JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2018
- Medium
- MCQ
The relative uncertainty in the period of a sateilite 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 is
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Answer & Solution
Correct Answer
(C) \(2\times 10^{-2}\)
Step-by-step Solution
Detailed explanation
From Kepler's law, time period of a satellite,
\(T = 2\pi \sqrt {\frac{{{r^3}}}{{Gm}}} \,\,\,\,\,\,\,{T^2} = \frac{{4{\pi ^2}}}{{GM}}{r^3}\)
\(M = \frac{4\pi^2 r^3}{GT^2}\)
\(\frac{\Delta M}{M} = 3 \frac{\Delta r}{r} + 2 \frac{\Delta T}{T}\)
\(\frac{\Delta M}{M} = 3(0) + 2(10^{-2})\)
\(\frac{\Delta M}{M} = 2 \times 10^{-2}\)
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