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The relative uncertainty in the period of a satellite orbiting around the earth is 10⁻². If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth I -

1. 3 × 10⁻²
2. 10⁻²
3. 2 × 10⁻²
4. 6 × 10⁻²

Correct answer: 2 × 10⁻²

Solution

From Kepler's third law T^2 ~ r^3/(GM), so M ~ r^3/T^2. With r uncertainty negligible, dM/M = 2 (dT/T) = 2 x 10^-2.

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