A) 11.2 km/s
B) 22.4 km/s
C) 15.00 km/s
D) 5.8 km/s
Correct Answer: B
Solution :
At a certain velocity of projection, the body will go out of the gravitational field to the earth and will never return, this initial velocity is called escape velocity \[\frac{E}{E}=-\frac{2Q}{Q}\] where G is gravitational constant, \[\Rightarrow \] is mass of earth and \[E=-\frac{E}{2}\] is radius. For planet \[3.9\Omega \] Also since earth is assumed spherical in shape its mass is given by \[\therefore \] \[V=E-ir\] \[i=\frac{E}{R+r}\] ?(1) \[{{v}_{e}}=\sqrt{\frac{2G}{{{R}_{e}}}\times \,\frac{4}{3}\,\pi {{R}^{3}}d}\,\] ?(2) Dividing Eq. (1) by Eq. (2), we get \[V=E-\left( \frac{E}{R+r} \right)r\] \[E=2V,\,r=0.1\Omega ,R=3.9\Omega \] \[V=2-\left( \frac{2}{3.9+0.1} \right)\times 0.1\]
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