Acceleration Due To Gravity

From NCERT Class 11, Chapter 8 (Gravitation), the acceleration due to gravity at a planet's surface is $g=\frac{GM}{R^2}$, obtained by equating the gravitational force on a small test mass to its weight $mg$. This expression shows that $g$ increases with planetary mass but decreases with the square of the radius, so the two effects compete and the radius term tends to dominate. Forming the ratio for the new planet relative to Earth, $\frac{g^{\prime }}{g}=\frac{M^{\prime }}{M}{\left(\frac{R}{R^{\prime }}\right)}^2$. Substituting $M^{\prime }=2M$ and $R^{\prime }=2R$ gives $\frac{g^{\prime }}{g}=2\times \frac{1}{4}=\frac{1}{2}$, so $g^{\prime }=\frac{g}{2}$. The answer $g$ wrongly assumes the doubled mass and doubled radius cancel one-for-one, ignoring that the radius is squared in the denominator. The answer $2g$ accounts only for the doubled mass and omits the radius term entirely. The answer $g/4$ accounts only for the radius and forgets that the mass has also doubled. As a plausibility check, because the radius enters as a square it outweighs a mere doubling of the mass, so this larger planet must have weaker surface gravity than Earth, and the factor of one-half is fully consistent with that physical expectation.