If \(A\) is the areal velocity of a planet of mass \(M,\) then its angular momentum is:

1.	\(\frac{M}{A}\)	2.	\(2MA\)
3.	\(A^2M\)	4.	\(AM^2\)

Magnitude of potential energy (U) and time period (T) of a satellite are related to each other as:

1. $T^2$ $\alpha$ $\frac{1}{U^3}$

2. $T$ $\alpha$ $\frac{1}{U^3}$

3. $T^2$ $\alpha$ $U^3$

4. $T^2$ $\alpha$ $\frac{1}{U^2}$

A point P lies on the axis of a ring of mass M and radius 'a' at a distance 'a' from its centre C. A small particle starts from P and reaches C under gravitational attraction. Its speed at C will be :

1. $\sqrt{\frac{2\mathrm{GM}}{\mathrm{a}}}$

2. $\sqrt{\frac{2\mathrm{GM}}{\mathrm{a}}\left(1-\frac{1}{\sqrt{2}}\right)}$

3. $\sqrt{\frac{2\mathrm{GM}}{\mathrm{a}}\left(\sqrt{2}-1\right)}$

4. zero

In planetary motion, the areal velocity of the position vector of a planet depends on the angular velocity ($\omega$) and the distance of the planet from the sun (r). The correct relation for areal velocity is:

1. $\frac{dA}{dt}$ $\alpha$ $\omega r$

2. $\frac{dA}{dt}$ $\alpha$ ${\omega }^{2}r$

3. $\frac{dA}{dt}$ $\alpha$ $\omega r^2$

4. $\frac{dA}{dt}$ $\alpha$ $\sqrt{\omega r}$

A planet is moving in an elliptical orbit. If T, V, E, and L stand, respectively, for its kinetic energy, gravitational potential energy, total energy and angular momentum about the center of the orbit, then:

A projectile is fired upwards from the surface of the earth with a velocity ${\mathrm{kv}}_{\mathrm{e}}$ where ${\mathrm{v}}_{\mathrm{e}}$ is the escape velocity and k < 1. If r is the maximum distance from the center of the earth to which it rises and R is the radius of the earth, then r equals:
1. \(\frac{R}{k^2}\)

2. \(\frac{R}{1-k^2}\)

3. \(\frac{2R}{1-k^2}\)

4. \(\frac{2R}{1+k^2}\)

A satellite is moving very close to a planet of density $\mathrm{\rho }$. The time period of the satellite is:

1.  $\sqrt{\frac{3\mathrm{\pi }}{\mathrm{\rho G}}}$

2.  ${\left(\frac{3\mathrm{\pi }}{\mathrm{\rho G}}\right)}^{3/2}$

3.  $\sqrt{\frac{3\mathrm{\pi }}{2\mathrm{\rho G}}}$

4.  ${\left(\frac{3\mathrm{\pi }}{2\mathrm{\rho G}}\right)}^{3/2}$

A projectile is fired from the surface of the earth with a velocity of 5 m/s and angle $\theta$ with the horizontal. Another projectile fired from another planet with a velocity of 3 m/s at the same angle follows a trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet (in ms^-2) is: (given, g=9.8 ms^-2)

1. 3.5

2. 5.9 

3. 16.3

4. 110.8

The escape velocity for a rocket from the earth is \(11.2\) km/s. Its value on a planet where the acceleration due to gravity is double that on the earth and the diameter of the planet is twice that of the earth (in km/s) will be:

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