If two planets are at mean distances $d_1$ and $d_2$ from the sun and their frequencies are n₁ and n₂ respectively, then:

1. ${n_1}^2{d_1}^2=n_2{d_2}^2$                                 

2. ${n_2}^2{d_2}^3={n_1}^2{d_1}^3$

3. $n_1{d_1}^2=n_2{d_2}^2$                                   

4. ${n_1}^2{d}_1={n_2}^2{d}_2$

Suppose the law of gravitational attraction suddenly changes and becomes an inverse cube law i.e., $F\propto 1/r^3$ but the force still remains a central force, then:

The distance of a planet from the sun is \(5\) times the distance between the earth and the sun. The time period of the planet is: 

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.  ${\left(\frac{R_A}{R_B}\right)}^2$

The orbital angular momentum of a satellite revolving at a distance \(r\)from the centre is \(L\). If the distance is increased to 16r, then the new angular momentum will be:

A body of mass m kg starts falling from a point 2R above the Earth’s surface. Its kinetic energy when it has fallen to a point ‘R’ above the Earth’s surface, is:
[R - Radius of Earth, M - Mass of Earth, G - Gravitational Constant] -

1. $\frac{1}{2}\frac{GMm}{R}$              

2. $\frac{1}{6}\frac{GMm}{R}$

3. $\frac{2}{3}\frac{GMm}{R}$               

4. $\frac{1}{3}\frac{GMm}{R}$

A body is projected vertically upwards from the surface of a planet of radius R with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is:

1.  R/3                            

2.  R/2

3.  R/4                            

4.  R/5

A satellite is launched into a circular orbit of radius R around the Earth while a second satellite is launched into an orbit of radius 1.02R. The percentage difference in the time periods of the two satellites is: 

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. \(1/R\)

If the acceleration due to gravity at a height 1 km above the earth is similar to a depth d below the surface of the earth, then: 

1. d=0.5 km

2. d=1 km

3. d=1.5 km

4. d=2 km