Gravitation Question 227
Question:A satellite moves eastwards very near the surface of the Earth in equatorial plane with speed ($ [v _{0}] $ ). Another satellite moves at the same height with the same speed in the equatorial plane but westwards. If $ [R] $ = radius of the Earth and $ [\omega ] $ ) be its angular speed of the Earth about its own axis. Then find the approximate difference in the two time period as observed on the Earth.
Options:
A) $ [\frac{4\pi \omega R^{2}}{v _{0}^{2}+R^{2}{{\omega }^{2}}}] $
B) $ [\frac{2\pi \omega R^{2}}{v _{0}^{2}-R^{2}{{\omega }^{2}}}] $
C) $ [\frac{4\pi \omega R^{2}}{v _{0}^{2}-R^{2}{{\omega }^{2}}}] $
D) $ [\frac{2\pi \omega R^{2}}{v _{0}^{2}+R^{2}{{\omega }^{2}}}] $
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Answer:
Correct Answer: C
Solution:
- $ [T _{west}=\frac{2\pi R}{v _{0}+R\omega }] $ and $ [T _{east}=\frac{2\pi R}{v _{0}-R\omega }\Rightarrow \Delta T=T _{east}] $ $ [\Rightarrow T _{east}-T _{west}=2\pi R[ \frac{2\pi R}{v^{2} _{0}-R^{2}{{\omega }^{2}}}]=\frac{4\pi \omega R^{2}}{v _{0}^{2}-R^{2}{{\omega }^{2}}}] $
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