If the gravitational force between two objects were proportional to \(\frac{1}{R}\) (and not as $1/R^2$) where \(R\) is the separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to:

1. $1/R^2$                        

2. $R^{\circ }$

3. $R^1$                             

4. \(1/R\)

For the moon to cease as the earth's satellite, its orbital velocity has to be increased by a factor of -

Rohini satellite is at a height of 500 km and Insat-B is at a height of 3600 km from the surface of the earth. The relation between their orbital velocity (\(v_R,~v_i\)) is:

1. \(v_R>v_i\)

2. \(v_R<v_i\)

3. \(v_R=v_i\)

4. No relation 

If $v_e$ is the escape velocity and $v_o$ is the orbital velocity of a satellite for orbit close to the earth's surface, then these  are related by:

Two satellites A and B go around the earth in circular orbits at heights of ${\mathrm{R}}_{\mathrm{A}} \mathrm{and} {\mathrm{R}}_{\mathrm{B}}$ respectively from the surface of the earth. Assuming earth to be a uniform sphere of radius $R_e$, the ratio of the magnitudes of their orbital velocities is:

1.  $\sqrt{\frac{R_B}{R_A}}$

2.  $\frac{R_B+R_e}{R_A+R_e}$

3.  $\sqrt{\frac{R_B+R_e}{R_A+R_e}}$

4.  ${\left(\frac{R_A}{R_B}\right)}^2$

Two particles of equal masses go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is:

1. $\mathrm{v}=\frac{1}{2\mathrm{R}}\sqrt{\frac{1}{\mathrm{Gm}}}$

2. $\mathrm{v}=\sqrt{\frac{\mathrm{Gm}}{2\mathrm{R}}}$

3. $\mathrm{v}=\frac{1}{2}\sqrt{\frac{Gm}{R}}$

4. $\mathrm{v}=\sqrt{\frac{4\mathrm{Gm}}{\mathrm{R}}}$

The radii of the circular orbits of two satellites A and B of the earth are \(4R\) and \(R,\) respectively. If the speed of satellite A is \(3v,\) then the speed of satellite B will be:

A satellite S is moving in an elliptical orbit around the earth. If the mass of the satellite is very small as compared to the mass of the earth, then:

A satellite is revolving around the earth with speed ${\mathrm{v}}_0$. If it is stopped suddenly, then with what velocity will the satellite hit the ground? (${\mathrm{v}}_{\mathrm{e}}$ = escape velocity from the earth's surface)

1.  $\sqrt{v_e^2 - v_0^2}$

2.  $\sqrt{v_e^2 - 2v_0^2}$

3.  $\sqrt{v_e^2 - 3v_0^2}$

4.  $\sqrt{v_e^2 - \frac{v_0^2}{2}}$

A remote sensing satellite of the earth revolves in a circular orbit at a height of 0.25 x 10⁶ m above the surface of the earth. If the earth’s radius is 6.38x10⁶ m and g=9.8ms^-1, then the orbital speed of the satellite is:

1.  7.76 kms^-1

2.  8.56 kms^-1

3.  9.13 kms^-1

4.  6.67 kms^-1