# Escape velocity of a rocket standing on Ganymede (Moon of Jupiter)

I want to calculate the escape velocity of a rocket, standing on the surface of Ganymede (moon of Jupiter) and trying to leave Ganymede.

My thinking was, the kinetic energy $E_{\text{KIN}}$ must be equal or bigger then the potential energy $E_{\text{POT}}$. So $E_{\text{KIN}}\geq E_{\text{POT}}$

Assuming $m$ is the mass of the rocket and $M_G$ is the mass of the moon Ganymede and $M_R$ is the radius of the moon and $v$ is the velocity of the rocket and $G$ is the Gravitational constant we can set

$$G \frac{mM_G}{R^2_G}=\frac{mv^2}{2R_G}$$ which results in

$$\sqrt{2G \frac{M_G}{R_G}}=v$$

The first question, that I have is: Is $G$ a constant only for earth or does it apply to all other planets, too? (If it does not apply, how do I calculate it?)

If you have a look at the following picture you can see, that the rocket does not only have to overcome Ganymede but Jupiter himself, too:

So I built the formula this way:

$$G \frac{M_J}{R^2} + G \frac{M_G}{R^2_G}=\frac{v^2}{2R_G}$$ where $M_J$ is the mass of Jupiter and $R$ is the distance from Jupiter's center to the starting point of the rocket.

Does this sound right?

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I am tired but this feels strange. The usual approach is to look at the involved forces, not kinetic energies. You also neglect how the rocket is propelling itself. Its mass is not constant after starting the engines. And yes, G is a universal constant. – Alexander Nov 18 '11 at 2:23
Thanks for your comment, Alexander. You are right about the changing of the mass of the rocket. But we only look at it as a model, so we neglect this. The rocket is heading straight to the right on one line with the two centers of the moon and the planet. (There is an arrow in the picture). I am tired, too. So sorry for any typos and such... :-) – Aufwind Nov 18 '11 at 2:29
Look for missing "2"s and you should be alright :-) – Alexander Nov 18 '11 at 2:49
Thanky you! :-) – Aufwind Nov 18 '11 at 3:03
No wonder you are tired if you do the problems so close to the due date ;-) – Martin Ueding Nov 19 '11 at 11:58

## 1 Answer

The answer is not correct, but only because the potential energy and kinetic energy formulas are written wrong.

To find the escape velocity, the change in kinetic plus potential energy of the rocket is set to zero. If the rocket is going at escape velocity, it isn't moving at the end, and it has no potential energy (in the usual convention where the potential energy between point gravitating masses vanishes at infinity). This gives the equation:

$${mv^2\over 2} -{G m M_J\over R_{JG}} - {G m M_G\over R_G} = 0$$

Where the left side involves $R_{JG}$, the distance between Jupiter and Ganymede, $R_G$ is the radius of Ganymede, $M_J$, the mass of Jupiter, $M_G$, the mass of Ganymede, and m, the mass of the rocket (which divides out of both sides). There are also small corrections depending on the rotational velocity of Ganymede and whether you are starting on the Jupiter side or the far-Jupiter side of the moon. Solving for v gives your equation. I don't know why you divided by R factors, the potential energy is just the product of the masses divided by the distance times G.

G is universal for all matter in the universe--- it is a universal constant which is the same on Earth, on Jupiter, or anywhere else. You can think of it as a constant the fixes the correct unit of mass for our universe as the Planck mass, once you first set the speed of light and Planck's constant (divided by 2pi) to one.

The answer for the escape velocity is

$$v= \sqrt{{2 GM_J \over R_{GJ}} + {2 M_G\over R_G} }$$

### EDIT: stupid omission

I left out something non-negligible, which is that Ganymede is orbiting Jupiter! So it has a large velocity already, enough to comepensate for half the negative potential energy of gravity. The rocket, if it takes off in the optimal direction--- going along with the orbit direction of Ganymede, will only need to increase its velocity to the escape velocity from the initial already large orbital velocity. The rocket, relative to surface Ganymede, will only need to get a velocity increment which equals v from above.

The starting velocity of the rocket is equal to the Ganymede orbital velocity around Jupiter $v_i$, which is found by making the gravitational force the centripetal force

$$v_i = \sqrt{GM_J\over R_{JG}}$$

And the correct answer for the escape velocity that a rocket would need to acquire relative to Ganymede is $v-v_i$.

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Thank you very much for this nice and understandable explanation! – Aufwind Nov 21 '11 at 23:20
One thing I don't understand. Is the distance $R_{JG}$ between the center of mass of Jupiter to the center of mass of Ganymede? Also, for the potential energy between the rocket and Jupiter, why didn't you include the radius of Ganymede? Shouldn't it be like this $$-{G m M_J\over R_{JG} + R_G}$$ – Nazaf Jun 24 '15 at 20:16
@Nazaf: I neglected R_G where it is negligible. – Ron Maimon Jun 24 '15 at 22:31