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

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. \(\frac{1}{8}~\text{rad/s}\)