Gravitational PE for a certain distance from the Sun

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SUMMARY

The discussion centers on calculating the escape velocity from a distance of 1 Astronomical Unit (AU) from the Sun, specifically addressing the formula for escape velocity, v_{esc} = \sqrt{2GM_S/r_0}. Participants clarify that the problem is about determining the velocity required to escape the Sun's gravitational influence, rather than the Earth. The confusion arises from misinterpreting the reference points for distance and velocity, emphasizing the need to understand gravitational potential energy and forces.

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simphys
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Homework Statement
Determine the escape velocity from the Sun for an
object (a) at the Sun’s surface (##r=7E5km, M=2E30kg##)
and (b) at the average distance of the
Earth(##1.5E8km##) Compare to the speed of the Earth
in its orbit.
Relevant Equations
##U = -GM_Sm/r_S##
hello guys, sims back again with another question..

I don't understand what is up with question (b)
cuz like.. to get ##v_esc## we assume that at ##r_0=\inf## ##v=0## but now if I assume at ##r=1.5E8## that ##v=0##.
And then find ##v_esc## from the following:
##\frac12*mv_{esc}^2 - \frac{GM_Sm}{r_S} = GM_Sm/r_0## is not correct?

HOWEVER IN THE STUPID SOLUTIONS IT SAYS THAT IT'S FOUND AS IF IT'S THE Escape velocity as is written for the sun but with ##r_0## instead of ##r_S##
i.e. ##v_{esc} = \sqrt{2GM_S/r_0}##
What I don't understand is... why?

Note: I am not acquainted with kepler's laws, not covered at univ (engineering) but I'll look into it myself a bit later.
I only know the ##U_grav## and ##F_grav## and that's about it.Thanks in advance
 
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(b) is about finding the velocity required to reach infinity from Earth’s distance to the Sun, not about reaching the Earth’s distance from the Sun’s surface.
 
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Orodruin said:
(b) is about finding the velocity required to reach infinity from Earth’s distance to the Sun, not about reaching the Earth’s distance from the Sun’s surface.
wow, can you elaborate on that please? 😬
 
do you mean it's like the initial velocity needed to get FROM the sun TO the earth?
 
if so, nope doesn't make sense to me in line with the eq. ##\frac12*mv_{esc}^2 - \frac{GM_Sm}{r_S} = \frac{GM_Sm}{r_0}##
 
simphys said:
do you mean it's like the initial velocity needed to get FROM the sun TO the earth?
No, that is essentially what you did. You are being asked about the escape velocity from 1 AU to infinity.
 
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Orodruin said:
No, that is essentially what you did. You are being asked about the escape velocity from 1 AU to infinity.
arghh, unfortunately, I don't get it.. but really thank you for your help!
edit: oh wait so you are saying that the initial velocity is at the 1AU (which is I presume the distance at the Earth's surface? And then what to escape what exactly?

f.e. for the Earth the escape velocity is to escape the Earth to not return so to say. but for this idk??
 
simphys said:
arghh, unfortunately, I don't get it.. but really thank you for your help!
edit: oh wait so you are saying that the initial velocity is at the 1AU (which is I presume the distance at the Earth's surface? And then what to escape what exactly?

f.e. for the Earth the escape velocity is to escape the Earth to not return so to say. but for this idk??
No, you are still misrepresenting the problem. The Earth is irrelevant apart from giving the relevant distance. The question is: ”You are 1 AU from the Sun. How fast do you need to move to escape to infinity?”

In other terms, how fast would the Earth need to move to escape Sun’s gravity?
 
okaaay, amazing, thanks a lot! that explains it!
I'll need to go through the gravity(+Newton's synthesis) chapter to understand it better then.

But thanks a lot for explaining this one, very much appreciated!