Q 11-07-194JEE MainJEE Main 2018 (16 Apr, Shift 1)Easy
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 is:
Answer: (A) $2\times10^{-2}$
Kepler's third law: $T^2 = \dfrac{4\pi^2r^3}{GM}$, so
$$M = \frac{4\pi^2r^3}{GT^2}$$
With $r$ exact,
$$\frac{\Delta M}{M} = 2\frac{\Delta T}{T} = 2\times10^{-2}$$
Solution by Sreeraj P, M.Sc Physics