Basic question on gravitational potential energy

Question

When a point mass $m$ is moved in the gravitational field of a central body $M$ from a point $P_1(r_1)$ to a point $P_2(r_2)$, its potential energy changes by $$ \Delta E = GmM \left( \frac{1}{r_1} - \frac{1}{r_2} \right). $$

From what I understand is that if $r_2 < r_1$, which means $m$ is moved closer to $M$, $\Delta E$ becomes negative.

1. My first question is: Where does this energy go?

Another thing is that one usually calculates the potential energy with reference to a point that is infinitely far away, i.e. $r_1 = \infty$.

2. Second question: Why does one usually do this? What are the benefits from this? A consequence is that potential energy is now always negative. Isn't it easier to use positive values for potential energy when calculating things?

Answer 1

Energy is always conserved, and the energy is normally the sum of the potential and kinetic energies. So for example consider a satellite in an elliptical orbit. When the satellite is at the greatest distance from the central body the PE is least negative and the KE is smallest i.e. at its farthest point the satellite is moving slowest. As the satellite falls inwards the PE becomes more negative and the KE increases to keep the sum constant.

The sum of the PE and KE can only change if there is some way to transfer energy away. For example by a slingshot manoeuvre or if two bodies collide some of the energy can be converted to heat.

There is a useful relationship between the PE and KE for gravitationally bound objects due to a theorem called the virial theorem. For gravitationally bound systems this tells us the PE and KE are related by:

$$ KE = -\tfrac{1}{2} PE $$

As for the zero of the potential energy, setting the zero to be at infinity makes it easy to tell if an object is gravitationally bound or not. If the total energy is negative then the object is gravitationally bound, if the total energy is positive the object is unbound (in a hyperbolic orbit) and if the total energy is zero the object has exactly the escape velocity.