A) \[\frac{-2\alpha {{v}^{2}}}{M}\]
B) \[\frac{-\alpha {{v}^{2}}}{M}\]
C) \[\frac{-\alpha {{v}^{2}}}{2M}\]
D) \[-\alpha {{v}^{2}}\]
Correct Answer: B
Solution :
[b] Thrust on the satellite, \[F=\frac{-vdM}{dt}=-v\left( \alpha v \right)=-\alpha {{v}^{2}}\] \[\text{Acceleration =}\frac{F}{M}=\frac{-\alpha {{v}^{2}}}{M}\]
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