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

  • Thread starter Thread starter Meistro
  • Start date Start date
  • Tags Tags
    Rocket
AI Thread Summary
To determine the escape speed of a rocket on the far side of Ganymede, the gravitational effects of both Ganymede and Jupiter must be considered. The potential energy is calculated using the formula U = -GMm/r, incorporating both Ganymede's mass and Jupiter's mass at the appropriate distances. The escape speed is derived by equating the kinetic energy needed to overcome the gravitational potential difference between the rocket's position and infinity. The mass of the rocket cancels out in the calculations, simplifying the equation. Ultimately, the escape speed can be calculated by solving for v in the context of the derived potential energies.
Meistro
Messages
5
Reaction score
0
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)
 
Physics news on Phys.org
Alright, if "r" is the distance of the rocket from the center of Ganymede, then the potential energy at position "r" is given by:
U=\frac{G*mass of Ganymede*mass of rocket}{r}+\frac{G*mass of Jupiter*mass of rocket}{r+distance between Jupiter and Ganymede}
=Gm(\frac{M_G}{r}+\frac{M_J}{r+1.071*10^9})
(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 \frac{1}{2}mv^2[/tex], and solve for v. The mass of the rocket will cancel from the equation.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top