JEE Main 2025 Jan 24 Shift 1, Physics Q14: Satellite, Planetary Motion And Kepler’s Law

JEE Main2025Jan 24, Shift 1Physics

A satellite is launched into a circular orbit of radius ' $R$ ' around the earth. A second satellite is launched into an orbit of radius $1.03 R$ . The time period of revolution of the second satellite is larger than the first one approximately by :-

A
$3%$
B
$4.5%$
C
$9%$
D
$2.5%$

Solution

Kepler's third law: $T^{2} \propto R^{3}$ . Take logarithmic differentials:

$2 \frac{Δ T}{T} = 3 \frac{Δ R}{R}$

$\frac{Δ T}{T} = \frac{3}{2} \cdot \frac{Δ R}{R} = \frac{3}{2} \times 0.03 = 0.045 = 4.5%$

$∴$ Option (2).

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JEE Main 2025 Jan 24 Shift 1, Physics Q14: Satellite, Planetary Motion And Kepler’s Law

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