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:

1.	\(11.2\)	2.	\(5.6\)
3.	\(22.4\)	4.	\(53.6\)

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

What should be the velocity of earth due to rotation about its own axis so that the weight at equator become 3/5 of initial value. Radius of earth on equator is 6400 km

 (a)  $7.4\times {10}^{-4}$ rad/sac             

 (b)  6.4$\times {10}^{-4}$ rad/sac

 (c)  $7.8\times {10}^{-4}$ rad/sac               

(d)  8.7$\times {10}^{-4}$ rad/sac 

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

The escape velocity of a projectile from the earth is approximately

(1)11.2 m/sec                            (2)112 km/sec

(3)11.2 km/sec                           (4)11200 km/sec

The escape velocity of a particle of mass m varies as:

(1) $m^2$                                   

(2) m

(3) $m^0$                                   

(4) $m^{-1}$

Acceleration due to gravity is ‘g’ on the surface of the earth. The value of acceleration due to gravity at a height of 32 km above earth’s surface is (Radius of the earth = 6400 km)

(1) 0.9 g           

(2) 0.99g

(3) 0.8 g           

(4) 1.01 g

The time period of a simple pendulum on a freely moving artificial satellite is

(1) Zero                           

(2) 2 sec

(3)  3 sec                          

(4)  Infinite

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

The height of a point vertically above the earth’s surface, at which the acceleration due to gravity becomes \(1\%\) of its value at the surface is: (Radius of the earth = \(R\))
1. \(8R\)
2. \(9R\)
3. \(10R\)
4. \(20R\)