The distance of a geostationary satellite from the centre of the earth (Radius R = 6400 km) is nearest to:

1. 5R                                     
2. 7R
3. 10R                                   
4. 18R

In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be: (\(g= 10~\text{ms}^{-2}\) and the radius of the earth is \(6400\) kms)
1. \(0~\text{rad/s}\)
2. \(\frac{1}{800}~\text{rad/s}\)
3. \(\frac{1}{80}~\text{rad/s}\)
4. \(\frac{1}{8}~\text{rad/s}\)

A body of mass \(m\) is taken from the earth's surface to the height \(h\) equal to the radius of the earth, the increase in potential energy will be:
1. \(mgR\)
2. \(\frac{1}{2}~mgR\)
3. \(2 ~mgR\)
4. \(\frac{1}{4}~mgR\)

Time period of a satellite revolving above Earth’s surface at a height equal to \(R\) (the radius of Earth) will be:
(\(g\) is the acceleration due to gravity at Earth’s surface)
1. \(2 \pi \sqrt{\frac{2 R}{g}}\)
2. \(4 \sqrt{2} \pi \sqrt{\frac{R}{g}}\)
3. \(2 \pi \sqrt{\frac{R}{g}}\)
4. \(8 \pi \sqrt{\frac{R}{g}}\)

An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy ${\mathrm{E}}_0$ . Its potential energy is:

1. $-{\mathrm{E}}_0$                       

2. $1.5 {\mathrm{E}}_0$

3. $2{\mathrm{E}}_0$                         

4. ${\mathrm{E}}_0$

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. $M\sqrt{\frac{GM}{R_0}}$