A) \[gx\]
B) \[\frac{gR}{R-x}\]
C) \[\frac{g{{R}^{2}}}{R+x}\]
D) \[{{\left( \frac{g{{R}^{2}}}{R+x} \right)}^{1/2}}\]
Correct Answer: D
Solution :
The gravitational force exerted on satellite at a height\[x\]is \[{{F}_{G}}=\frac{G{{M}_{e}}m}{{{(R+x)}^{2}}}\] where\[Mg=\]mass of earth Since, gravitational force provides the necessary centripetal force, so, \[\frac{G{{M}_{e}}m}{{{(R+x)}^{2}}}=\frac{mv_{o}^{2}}{(R+x)}\] where\[{{v}_{o}}\]is orbital speed of satellite, \[\Rightarrow \] \[\frac{G{{M}_{e}}m}{(R+x)}=mv_{o}^{2}\] \[\Rightarrow \] \[\frac{g{{R}^{2}}m}{(R+x)}=mv_{o}^{2}\] \[\left( \because g=\frac{G{{M}_{e}}}{{{R}^{2}}} \right)\] \[\Rightarrow \] \[{{v}_{o}}=\sqrt{\left[ \frac{g{{R}^{2}}}{(R+x)} \right]}={{\left[ \frac{g{{R}^{2}}}{(R+x)} \right]}^{1/2}}\]
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