Acceleration Due To Gravity On Other Planets

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An exoplanet discovered by an observatory has twice the mass of the Earth and also twice the Earth's radius. What is the acceleration due to gravity on its surface, taking Earth's surface gravity as 9.8 m/s^2?

RankGuru · Accepted Answer

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, $\frac{g_p}{g_e} = \frac{M_p}{M_e}	imes\left(\frac{R_e}{R_p}ight)^2 = 2 	imes \left(\frac{1}{2}ight)^2 = 2	imes\frac{1}{4} = \frac{1}{2}$. 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.

Acceleration Due To Gravity On Other Planets

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