What is the escape speed of a rocket on the far side of Ganymede?

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SUMMARY

The escape speed of a rocket on the far side of Ganymede, Jupiter's largest moon, can be calculated using gravitational potential energy equations. The radius of Ganymede is 2.64 x 106 m, and its mass is 1.495 x 1023 kg. The mass of Jupiter is 1.90 x 1027 kg, with a distance of 1.071 x 109 m between Jupiter and Ganymede. The escape velocity is derived from the equation U = -GMm/r, where the total potential energy combines the gravitational effects of both Ganymede and Jupiter.

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A friend asked me for help on this problem. I don't know much about physics but I did find this place. Any chance you guys could help me?

1: Determine the escape speed of a rocket on the far side of Ganymede, the largest of Jupiter's moons. The radius of Ganymede is 2.64 X 10^6m, and its mass is 1.495 X 10^23 kg. The mass of Jupiter is 1.90 x 10^27 kg, and the distance between Jupiter and Ganymede is 1.071 X 10^9m. Be sure to include the gravitational effect due to jupiter, but you may ignore the motion of Jupiter and Ganymede as they revolve about their center of mass. (U = -GMm/r)
 
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Alright, if "r" is the distance of the rocket from the center of Ganymede, then the potential energy at position "r" is given by:
[tex]U=\frac{G*mass of Ganymede*mass of rocket}{r}+\frac{G*mass of Jupiter*mass of rocket}{r+distance between Jupiter and Ganymede}[/tex]
[tex]=Gm(\frac{M_G}{r}+\frac{M_J}{r+1.071*10^9})[/tex]
(You can just add the distance between Ganymede and Jupiter to "r" because the rocket is on the far side)

So in order to escape, you need to have enough kinetic energy to overcome the potential difference. In our case the potential difference is the difference between U at r=infinity and U at r=radius of Ganymede. So take the difference of the two potentials and set them equal to [itex]\frac{1}{2}mv^2[/tex], and solve for v. The mass of the rocket will cancel from the equation.[/itex]
 

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