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 

(a) $4 V_0$                    (b) $2 V_0$

(c) $0.5V_0$                   (d) $4 V_0$

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