is $R.$ The duration of year if the distance between Sun and Earth becomes $3R$ will be 

A geostationary satellite is orbiting around an arbitary planet $^{\prime} P ^{\prime}$ at a height of $11 R$ above the surface of $^{\prime} P ^{\prime} ,$ $R$ being the radius of $^{\prime} P .^{\prime}$ The time period of another satellite in hours at a height of $2R$ from the surface of $^{\prime} P ^{\prime}$ is $........$.$^{\prime} P ^{\prime}$ has the time period of $24\, hours.$

The relative uncertainty in the period of a sateilite orbiting around the earth is $10^{-2}$. If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is

Kepler's second law (law of areas) is nothing but a statement of