-1
$\begingroup$

why is $r_2$ striving against infinity in the formula $π‘Š = πΊπ‘šπ‘€(\frac{1}{π‘Ÿ_1}βˆ’\frac{1}{π‘Ÿ_2})$, so its often simplified to $π‘Š = \frac{πΊπ‘šπ‘€}{r}$ ?

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 $E_{pot} = πΊπ‘šπ‘€(\frac{1}{π‘Ÿ_1}βˆ’\frac{1}{π‘Ÿ_2})$ and equate it to $E_{kin} = \frac{1}{2}π‘šv^2$ instead of just calculating weight force and then decide if a human can bring up this force?

$\endgroup$
1
  • $\begingroup$ where did you get that extremely incorrect expression for weight? $\endgroup$ – OVERWOOTCH Jan 9 at 15:52
1
$\begingroup$

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.)

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$ – insertRandomName Jan 9 at 23:50

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.