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. \(20R\)

The escape velocity of an object from the earth depends upon the mass of the earth (M), its mean density, its radius (R) and the gravitational constant (G). Thus the formula for escape velocity is:

(1) $v=R\sqrt{\frac{8\mathrm{\pi }}{3}Gp}$                                         

(2) $v=M\sqrt{\frac{8\mathrm{\pi }}{3}GR}$

(3) $v=\sqrt{2GMR}$                                             

(4) $v=\sqrt{\frac{2GM}{R^2}}$