Gravitation Question 347
Question: A satellite is launched in the equatorial plane in such a way that it can transmit signals up to $ [60{}^\circ ] $ latitude on the earth. The angular velocity of the satellite is
Options:
A) $ [\sqrt{\frac{GM}{8R^{3}}}] $
B) $ [\sqrt{\frac{GM}{2R^{3}}}] $
C) $ [\sqrt{\frac{GM}{4R^{3}}}] $
D) $ [\sqrt{\frac{3\sqrt{3}GM}{8R^{3}}}] $
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Answer:
Correct Answer: A
Solution:
In $ [\Delta AOB: \cos 60{}^\circ= \frac{R}{OB}\RightarrowOB = 2R] $
Here gravitational force will provide the required centripetal force.
Hence.$ \[\frac{GMm}{{{\left( OB \right)}^{2}}}=m\left( OB \right){{\omega }^{2}}\] $ $ \[\Rightarrow \omega =\sqrt{\frac{GM}{{{(OB)}^{3}}}}=\sqrt{\frac{GM}{{{(2R)}^{3}}}}\Rightarrow \omega =\sqrt{\frac{GM}{8R^{3}}}\] $
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