JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Gravitation

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

Answer: 2.33

Solution

Using Kepler's third law:

\frac{T _{2}}{T _{1}} = (\frac{r _{2}}{r _{1}})^{3/2} ⟹ \frac{ω _{2}}{ω _{1}} = (\frac{r _{1}}{r _{2}})^{3/2} = (1.21)^{3/2} = 1.331

ω_{2} = 1.331 ω_{1}

.

The relative angular velocity is:

ω_{rel} = ω_{2} + ω_{1} = (1.331 + 1) ω_{1} = 2.331 ω_{1}

Time for one complete relative revolution:

t_{0} = \frac{2 π}{ω _{rel}} = \frac{2 π}{2.331 ω _{1}} = \frac{T _{1}}{2.331} = \frac{24}{2.331} hours

Therefore

p = 2.33

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer