A geostationary satellite above the equator is orbiting around the Earth at a fixed distance from the center of the Earth. [2025]

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 $r_{1} = 1.21 r_{2}$ . The time period of the second satellite as measured from the geostationary satellite is $\frac{24}{p}$ hours. The value of $p$ is _______. [2025]

Smart Achievers · Accepted Answer

(2.33)

From Kepler's law $T^{2} \propto R^{3}$

$\frac{T_{2}}{T_{1}} = {(\frac{r_{2}}{r_{1}})}^{3 / 2}$

$Given r_{1} = 1.21 r_{2}, T_{1} = 24 hr$

$\frac{T_{2}}{24} = {(\frac{100}{121})}^{3 / 2}$

$T_{2} = {(\frac{10}{11})}^{3} \times 24 = \frac{24}{{(1.1)}^{3}} hr \dots (i)$

$T = \frac{2 π}{ω_{1} + ω_{2}} = \frac{2 π}{\frac{2 π}{T_{1}} + \frac{2 π}{T_{2}}}$

$\Rightarrow \frac{1}{T} = \frac{1}{T_{1}} + \frac{1}{T_{2}}$ $= \frac{1}{24} + \frac{1}{24} {(1.1)}^{3} [from eq. (i)]$

$= \frac{1}{24} [1 + {(1.1)}^{3}]$

$\Rightarrow T = \frac{24}{1 + {(1.1)}^{3}} = \frac{24}{2.331} = \frac{24}{P}$

$∴$ $P = 2.331 hr$

Solution

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Solution

Ans. (2.33) From Kepler's law T2∝R3 T2T1=(r2r1)3/2 Given r1=1.21 r2, T1=24 hr T224=(100121)3/2 T2=(1011)3×24=24(1.1)3hr ⋯(i) T=2πω1+ω2=2π2πT1+2πT2 ⇒1T=1T1+1T2=124+124(1.1)3 [from eq. (i)] =124[1+(1.1)3] ⇒T=241+(1.1)3=242.331=24P ∴ P=2.331 hr

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