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$

A rocket of mass \(M\) is launched vertically from the surface of the earth with an initial speed \(v\). Assuming the radius of the earth to be \(R\) and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is:
1. \(\frac{R}{\left(\frac{gR}{2v^2}-1\right)}\)
2. \(R\left({\frac{gR}{2v^2}-1}\right)\)
3. \(\frac{R}{\left(\frac{2gR}{v^2}-1\right)}\)
4. \(R{\left(\frac{2gR}{v^2}-1\right)}\)

Two bodies of masses ${\mathrm{m}}_1$ and ${\mathrm{m}}_2$  are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is

1. ${\left[2\mathrm{G}\frac{({\mathrm{m}}_1-{\mathrm{m}}_2)}{\mathrm{r}}\right]}^{1/2}$                         

2. ${\left[\frac{2\mathrm{G}}{\mathrm{r}}({\mathrm{m}}_1+{\mathrm{m}}_2)\right]}^{1/2}$

3. ${\left[\frac{\mathrm{r}}{2\mathrm{G}({\mathrm{m}}_1{\mathrm{m}}_{2)}}\right]}^{1/2}$                             

4. ${\left[\frac{2\mathrm{G}}{\mathrm{r}}{\mathrm{m}}_1{\mathrm{m}}_2\right]}^{1/2}$

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

A satellite of mass m is circulating around the earth with constant angular velocity. If the radius of the orbit is $R_0$ and mass of the earth M, the angular momentum about the centre of the earth is:

1.$\mathrm{ m}\sqrt{GM{R}_0}$                                         

2. $M\sqrt{Gm{R}_0}$

3.$\mathrm{ m}\sqrt{\frac{GM}{R_0}}$                                           

4. $M\sqrt{\frac{GM}{R_0}}$

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

A geostationary satellite orbits around the earth in a circular orbit of radius \(36000~\text{km}\). Then, the time period of a satellite orbiting a few hundred kilometres above the earth’s surface \((R_{earth}= 6400~\text{km})\) will approximately be:
1. \(\frac{1}{2}~\text{hrs}\)
2. \(1~\text{hrs}\)
3. \(2~\text{hrs}\)
4. \(4~\text{hrs}\)

A planet revolves around the sun whose mean distance is 1.588 times the mean distance between earth and the sun. The revolution time of the planet will be:

1. 1.25 years                               

2. 1.59 years

3. 0.89 years                               

4. 2 years

The earth E moves in an elliptical orbit with the sun S at one of the foci as shown in figure. Its speed of motion will be maximum at the point

1. C

2. A

3. B

4. D

The period of revolution of planet A around the sun is 8 times that of B. The distance of A from the sun is how many times greater than that of B from the sun ?

1. 2                                         

2. 3                                      

3. 4                                        

4. 5