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Escaping the solar system from Earth's orbit

by irishbob
Tags: earth, escaping, orbit, solar
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irishbob
#1
Nov28-10, 09:29 PM
P: 22
1. The problem statement, all variables and given/known data
(a) What is the minimum speed, relative to the Sun, necessary for a spacecraft to escape the solar system if it starts at the Earth's orbit?

2. Relevant equations
Kinetic Energy=Potential Energy
0.5mv^2=rGM/(r^2)
0.5v^2=GM/r
Where M is the mass of the sun, and r is the radius of the solar system

3. The attempt at a solution
v=[tex]\sqrt{\frac{2GM}{r}}[/tex]
v=[tex]\sqrt{\frac{2(6.673E-11)(1.98892E30)}{7.5E15}}[/tex]
v=188.1281 m/s
v=0.18813 km/s
So I'm pretty sure I'm using the right methods and variables, but I'm not sure what's going wrong. My only thought is that I'm using the wrong value for the radius, but I can't seem to find 1 value that's agreed upon for the radius of the solar system. Help?
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Delphi51
#2
Nov28-10, 11:16 PM
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There is a strange thing to understand on this E = GMm/R formula, Bob.
When R is infinitely large, E = 0. That means the makers of the formula defined zero potential energy to be when the m is infinitely far from the M. Any finite value of R gives a larger value for E which doesn't make sense. Of course you have to fire your rockets to boost m from any finite value of R to infinity. It DOES make sense if you put a minus sign on it: E = -GMm/R. Then your m has negative energy until you boost it way out there, where it then has zero energy. To make the long story short, energy E = GMm/R must be added to the negative energy it has at radius R in order to move it out of the solar system. And the R you use is not the radius of the solar system, but the starting radius - the radius of the Earth's orbit in this case.
irishbob
#3
Nov29-10, 12:00 AM
P: 22
I get the first part (number/large number=small number, lim as x->infinity of 1/x=0), but I'm having a little trouble following the second part. Would it be like this?
sqrt(2gm/r)=-GM(earth)m(rocket)/R(earth)+GM(sun)m(rocket)/R(earth)?

tiny-tim
#4
Nov29-10, 03:44 AM
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Escaping the solar system from Earth's orbit

hi irishbob!

(have a square-root: √ and try using the X2 tag just above the Reply box )
Quote Quote by irishbob View Post
I get the first part (number/large number=small number, lim as x->infinity of 1/x=0), but I'm having a little trouble following the second part. Would it be like this?
sqrt(2gm/r)=-GM(earth)m(rocket)/R(earth)+GM(sun)m(rocket)/R(earth)?
no, the mass of the rocket is irrelevant (because it doesn't affect the escape velocity), so is the mass of the Earth (because the question says "from the Earth's orbit", which needn't be anywhere near the Earth! )

go back to your √(2GMSun/R)
irishbob
#5
Nov29-10, 07:35 AM
P: 22
what do I use for R though? It's still not coming up right...
tiny-tim
#6
Nov29-10, 07:48 AM
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R is distance to the Sun
if it still doesn't come out right, show us what you've done
irishbob
#7
Nov29-10, 11:27 AM
P: 22
That was the problem! it was 3 a.m. my time when I was doing this, and for some reason I thought it would be a good idea to use the radius of the sun for R. Thanks!
Delphi51
#8
Nov29-10, 11:34 AM
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Glad to help! Hope you don't have to work through the night again.
tiny-tim
#9
Nov29-10, 12:29 PM
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