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. \(\frac{dA}{dt}\propto \omega r\)
2. \(\frac{dA}{dt}\propto \omega^2 r\)
3. \(\frac{dA}{dt}\propto \omega r^2\)
4. \(\frac{dA}{dt}\propto \sqrt{\omega r}\)

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

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

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

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 -