- #1

iScience

- 466

- 5

in physics II, i learned about how the electric potential goes to zero as r goes to infinity. Okay, well, this was when we were dealing with two positive charges. ie a repulsive force. now i'm learning about gravitational potential now and i see this in my notes: [itex]\Phi[/itex][itex]\rightarrow[/itex]0, r[itex]\rightarrow[/itex]∞.

what?? why? now we're dealing with an attractive force..

mathematically, i can see that yes, [itex]\Phi[/itex]=-GM/r and so they are inversely proportional. but conceptually, let's start with the two positive charges from physics II. because the force is repulsive, the closer the test charge is to the source charge, the more potential energy it should have. because, if we consider the points radially extending from the source charge, there is an electric field value/strength associated with each point. since the force is repulsive, the closer the test charge is to the source charge, the more electric field it will be exposed to as it tends toward infinity and therefore the more total force it should experience right?

equivalently, for the case of an attractive force (gravitational case), if the field is [itex]\vec{g}[/itex], the gravitational acceleration, then the farther the test mass is from the source mass, the more total [itex]\vec{g}[/itex] values the test mass should be exposed to as it tends toward the source body (r=0) and therefore the more acceleration the test mass should have by the time it gets to r=0 right? so why in the world is [itex]\Phi[/itex] inversely proportional to r??

well.. we did the derivation in class.

[itex]\vec{g}[/itex]=[itex]\nabla[/itex][itex]\Phi[/itex]

[itex]\vec{g}[/itex]=-[itex]\frac{GM}{|r|^2}[/itex][itex]\hat{r}[/itex]=-[itex]\frac{dPhi}{d|r|}[/itex][itex]\hat{r}[/itex]

[itex]\frac{dPhi}{d|r|}[/itex]=[itex]\frac{GM}{|r|^2}[/itex]

[itex]\Phi[/itex]=-[itex]\frac{GM}{|r|}[/itex]

i just don't understand why r is inversely proportional to phi conceptually.

what?? why? now we're dealing with an attractive force..

mathematically, i can see that yes, [itex]\Phi[/itex]=-GM/r and so they are inversely proportional. but conceptually, let's start with the two positive charges from physics II. because the force is repulsive, the closer the test charge is to the source charge, the more potential energy it should have. because, if we consider the points radially extending from the source charge, there is an electric field value/strength associated with each point. since the force is repulsive, the closer the test charge is to the source charge, the more electric field it will be exposed to as it tends toward infinity and therefore the more total force it should experience right?

equivalently, for the case of an attractive force (gravitational case), if the field is [itex]\vec{g}[/itex], the gravitational acceleration, then the farther the test mass is from the source mass, the more total [itex]\vec{g}[/itex] values the test mass should be exposed to as it tends toward the source body (r=0) and therefore the more acceleration the test mass should have by the time it gets to r=0 right? so why in the world is [itex]\Phi[/itex] inversely proportional to r??

well.. we did the derivation in class.

[itex]\vec{g}[/itex]=[itex]\nabla[/itex][itex]\Phi[/itex]

[itex]\vec{g}[/itex]=-[itex]\frac{GM}{|r|^2}[/itex][itex]\hat{r}[/itex]=-[itex]\frac{dPhi}{d|r|}[/itex][itex]\hat{r}[/itex]

[itex]\frac{dPhi}{d|r|}[/itex]=[itex]\frac{GM}{|r|^2}[/itex]

[itex]\Phi[/itex]=-[itex]\frac{GM}{|r|}[/itex]

i just don't understand why r is inversely proportional to phi conceptually.

Last edited: