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

Question

why is $r_2$ striving against infinity in the formula $? = ???(frac{1}{?_1}−frac{1}{?_2})$, so its often simplified to $? = frac{???}{r}$ ?
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 $E_{pot} = ???(frac{1}{?_1}−frac{1}{?_2})$ and equate it to $E_{kin} = frac{1}{2}?v^2$
instead of just calculating weight force and then decide if a human can bring up this force?

R.W. Bird · Answer

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.)

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

One Answer

Add your own answers!

Ask a Question