A) \[\frac{10}{9}R\]
B) \[\frac{9}{7}R\]
C) \[\frac{9}{8}R\]
D) \[\frac{10}{3}R\]
Correct Answer: B
Solution :
[b]: Let v be the velocity of projection of the body from the surface of the earth. According to conservation of energy, we get \[\frac{1}{2}m{{v}^{2}}-\frac{GMm}{R}=-\frac{GMm}{(R+h)}\] \[\Rightarrow \]\[{{v}^{2}}-\frac{2GM}{R}=-\frac{2GM}{(R+h)}\] As per question, \[v=\frac{3}{4}{{v}_{e}}=\frac{3}{4}\sqrt{\frac{2GM}{R}}\]\[\left( \because {{v}_{e}}=\sqrt{\frac{2GM}{R}} \right)\] \[\therefore {{\left( \frac{3}{4}\sqrt{\frac{2GM}{R}} \right)}^{2}}-\frac{2GM}{R}=-\frac{2GM}{(R+h)}\] \[\Rightarrow \]\[\frac{9}{16}\left( \frac{2GM}{R} \right)-\frac{2GM}{R}=-\frac{2GM}{(R+h)}\] \[\Rightarrow \]\[\frac{9}{16R}-\frac{1}{R}=-\frac{1}{(R+h)}\] \[\Rightarrow \]\[\frac{7}{16R}=-\frac{1}{(R+h)}\Rightarrow 16R=7R+7h\] \[\therefore \]\[h=\frac{9}{7}R\]
You need to login to perform this action.
You will be redirected in
3 sec
