$A$ satellite moves in a circle around the earth. The radius of this circle is equal to one half of the radius of the moon's orbit. The satellite completes one revolution in

Which quantity remains constant in the orbital motion of the Earth around the Sun?

Kepler's third law states that the square of the period of revolution $(T)$ of a planet around the sun is proportional to the cube of the average distance $(r)$ between the sun and the planet,i.e.,$T^2 = Kr^3$,where $K$ is a constant. If the masses of the sun and the planet are $M$ and $m$ respectively,then according to Newton's law of gravitation,the force of attraction between them is $F = \frac{GMm}{r^2}$,where $G$ is the gravitational constant. The relation between $G$ and $K$ is:

The figure shows an elliptical orbit of a planet of mass $m$ about the sun $S.$ The shaded area $SCD$ is twice the shaded area $SAB.$ If $t_1$ is the time taken for the planet to move from $C$ to $D$ and $t_2$ is the time taken to move from $A$ to $B,$ then:

If the earth is at one-fourth of its present distance from the sun,the duration of the year will be

$A$ planet is revolving around the sun in an elliptical path as shown in the figure. Which of the following is the correct option?