The time period of a geostationary satellite is 24 hr at a height 6R_E ( R_E is the radius of the Earth) from the surface of the earth. The time perio - PHYSICS MCQ NEET 2019 2026 | ExamTest.live

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The time period of a geostationary satellite is 24 hr at a height 6 R_{E} (R_{E} is the radius of the Earth) from the surface of the earth. The time period of another satellite whose height is 2.5 R_{E} from the surface will be:

Solution & Step-by-step Explanation

According to Kepler's Third Law, T^{2} \propto r^{3}, where r is the distance from the center of the Earth.For satellite 1: r_{1} = R_{E} + 6 R_{E} = 7 R_{E}, T_{1} = 24 hr .For satellite 2: r_{2} = R_{E} + 2.5 R_{E} = 3.5 R_{E} . \frac{T _{2}^{2}}{T _{1}^{2}} = (\frac{r _{2}}{r _{1}})^{3} \frac{T _{2}^{2}}{2 4 ^{2}} = (\frac{3.5 R _{E}}{7 R _{E}})^{3} = (\frac{1}{2})^{3} = \frac{1}{8} T_{2}^{2} = \frac{24 \times 24}{8} = 72 T_{2} = 72 = 62 hr .

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The time period of a geostationary satellite is 24 hr at a height 6 R_{E} (R_{E} is the radius of the Earth) from the surface of the earth. The time period of another satellite whose height is 2.5 R_{E} from the surface will be:

62 hr

122 hr

\frac{24}{2.5} hr

\frac{12}{2.5} hr

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