A satellite whose mass is $m$, is revolving in a circular orbit of radius $r$, around the earth of mass $M$. Time of revolution of the satellite is:
1. $T \propto \frac{r^5}{GM}$
2. $T \propto \sqrt{\frac{r^3}{GM}}$
3. $T \propto \sqrt{\frac{r}{\frac{GM^2}{3}}}$
4. $T \propto \sqrt{\frac{r^3}{\frac{GM^2}{4}}}$

The distance of a planet from the sun is $5$ times the distance between the earth and the sun. The time period of the planet is: 

A planet moves around the sun. At a point $P,$ it is closest to the sun at a distance $d_1$ and has speed $v_1.$ At another point $Q,$ when it is farthest from the sun at distance $d_2,$ its speed will be:

A satellite that is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of the earth in the new orbit is $2$ times that of the earlier orbit. The time period in the second orbit is:
1. $48$$\sqrt{2}$ hr
2. $48$ hr
3. $24$$\sqrt{2}$ hr
4. $24$ hr

If $A$ is the areal velocity of a planet of mass $M,$ then its angular momentum is:

Kepler's third law states that the square of the period of revolution ($T$) of a planet around the sun, is proportional to the third power of average distance $r$ between the sun and planet i.e. $T^2 = Kr^3$, here $K$ is constant. If the masses of the sun and planet are $M$ and $m$ respectively, then as per Newton's law of gravitation, the force of attraction between them is $F = \frac{GMm}{r^2},$ here $G$ is the gravitational constant. The relation between $G$ and $K$ is described as:
1. $GK = 4\pi^2$
2. $GMK = 4\pi^2$
3. $K =G$
4. $K = \frac{1}{G}$

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

If $R$ is the radius of the orbit of a planet and $T$ is the time period of the planet, then which of the following graphs correctly shows the motion of a planet revolving around the sun?

1.		2.
3.		4.

If two planets are at mean distances $d_1$ and $d_2$ from the sun and their frequencies are $n_1$ and $n_2$ respectively, then:
1. $n^2_1d^2_1= n_2d^2_2$
2. $n^2_2d^3_2= n^2_1d^3_1$
3. $n_1d^2_1= n_2d^2_2$
4. $n^2_1d_1= n^2_2d_2$

The time period of a geostationary satellite is $24$ h, at a height of $6R_E$$(R_E$ is the radius of the earth) from the surface of the earth. The time period of another satellite whose height is $2.5R_E$ from the surface will be:
1. $6\sqrt{2}~\text{h}$
2. $12\sqrt{2}~\text{h}$
3. $\frac{24}{2.5}~\text{h}$
4. $\frac{12}{2.5}~\text{h}$