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 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. \(\frac{2R}{1+k^2}\)

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

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

Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

1.  $-\frac{5\mathrm{Gm}}{\mathrm{r}}$

2.  $-\frac{6\mathrm{Gm}}{\mathrm{r}}$

3.  $-\frac{9\mathrm{Gm}}{\mathrm{r}}$

4.  0

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 projectile fired vertically upwards with a speed v escapes from the earth. If it is to be fired at 45$°$ to the horizontal, what should be its speed so that it escapes from the earth?

1.  v

2.  $\frac{\mathrm{v}}{\sqrt{2}}$

3.  $\sqrt{2}\mathrm{v}$

4.  2v

Kepler's second law regarding constancy of the areal velocity of a planet is a consequence of the law of conservation of:

1. Energy

2. Linear momentum

3. Angular momentum

4. Mass

A body of super dense material with mass twice the mass of the earth but size very small compared to size of the earth starts from rest from h<<R above the Earth's surface. It reaches earth in time t:

1. $t=\sqrt{\frac{h}{g}}$

2. $t=\sqrt{\frac{2h}{g}}$

3. $t=\sqrt{\frac{2h}{3g}}$

4. $t=\sqrt{\frac{4h}{3g}}$

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