In planetary motion, the areal velocity of the position vector of a planet depends on the angular velocity ($\omega$) and the distance of the planet from the sun (r). The correct relation for areal velocity is:

1. $\frac{dA}{dt}$ $\alpha$ $\omega r$

2. $\frac{dA}{dt}$ $\alpha$ ${\omega }^{2}r$

3. $\frac{dA}{dt}$ $\alpha$ $\omega r^2$

4. $\frac{dA}{dt}$ $\alpha$ $\sqrt{\omega r}$

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

Kepler's second law regarding constancy of the areal velocity of a planet is a consequence of the law of conservation of:

1. Energy

2. Linear momentum

3. Angular momentum

4. Mass

The distance of neptune and saturn from sun are nearly ${10}^{13}$ and ${10}^{12}$ meters respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio

(1) $\sqrt{10}$                                               

(2) 100

(3) $10\sqrt{10}$                                             

(4) $1/\sqrt{10}$ 

The period of a satellite in a circular orbit of radius R is T, the period of another satellite in a circular orbit of radius 4R is

(1)4T                                                   

(2)$\frac{T}{4}$

(3)8T                                                   

(4)$\frac{T}{8}$

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

(1)$\frac{{d_1}^2{v}_1}{{d_2}^2}$                                             

(2)$\frac{d_2{v}_1}{d_1}$

(3)$\frac{d_1{v}_1}{d_2}$                                               

(4)$\frac{{d_2}^2{v}_1}{{d_1}^2}$

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. $\mathrm{T}\propto \frac{{\mathrm{r}}^5}{\mathrm{GM}}$

2. $\mathrm{T}\propto \sqrt{\frac{{\mathrm{r}}^3}{\mathrm{GM}}}$

3. $\mathrm{T}\propto \sqrt{\frac{\mathrm{r}}{{\mathrm{GM}}^2/3}}$

4. $\mathrm{T}\propto \sqrt{\frac{{\mathrm{r}}^3}{{\mathrm{GM}}^2/4}}$

Given the radius of Earth ‘R’ and length of a day ‘T’, the height of a geostationary satellite is:

[G–Gravitational Constant, M–Mass of Earth] 

(a) ${\left(\frac{4{\mathrm{\pi }}^2\mathrm{GM}}{T^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$                      (b) ${\left(\frac{4\mathrm{\pi GM}}{R^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-R$

(c) ${\left(\frac{{\mathrm{GMT}}^2}{4{\mathrm{\pi }}^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-R$                   (d) ${\left(\frac{{\mathrm{GMT}}^2}{4{\mathrm{\pi }}^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+R$

The rotation period of an earth satellite close to the surface of the earth is 83 minutes. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be

(1) 83 minutes                                   

(2) $83\times \sqrt{8}$ minutes

(3) 664 minutes                                 

(4) 249 minutes

If the distance between the earth and the sun becomes half its present value, the number of days in a year would have been
(1) 64.5                                     

(2) 129

(3) 182.5                                   

(4) 730