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Physics · PYQ NEET UG 2019(2)
हिन्दीThe time period of a geostationary satellite is \( 24\text{ h} \), at a height \( 6R_E \) (\( R_E \) is radius of earth) from surface of earth. The time period of another satellite whose height is \( 2.5 R_E \) from surface will be,
Correct answer · Option B
\( 6\sqrt{2}\text{ h} \)
Explanation
Using Kepler's third law \( T^2 \propto r^3 \), where \( r = R_E + h \). For satellite 1, \( r_1 = R_E + 6R_E = 7R_E \), \( T_1 = 24\text{ h} \). For satellite 2, \( r_2 = R_E + 2.5R_E = 3.5R_E \). Thus, \( T_2 = T_1 (r_2/r_1)^{3/2} = 24 \times (1/2)^{3/2} = 6\sqrt{2}\text{ h} \).
