Understanding Gravitational Potential: Negative Values and Decreasing Trends

[Knightycloud]

Gravitational potential is always a negative value according to the theory.
As per to the equation V = - \frac{GM}{r}; when the r (distance) increases the potential decreases. But considering the potential at infinity as zero and since this a negative value, on what basis do we consider the potential is decreasing, not increasing (When the distance is increasing)?

 

[jtbell]

As per to the equation V = - \frac{GM}{r}; when the r (distance) increases the potential decreases.

No. As r increases, 1/r decreases, but -1/r increases.

 

[sophiecentaur]

No. As r increases, 1/r decreases, but -1/r increases.

i.e. it's less negative.

 

[jtbell]

...or slopes upward (has a positive slope) on a graph of V versus r.

 

[sophiecentaur]

'Everywhere' is the equivalent of 'underground', effectively and up is up, however deep or high you are.

 

[Knightycloud]

No. As r increases, 1/r decreases, but -1/r increases.

So as the negative value decreases, potential is increasing accordingly?

Let's say that the potential on Earth surface is V_a = -\frac{GM}{R} and if we move to a higher place where the distance is twice, the potential is V_b = -\frac{GM}{2R}. But at V_b, the negative value is smaller than V_a, V_b has a higher potential. Right?

 

[sophiecentaur]

So as the negative value decreases, potential is increasing accordingly?

Let's say that the potential on Earth surface is V_a = -\frac{GM}{R} and if we move to a higher place where the distance is twice, the potential is V_b = -\frac{GM}{2R}. But at V_b, the negative value is smaller than V_a, V_b has a higher potential. Right?

It's exactly the same as when you deal with xy co ordinates. Moving tp the right is increasing the x co ordinate, whether you start on the right or to the left of the origin. Just let the Maths work for you.
And you really mean 'magnitude'.

 

[Knightycloud]

Yep. I got answers. Thank you all! :D