A satellite $A$ of mass $m$ is at a distance of $r$ from the centre of the earth. Another satellite $B$ of mass $2m$ is at a distance of $2r$ from the earth's centre. Their time periods are in the ratio of

According to Kepler, the period of revolution of a planet $(T)$ and its mean distance from the sun $(r)$ are related by the equation

Mean solar day is the time interval between two successive noon when sun passes through zenith point (meridian). Sidereal day is the time interval between two successive transit of a distant star through the zenith point (meridian). By drawing appropriate diagram showing the earth's spin and orbital motion, show that mean solar day is $4\,\min$ longer than the sidereal day. In other words, distant stars would rise $4\,\min$ early every successive day

If the radius of earth's orbit is made $1\frac{1}{4}$, the duration of an year will become

Aplanet of mass $m$ is in an elliptical orbit about the sun $(m < < M_{sun})$ with an orbital period $T.$ If $A$ be the area of orbit, then its angular momentum would be :

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 :