JEE Advanced 2025 — Paper 2 — PHY-III solution

JEE Advanced2025Paper 2PHY-III

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 _____

Your answer

Solution

By Kepler's third law $T \propto r^{3/2}$ , so $ω \propto r^{- 3/2}$ . For the two satellites:

$\frac{ω _{2}}{ω _{1}} = (\frac{r _{1}}{r _{2}})^{3/2} = (1.21)^{3/2} = (1.1)^{3} = 1.331 = \frac{4}{3}$

The two satellites move in opposite senses, so their relative angular speed (as seen from one to the other) is $ω_{1} + ω_{2}$ . The relative period is

$t_{0} = \frac{2 π}{ω _{1} + ω _{2}} = \frac{2 π}{ω _{1} ( 1 + 4/3 )} = \frac{2 π}{( 7/3 ) ω _{1}} = \frac{3}{7} \cdot T_{1}$

With $T_{1} = 24$ hours (geostationary):

$t_{0} = \frac{3 \times 24}{7}$ hours $= \frac{24}{7/3}$ hours $\Rightarrow p = \frac{7}{3} \approx 2.33$ .

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