# Why does Titan have lower surface gravity than the Moon when Titan is more massive?

Reading on Wikipedia I saw that Titan is 80% more massive than the earth's moon but has only 85% the surface gravity. Why is this?

• Titan is less dense than the moon? – user151558 Apr 25 '16 at 22:28
• As a more extreme example, Uranus' mass is over 14.5 times that of the Earth, but Uranus' "surface gravity" is about 89% of Earth's surface gravity. – David Hammen Apr 26 '16 at 0:46
• The Moon is is basically rock, Titan is made up of about 50% ices. Most of the outer moons and Ceres contain a significant amount of ices/water, so they have much lower density than the inner planets and our moon. – userLTK Apr 26 '16 at 5:14
• @user151558 While this is generally correct, it's a very brief and unsatisfying answer. Surface gravity is a function of mass and radius (and sometimes speed of rotation). Density is relevant, but it's not a complete answer. – userLTK Apr 26 '16 at 5:20
• @userTLK The comment is not generally correct. It is the specifically correct answer to this very elementary question. $g \propto \rho^{2/3} M^{1/3}$. Though saying that the radius of Titan is much bigger would have been equally valid. – ProfRob Apr 27 '16 at 6:42

Surface gravitational acceleration on an object with mass $$M$$ and radius $$R$$ is given by $$g = \frac{GM}{R^2} \propto G\rho R$$ where $$\rho \propto M/R^3$$ is the density of the object. If one body has smaller surface $$g$$ than another, it must have smaller density $$\rho$$, smaller radius $$R$$, or both. Titan is larger than Earth's Moon, so your observation about its surface gravity means Titan must be less dense than the Moon. Wikipedia confirms:
• $$R_\text{Titan} = 1.5 R_\text{Moon}$$, but
• $$\rho_\text{Moon} = 3.34\rm\,g/cm^3$$ while Titan has only $$\rho_\text{Titan} = 1.88\rm\,g/cm^3$$.