# Escape velocity

1. Apr 23, 2005

### UrbanXrisis

The question is here

$$0.5mv^2=\frac{GmM_{moon}}{R_{moon}}+\frac{GmM_{jupiter}}{R_{jupiter}+x}$$
x=distance between jupiter and the moon
$$v=\sqrt{\frac{2GM_{moon}}{R_{moon}}+\frac{2GM_{jupiter}}{R_{jupiter}+x}}$$

$$v=\sqrt{\frac{2*6.673x10^{-11}*1.495E23kg}{2.64E6m}+\frac{2*6.673x10^{-11}*1.9E27kg}{6.99E7+1.071E9}}$$

$$v=17.657km/s$$

when I subbed in the numbers, I get about 17km/s. The books gives 15km/s, am I doing this right?

Last edited: Apr 23, 2005
2. Apr 23, 2005

### whozum

I think you should consider the planets as point masses, particularly jupiter on the right side of your first line. I would think that the distance 'x' is from center to center.

3. Apr 23, 2005

### UrbanXrisis

that would make the velocity bigger, not smaller, I already tired that but got +18km/s

4. Apr 23, 2005

### whozum

Take into account that the rocket is on the opposite side of the moon from jupiter. so the distance between jupiter and the rocket would be the distance between them plus a radius of the moon

5. Apr 23, 2005

### Gokul43201

Staff Emeritus
Nevertheless, this is the correct answer (~18 km/sec).

You can neglect the radius of Ganymede (it's 3 orders of magnitude smaller) when calculating the distance from Jupe's center.

6. Apr 23, 2005

### UrbanXrisis

I already tried that too.. the radius of the moon is too small to decrease the escape velocity by 2km/s

my book tells me an answer of 15.6 km/s, unless you're saying that it was a missprint?

Last edited: Apr 23, 2005
7. Apr 23, 2005

### UrbanXrisis

$$v=\sqrt{\frac{2GM_{moon}}{R_{moon}}+\frac{2GM_{jupiter}}{R_{jupiter}}}$$

$$v=\sqrt{\frac{2*6.673x10^{-11}*1.495E23kg}{2.64E6m}+\frac{2*6.673x10^{-11}*1.9E27kg}{1.071E9}}$$

$$v=18136m/s$$

Last edited: Apr 23, 2005
8. Apr 23, 2005

### Gokul43201

Staff Emeritus
Either that, or perhaps, one of the numbers provided in the problem is a misprint.

9. Apr 23, 2005

### UrbanXrisis

so you would agree that the escape velocity is indeed 18.136km/s?

10. Apr 23, 2005

### Gokul43201

Staff Emeritus
No, I just redid the calculation and got 15.6 km/sec. Recheck your calculation - step by step. Post numbers here if necessary.

11. Apr 23, 2005

### VietDao29

Distance = Radius of Jupiter + Distance between 2 planets + 2* Radius of Moon.
Viet Dao,

12. Apr 23, 2005

### Gokul43201

Staff Emeritus
I don't think so. The specified distance between Jupe and Gany is the distance between their centers.

13. Apr 23, 2005

### UrbanXrisis

what this the right path? I've been punching numbers into my calc for a while now, I dont see how you got 15.6 km/s

14. Apr 23, 2005

### whozum

You're punching numbers wrong, if you have maple, you can do the following:

>v=sqrt((2*G*M)/R+(2*G*N/P));

> subs(G=6.67*10^(-11),M=1.495*10^23,N=1.9*10^27,R=2.64*10^6,P=1.071*10^9,%);

15. Apr 23, 2005

### UrbanXrisis

$$v=\sqrt{\frac{2GM_{moon}}{R_{moon}}+\frac{2GM_{jupiter}}{R_{jupiter}}}$$

$$v=\sqrt{\frac{2*6.673x10^{-11}*1.495E23kg}{2.64E6m}+\frac{2*6.673x10^{-11}*1.9E27kg}{1.071E9}}$$

$$v=\sqrt{\frac{2*6.673x10^{-11}*1.495E23kg}{2.64E6m}}+\sqrt{\frac{2*6.673x10^{-11}*1.9E27kg}{1.071E9}}$$

$$v=\sqrt{\frac{1.995227*10^{13}}{2.64x10^6}}+\sqrt{\frac{2.5357*10^{17}}{1.071x10^9}}$$

$$v=2.749x10^3 + 1.5387x10^4 = 1.8156x10^4$$

Last edited: Apr 23, 2005
16. Apr 23, 2005

### VietDao29

$$\sqrt{a + b} = \sqrt{a} + \sprt{b}$$
Are you sure??????
Viet Dao,

17. Apr 23, 2005

### UrbanXrisis

I see my mistake... thanks

that took me way too long :grumpy:

18. Apr 23, 2005

### whozum

You can't split up the square root.

19. Apr 23, 2005

### UrbanXrisis

yeah... what a stupid mistake... arggg :grumpy:

20. Apr 23, 2005

### whozum

dont worry, i once embarassed myself in a class by telling the teacher to split up 1/(x^2+x) into 1/x^2 + 1/x.

Atleast your mistake was infront of 3 and not 40.