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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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