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 gravitational potential difference between the surface of a planet and 10 m above is 5 J/kg. If the gravitational field is supposed to be uniform, the work done in moving a 2 kg mass from the surface of the planet to a height of 8 m is

1.  2J

2.  4J

3.  6J

4.  8J

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

The satellite of mass m orbiting around the earth in a circular orbit with a velocity v. The total energy will be:

1.  $\frac{3}{4}{\mathrm{mv}}^2$

2.  $\frac{1}{2}{\mathrm{mv}}^2$

3.  $-\frac{1}{2}{\mathrm{mv}}^2$

4.  ${\mathrm{mv}}^2$

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 thin rod of length L is bent to form a semicircle. The mass of the rod is M. The gravitational potential at the centre of the circle is :

1. $-\frac{\mathrm{GM}}{\mathrm{L}}$

2. $-\frac{\mathrm{GM}}{2\mathrm{\pi L}}$

3. $-\frac{\mathrm{\pi GM}}{2\mathrm{L}}$

4. $-\frac{\mathrm{\pi GM}}{\mathrm{L}}$

Weightlessness experienced while orbiting the earth in space-ship is the result of

(1)   Inertia                        

(2)  Acceleration

(3)  Zero gravity

(4)  Freefall towards the earth

The escape velocity from the earth is about 11 km/second. The escape velocity from a planet having twice the radius and the same mean density as the earth is

(1) 22 km/sec                               (2) 11 km/sec

(3) 5.5 km/sec                               

(4) 15.5 km/sec

If g is the acceleration due to gravity at the earth's surface and r is the radius of the earth, the escape velocity for the body to escape out of the earth's gravitational field is:

(1) gr                                       

(2) $\sqrt{2gr}$

(3) g/r                                       

(4) r/g