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A satellite A of mass m is at a distance of r from the surface of the earth. Another satellite B of mass 2m is at a distance of 2r from the earth’s centre. Their time periods are in the ratio of

1. 1 : 2
2. 1 : 16
3. 1 : 32
4. 1 : 2√2

Correct answer: 1 : 2√2

Solution

The time period of a satellite is given by Kepler's third law, T² ∝ R³, where R is the distance from the center of the Earth. For satellite A, R = r + Rₑ (Earth's radius), and for satellite B, R = 2r. Taking the ratio of their time periods and simplifying gives T_A / T_B = 1 / 2√2.

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