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. \(\frac{2R}{1+k^2}\)

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

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:

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

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) $m^{-1}$

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) $v=\sqrt{\frac{2GM}{R^2}}$

Escape velocity on a planet is $v_e$. If radius of the planet remains same and mass becomes 4 times, the escape velocity becomes

(1)$4v_e$                                               

(2)$2v_e$

(3)$v_e$                                                 

(4)$\frac{1}{2}{v}_e$

The mass of the earth is 81 times that of the moon and the radius of the earth is 3.5 times that of the moon. The ratio of the escape velocity on the surface of earth to that on the surface of moon will be

(1)0.2                                                             

(2)2.57

(3)4.81                                                             

(4)0.39