A) \[\frac{1}{2}m{{v}^{2}}\]
B) \[-\frac{1}{2}m{{v}^{2}}\]
C) \[-m{{v}^{2}}\]
D) \[\frac{3}{2}m{{v}^{2}}\]
Correct Answer: B
Solution :
Kinetic energy of satellite\[KE=\frac{1}{2}m{{v}^{2}}\] Potential energy of satellite \[PE=\frac{-GMm}{r}=-m{{v}^{2}}\] where \[v=\sqrt{\frac{GM}{r}}\] \[\therefore \] Total energy \[=KE+PE\] \[=\frac{1}{2}m{{v}^{2}}-m{{v}^{2}}=\frac{1}{2}m{{v}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
