A projectile is fired upwards from the surface of the earth with a velocity \(kv_e\) where \(v_e\) 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