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?

  • $\begingroup$ Titan is less dense than the moon? $\endgroup$ – user151558 Apr 25 '16 at 22:28
  • 3
    $\begingroup$ 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. $\endgroup$ – David Hammen Apr 26 '16 at 0:46
  • $\begingroup$ 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. $\endgroup$ – userLTK Apr 26 '16 at 5:14
  • $\begingroup$ @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. $\endgroup$ – userLTK Apr 26 '16 at 5:20
  • 1
    $\begingroup$ @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. $\endgroup$ – Rob Jeffries 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. Titan is larger than Earth's Moon, so it must be less dense. Wikipedia confirms:

  • $R_\text{Titan} = 1.5 R_\text{Moon}$, but
  • $\rho_\text{Moon} = 3.34\rm\,g/cm^3$ while Titan has only $\rho_\text{Titan} = 1.88\rm\,g/cm^3$.
| improve this answer | |
  • $\begingroup$ So by that logic, Ganymede, the largest of our moons, is also the lightest? $\endgroup$ – JohnWDailey Jan 23 '17 at 0:40
  • $\begingroup$ I don't have any information about Ganymede's density or surface gravity --- I'd have to look it up. $\endgroup$ – rob Jan 23 '17 at 0:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.