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Escaping the solar system from Earth's orbit 
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#1
Nov2810, 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(6.673E11)(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? 


#2
Nov2810, 11:16 PM

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P: 3,394

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. 


#3
Nov2910, 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)? 


#4
Nov2910, 03:44 AM

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Escaping the solar system from Earth's orbit
hi irishbob!
(have a squareroot: √ and try using the X_{2} tag just above the Reply box ) go back to your √(2GM_{Sun}/R) 


#5
Nov2910, 07:35 AM

P: 22

what do I use for R though? It's still not coming up right...



#6
Nov2910, 07:48 AM

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P: 26,157

R is distance to the Sun …
if it still doesn't come out right, show us what you've done 


#7
Nov2910, 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!



#8
Nov2910, 11:34 AM

HW Helper
P: 3,394

Glad to help! Hope you don't have to work through the night again.



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