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, \[{{M}_{e}}=\]mass of the earth. Since, gravitational force provides the necessary centripetal force, so \[\frac{G{{M}_{e}}m}{{{(R+x)}^{2}}}=\frac{mv_{0}^{2}}{(R+x)}\] (where,\[{{v}_{0}}\]is orbital speed of satellite) \[\Rightarrow \]\[\frac{G{{M}_{e}}m}{(R+x)}=mv_{0}^{2}\] \[\Rightarrow \]\[\frac{g{{R}^{2}}m}{(R+x)}=mv_{0}^{2}\] \[\left( \because g=\frac{G{{M}_{e}}}{{{R}^{2}}} \right)\] \[\Rightarrow \]\[{{v}_{0}}=\sqrt{\left[ \frac{g{{R}^{2}}}{(R+x)} \right]}={{\left[ \frac{g{{R}^{2}}}{(R+x)} \right]}^{1/2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
