# How to find the escape speed of moon Titan?

1. Nov 17, 2013

### Remon

1. The problem statement, all variables and given/known data

Saturn’s moon Titan is the only moon in the solar system with an atmosphere, which is 95% nitrogen molecules (N2), similar to Earth’s atmosphere. At Saturn’s distance from the Sun, the temperature of the atmosphere is only 95 K (−180 C), and the molecules have an average speed of about 0.4 km/s. How does this speed compare with the speed needed for the molecules to escape from Titan and evaporate into space? Show how you got your answer.

(A) 0.01 × Titan’s escape speed
(B) 0.07 × Titan’s escape speed
(C) 0.15 × Titan’s escape speed
(D) 0.5 × Titan’s escape speed
(E) 0.95 × Titan’s escape speed

2. Relevant equations

Sorry, there are no equations in the text or the notes... the professor simply gave us nothing to work with which is why I'm asking you guys
But I looked up online and found the formula for escape speed: Vescape = √(2GM/R), although I don't think it even applies to "atmospheres" since their mass is almost nonexistent

3. The attempt at a solution

I know that titan's escape speed is 2.65 km/sec so I divided 0.4 km/s by 2.65 km/sec and got ≈ 0.15 but I obviously need to show my work so how can I come up with Titan's escape velocity so I can compare it to its atmosphere's speed?

2. Nov 17, 2013

### cepheid

Staff Emeritus
Remon,

That is the correct equation for escape speed (at a distance R from the centre of the planet). You can derive it by saying kinetic energy = gravitational potential energy, and solving for v.

If you do that, you will see that the "M" that appears in that equation is the mass of the planet or moon, not the mass of the escaping object. It's the planet or moon's gravity (potential well) that the object has to escape from, which is why its mass comes in.

3. Nov 17, 2013

### Remon

Hi, thank you for replying back. I tried using Vescape = √(2GM/R) but I got much bigger numbers (in the 1000s) when I'm actually suppose to get ≈ 2.65 km/sec, I also tried to play around with the units a bit. Here's what I did:
I know that G = 6.673 × 10-11m3kg-1s-2, MTitan is supposedly 1.3452 x 1023 kg and R is Titan's radius (2,575 km) + 975 km (because its atmosphere is ≈ 975 km above its surface), therefore R is 3550 km (I also tried using 3550000 m in the equation instead to see if I get the right answer... but I didn't).
Most of these numbers I found online because once again, there's absolutely nothing in the textbook and my lecture notes to help me with this question (great education system right?) aside from a small fact stated in the textbook: Titan's atmosphere is "extending hundreds of kilometers above the surface." (which is I guess somewhat close to the 975 km I found on wikipedia... yeah, wikipedia). Any further help would be greatly appreciated, thank you again.

Last edited: Nov 17, 2013
4. Nov 17, 2013

### SteamKing

Staff Emeritus
You must understand and pay attention to the units in your calculations. They are not ornamental irritants intended to confuse the unwary physics student, but important parts of the calculation.

If you had done a dimensional analysis of your formula with the data used in your calculations, you would have seen that the escape velocity obtained is in meters/sec, not km/sec. Also, the magnitude of the escape velocity depends on the distance from the body. When you see that such and such a body has x escape velocity, it means the escape velocity at the surface of the body, unless specified otherwise.

5. Nov 17, 2013

### haruspex

Sorry to be pedantic but I don't think dimensional analysis cares about units, only dimensions, like mass, length, time, charge... But I wholly support your exhortation to track units through the equations.

6. Nov 17, 2013

### Remon

Ok, after a full page of calculations and explanations, I finally reached an answer of ≈ 0.18, which I'm hoping for it to be close enough to 0.15 because that's the answer that made sense in my first post, so I chose 0.15 (the difference was that the escape speed for Titan that I got was 2.249 km/s where the actual escape speed is 2.65 km/sec, I'm hoping that this difference is simply caused by the approximation of decimals, etc. and not a significant one). Hopefully this is right and thank you guys

7. Nov 17, 2013

### haruspex

Fwiw, I get 2640m/s (so 0.15) using the potential at the surface. If I were to take the top of atmosphere I would obviously get less, so I don't know how you got 0.18 as the fraction. Sounds like a bit more than just rounding error.