<p>Surface gravitational acceleration on an object with mass <span class="math-container">$M$</span> and radius <span class="math-container">$R$</span> is given by
<span class="math-container">$$
g = \frac{GM}{R^2} \propto G\rho R
$$</span>
where <span class="math-container">$\rho \propto M/R^3$</span> is the density of the object.
If one body has smaller surface <span class="math-container">$g$</span> than another, it must have smaller density <span class="math-container">$\rho$</span>, smaller radius <span class="math-container">$R$</span>, 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:</p>
<ul>
<li><span class="math-container">$R_\text{Titan} = 1.5 R_\text{Moon}$</span>, but</li>
<li><span class="math-container">$\rho_\text{Moon} = 3.34\rm\,g/cm^3$</span> while Titan has only <span class="math-container">$\rho_\text{Titan} = 1.88\rm\,g/cm^3$</span>.</li>
</ul>

CC BY-SA 3.0
2017-01-23T00:40:27.420Z
So by that logic, Ganymede, the largest of our moons, is also the lightest?
JohnWDailey
CC BY-SA 3.0
2017-01-23T00:59:49.557Z
I don't have any information about Ganymede's density or surface gravity --- I'd have to look it up.
rob
CC BY-SA 4.0
2020-10-18T12:43:11.950Z
No, the problem is you said "larger therefore less dense". This does not always hold
Tosic
CC BY-SA 4.0
2020-10-18T20:57:53.377Z
@Tosic Oh! That wasn't what I meant (note that Earth is larger than and denser than either the Moon or Titan). I've added a sentence which hopefully clarifies.
rob
CC BY-SA 4.0
2016-04-25T22:34:51.360Z
2020-10-18T20:56:13.947Z
2020-10-18T20:56:13.947Z
rob
<p>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?</p>

CC BY-SA 3.0
2016-04-25T22:28:16.917Z
Titan is less dense than the moon?
user151558
CC BY-SA 3.0
2016-04-26T00:46:14.367Z
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
CC BY-SA 3.0
2016-04-26T05:14:29.573Z
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
CC BY-SA 3.0
2016-04-26T05:20:08.060Z
@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
CC BY-SA 3.0
2016-04-27T06:42:22.957Z
@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
<gravity><titan>
Why does Titan have lower surface gravity than the Moon when Titan is more massive?
Ceribia
# Why does Titan have lower surface gravity than the Moon when Titan is more massive?

(Asked by: username_0 on 2016-04-25)

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?

**Tags:** gravity, titan

> **Comments:**
> **username_1**: Titan is less dense than the moon?  
> **username_2**: 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.  
> **username_3**: 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.  
> **username_3**: @username_1 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.  
> **username_4**: @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.  

---

## 1 answers

### Answer by username_5 on 2016-04-25. Score: 16 | Accepted answer

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 $\rho\_\text{Titan} = 1.88\rm\,g/cm^3$.

> **Comments:**
> **username_6**: So by that logic, Ganymede, the largest of our moons, is also the lightest?  
> **username_5**: I don't have any information about Ganymede's density or surface gravity --- I'd have to look it up.  
> **username_7**: No, the pusername_5lem is you said "larger therefore less dense". This does not always hold  
> **username_5**: @username_7 Oh! That wasn't what I meant (note that Earth is larger than and denser than either the Moon or Titan). I've added a sentence which hopefully clarifies.  

---

