JEE PYQ: Units & Measurements - Question ID ea290c434aa5 (JEE Main 2018)

The question revolves around the Keplerian orbit of a satellite around the Earth. The fundamental relation we use is the period–radius relation for a circular orbit:

where

• $T$ = orbital period,
• $r$ = radius of the orbit,
• $G$ = universal gravitational constant,
• $M$ = mass of the Earth.

We are given that the relative uncertainty (fractional error) in $T$ is $10^{-2}$ and that the relative uncertainty in $r$ is negligible. We must propagate this uncertainty to find the relative uncertainty in $M$.

Step 1: Express the relation in logarithmic form

Take natural logarithms on both sides of the period–radius relation:

Simplify:

Step 2: Differentiate to obtain relative uncertainties

Differentiate with respect to each variable, treating $G$ and $\pi$ as exact constants:

Convert differentials to uncertainties (absolute values):

Step 3: Substitute given relative uncertainties

We are told $\displaystyle \frac{\Delta T}{T} = 10^{-2}$ and $\displaystyle \frac{\Delta r}{r}\approx 0$. Hence

Therefore

Step 4: Match with the given options

The relative uncertainty in the mass of the Earth is $2\times10^{-2}$, which corresponds to option B.

1. Sign error in differentiation: Many students forget the minus sign in front of $dM/M$, leading to an incorrect magnitude.
2. Neglecting the factor of 2: The period appears squared, so its relative uncertainty must be doubled when propagated.
3. Assuming $r$ uncertainty matters: The problem states that $\Delta r/r$ is negligible; including it would give a wrong result.