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. \(\frac{dA}{dt}\propto \omega r\)
2. \(\frac{dA}{dt}\propto \omega^2 r\)
3. \(\frac{dA}{dt}\propto \omega r^2\)
4. \(\frac{dA}{dt}\propto \sqrt{\omega r}\)

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. \(T \propto \frac{r^5}{GM}\)
2. \(T \propto \sqrt{\frac{r^3}{GM}}\)
3. \(T \propto \sqrt{\frac{r}{\frac{GM^2}{3}}}\)
4. \(T \propto \sqrt{\frac{r^3}{\frac{GM^2}{4}}}\)

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] 

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