The time period of a geostationary satellite is

1. 24 hours                         

2. 12 hours

3. 365 days                         

4. One month

The orbital velocity of Earth's satellite near the surface is 7 km/s. When the radius of the orbit is 4 times more than that of Earth's radius, then orbital velocity in that orbit is:

1. 3.5 km/s                         

2. 7 km/s

3. 72 km/s                           

4. 14 km/s

A mass M is split into two parts, m and (M–m), which are then separated by a certain distance. What ratio of m/M maximizes the gravitational force between the two parts
1. 1/3                           

2. 1/2

3. 1/4                           

4. 1/5

Two identical satellites are at R and 7R away from earth surface, the wrong statement is (R = Radius of earth)

1. Ratio of total energy will be 4

2. Ratio of kinetic energies will be 4

3. Ratio of potential energies will be 4

4. Ratio of total energy will be 4 but ratio of potential and kinetic energies will be 2

For a satellite, the escape velocity is 11 km/s. If the satellite is launched at an angle of 60° with the vertical, then escape velocity will be: 

1. 11 km/s                        

2. $11\sqrt{3} km/s$

3. $\frac{11}{\sqrt{3}} km/s$                     

4. 33 km/s

The mean radius of the earth is R, its angular speed on its own axis is $\omega$ and the acceleration due to gravity at the earth's surface is g. The cube of the radius of the orbit of a geostationary satellite will be -

1. $R^{2}g/\omega$                   

2. $R^2{\omega }^2/g$

3. $Rg/{\omega }^2$                    

4. $R^{2}g/{\omega }^2$

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

Suppose the gravitational force varies inversely as the ${\mathrm{}}^{}$\(n^{th}\) power of distance then the time period of a planet in circular orbit of radius \(R\) around the sun will be proportional to:
1. \(R^{\left(\frac{n+1}{2}\right)}\)
2. \(R^{\left(\frac{n-1}{2}\right)}\)
3. \(R^n\)
4. \(R^{\left(\frac{n-2}{2}\right)}\)

The orbital speed of an artificial satellite very close to the surface of the earth is $V_o$. Then the orbital speed of another artificial satellite at a height equal to three times the radius of the earth is 

1. $4 V_0$                    
2. $2 V_0$
3. $0.5V_0$                  
4. $4 V_0$
 

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