Acceleration Due To Gravity On Other Planets

Surface gravity of a planet is found by applying the universal law to a unit mass at its surface, giving $g=\frac{GM}{R^2}$, so $g$ is directly proportional to mass and inversely proportional to the square of radius. Comparing the exoplanet with Earth, $\frac{g_p}{g_e}=\frac{M_p}{M_e}\times {\left(\frac{R_e}{R_p}\right)}^2=2\times {\left(\frac{1}{2}\right)}^2=2\times \frac{1}{4}=\frac{1}{2}$. Therefore $g_p=\frac{9.8}{2}=4.9$ m/s^2. The option $19.6$ m/s^2 wrongly multiplies the mass factor by the radius factor as if both increased gravity. The option $9.8$ m/s^2 assumes the two effects cancel exactly, which they do not. The option $2.45$ m/s^2 over-reduces by using radius to the fourth power. This is the standard NCERT comparison of surface gravity between planets. The key insight is that radius enters the denominator as a square while mass enters the numerator only linearly, so equal fractional changes in mass and radius do not cancel; the radius effect always wins. This is also why a very dense, compact planet can have far stronger surface gravity than a larger but more diffuse one of equal mass. A plausibility check confirms that the radius increase, entering as a square, outweighs the mass increase, so gravity is reduced overall.