Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
A satellite close to the earth is in orbit above the equator with a period of rotation of 1.5 h. If it is above a point P on the equator at some time, it will be above P again after time (1) 1.5 h (2) 1.6 h if it is rotating from west to east (3)\[\frac{24}{17}\]h if it is rotating from west to east (4)\[\frac{24}{17}\]h if it is rotating from east to westA) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: C
Solution :
Let\[{{\omega }_{0}}\]be the angular velocity of earth about its axis \[{{\omega }_{0}}=\frac{2\pi }{T}=\frac{2\pi }{24}\] Let\[\omega \]be the angular velocity of the satellite \[\therefore \] \[{{\omega }_{0}}=\frac{2\pi }{1.5}\] For a satellite rotating from west to east (the same as the earth), the relative angular velocity is \[{{\omega }_{1}}=\omega -{{\omega }_{0}}\] \[=\frac{2\pi }{1.5}-\frac{2\pi }{24}=2\pi \times \frac{15}{24}\] The time period of rotation of satellite relative to earth is \[{{T}_{1}}=\frac{2\pi }{{{\omega }_{1}}}=\frac{2\pi \times 24}{2\pi \times 15}=1.6\,h\] For a satellite rotating from east to west (opposite to the other), the relative velocity of satellite is \[{{\omega }_{2}}=\omega +{{\omega }_{0}}=\frac{2\pi }{1.5}+\frac{2\pi }{24}\] \[=2\pi \times \frac{17}{24}\] The time period of rotation of satellite relative to the earth is \[{{T}_{2}}=\frac{2\pi }{{{\omega }_{2}}}=\frac{2\pi \times 24}{2\pi \times 17}=\frac{24}{17}h\] Hence, option is correct.
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