# Potential energy in the gravitational field - Why is $r_2$ striving against infinity?

why is $$r_2$$ striving against infinity in the formula $$π = πΊππ(\frac{1}{π_1}β\frac{1}{π_2})$$, so its often simplified to $$π = \frac{πΊππ}{r}$$ ?

I know that in the final formula, r is the distance between two masses, but what is $$r_1$$ and $$r_2$$ and how do they get simplified?

Also, if I want to see, if I could escape the gravitational field, why would I choose the Potential Energy $$E_{pot} = πΊππ(\frac{1}{π_1}β\frac{1}{π_2})$$ and equate it to $$E_{kin} = \frac{1}{2}πv^2$$ instead of just calculating weight force and then decide if a human can bring up this force?

• where did you get that extremely incorrect expression for weight? – OVERWOOTCH Jan 9 at 15:52

Your first formula represents the work done (by integrating the force over distance) to move a small mass from radius 1 to a different radius 2 relative to a much larger mass. This gives the change in potential energy. To define the potential energy at a point, you must choose a reference point where it is zero. In this case, chosing radius 1 as the reference point at infinity, gives a simpler (negative) result. (The potential rises from a negative value toward zero as you go up.)

• This is the thing I did not get in class and still don't get. This definition of the potential energy with the reference point... I understand that the further away your position is from a huge mass, the smaller is the gravitational force and thus: if the distance is infinity the potential is getting close to 0 from a negative value. But what about this reference point? Don't you always take infinity as reference where there's no Potential energy? – insertRandomName Jan 9 at 23:50