# Gravitational PE for a certain distance from the Sun

• simphys
In summary, the conversation was about finding the escape velocity from 1 AU to infinity, with the question being how fast one would need to move to escape the Sun's gravity. The confusion was around the use of the equation for finding escape velocity and the role of the Earth in the problem. It was clarified that the Earth is only relevant in providing the distance, and the question is about escaping from the Sun's gravity to infinity. The individual expressed their gratitude for the explanation and mentioned they would need to review the gravity and Newton's synthesis chapter to better understand the concept.
simphys
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

(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.

simphys
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.

simphys
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!

## 1. What is gravitational potential energy?

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. It is the potential for an object to do work as a result of its position relative to another object with mass.

## 2. How is gravitational potential energy calculated?

Gravitational potential energy is calculated using the equation U = mgh, where U is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or distance from the center of the gravitational field.

## 3. How does distance from the Sun affect gravitational potential energy?

The closer an object is to the Sun, the higher its gravitational potential energy will be. This is because the force of gravity between the Sun and the object increases as the distance between them decreases, resulting in a higher potential for work to be done.

## 4. What is the relationship between gravitational potential energy and gravitational force?

Gravitational potential energy and gravitational force are directly proportional to each other. As gravitational potential energy increases, so does the force of gravity between two objects. This means that the closer an object is to the Sun, the stronger the force of gravity and the higher the potential energy.

## 5. How does gravitational potential energy affect the motion of objects?

Gravitational potential energy is converted into kinetic energy as an object moves closer to the Sun. This means that as an object falls towards the Sun, its potential energy decreases while its kinetic energy increases. This is what causes objects to accelerate towards the Sun due to the force of gravity.