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Physics

Mechanics

Easy

Question

A satellite moves eastwards very near the surface of the Earth in equatorial plane with speed $open parentheses V subscript 0 close parentheses$ . 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.

$\frac{4 π ω R^{2}}{v_{0}^{2} + R^{2} ω^{2}}$
$\frac{2 π ω R^{2}}{v_{0}^{2} - R^{2} ω^{2}}$
$\frac{4 π ω R^{2}}{v_{0}^{2} - R^{2} ω^{2}}$
$\frac{2 π ω R^{2}}{v_{0}^{2} + R^{2} ω^{2}}$

The correct answer is: $fraction numerator 4 pi omega R squared over denominator v subscript 0 superscript 2 minus R squared omega squared end fraction$

$T subscript text west end text end subscript equals fraction numerator 2 pi R over denominator v subscript 0 plus R omega end fraction text and end text T subscript text east end text end subscript equals fraction numerator 2 pi R over denominator v subscript 0 minus R omega end fraction not stretchy rightwards double arrow straight capital delta T equals T subscript text east end text end subscript$

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