Gravitional potential energy problem

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Homework Help Overview

The problem involves calculating the minimum speed required for a probe to escape the solar system when fired from a space station orbiting the sun at the same distance as Earth. The discussion centers around gravitational potential energy and kinetic energy in the context of escape velocity.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • The original poster attempts to apply conservation of energy principles but questions the reasoning behind the assumption that both potential and kinetic energy are zero after the probe escapes. Other participants clarify that escaping the solar system implies reaching a point where potential energy is zero at infinity, leading to a total energy of zero.

Discussion Status

Participants are exploring the implications of energy conservation and the definitions of potential and kinetic energy in the context of the problem. Some guidance has been provided regarding the relationship between kinetic energy and escape velocity, but multiple interpretations of the probe's launch conditions and velocities are being discussed.

Contextual Notes

There is an ongoing discussion about the assumptions regarding the probe's velocity relative to the space station and the conditions necessary for escape from the solar system. The original poster and others are considering different scenarios for the probe's launch direction and the implications for its required speed.

frostking
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Homework Statement


A space station orbits the sun at the same distance as the Earth but on the opposite side of the sun. A small probe is fired away from the station. What minimum speed does the probe need to escape the solar system?


Homework Equations


Usubg(initial) + Kinitial = Usubg(after probe fired) + K (after)



The Attempt at a Solution


I set - GmM(sun)/ r(sun) + 1/2mv(inital)^2 = 0 + 1/2 mv(after)^2

I then solve for v initial. My problem is that the explanation fro this problem shows that the value of the right side of the equation is 0---that there is 0 potential and 0 kinetic after the probe escapes and I do not understand why, how, etc. the kinetic after is 0! Can someone please tell me why this can be? Thanks very much Frostking
 
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To escape from the solar system means that the body has just enough kinetic energy to go to infinity, where its potential energy would be zero. To reach infinity, it has to consume all its kinetic energy, therefore the total energy is also 0 at infinity. As energy is conserved, the total energy of the probe is also zero when it is fired from the station. Do not forget that the potential energy is negative. The gravitational potential at the station is the same as at the position of the Earth.

ehild
 
Thanks very much! It is a bit abstract but I can see the logic of assuming all of the kinetic energy is used up to escape. I will remember that potential energy due to gravity is negative. Thanks again for the assist. Frostking
 
If the probe is launched forward so that it's velocity is added to the space station velocity, it can escape the solar system. If it is launched backwards it will have to reach the station's negative speed and then exceed it. If the station's orbital velocity is increased, it will leave the solar system. If the station's orbital velocity is decreased, it will fall into the sun.
To calculate how long it will take to leave the solar system you will need to know 3 things. The first thing you need to know is the probe velocity relative to the sun. For the second thing you must decide where the solar system ends. The final thing you need to know is the distance of the arc path out of the solar system.
Your minimum speed will depend on when you want the probe to leave the solar system. Just divide the arc distance out by the time to get your velocity. This only works for velocities greater than the station's velocity.
 

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