A geostationary satellite above the equator is orbiting around the earth at a fixed distance r 1 from the center of the earth. A second satellite is orbiting in the equatorial plane in the opposite

Question

A geostationary satellite above the equator is orbiting around the earth at a fixed distance $r_{1}$ from the center of the earth. A second satellite is orbiting in the equatorial plane in the opposite direction to the earth's rotation, at a distance $r_{2}$ from the center of the earth, such that $r_{1} = 1.44 r_{2}$ . The time period of the second satellite as measured from the geostationary satellite is $\frac{24}{p}$ hours. The value of p is :-

Answer:

2.70

Solution:

$T \propto r^{3/2}$

$ω_{2} = (1.44)^{3/2} ω_{1} = 1.728 ω_{1}$

$T_{re l a t i v e} = \frac{2 π}{ω _{1} + ω _{2}} = \frac{24}{2.728} h$

$p = 2.728 \approx 2.73$

TestHub · Accepted Answer

T ∝ r 3/2 ω 2 ​ = ( 1.44 ) 3/2 ω 1 ​ = 1.728 ω 1 ​ T re l a t i v e ​ = ω 1 ​ + ω 2 ​ 2 π ​ = 2.728 24 ​ h p = 2.728 ≈ 2.73

Physics - Gravitation Question with Solution | TestHub

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