- #1
rogerk8
- 288
- 1
Hi!
I just wish you to help me derive two types of potential energy.
A) Gravitational Potential Energy
B) Electrical Potential Energy
A useful equation is
[tex]W=\int Fdr[/tex]
Which I consider is the work that has to be done to the particle to move it a certain distance from the attractive centre of force (which then becomes it's PE, is my thought).
For A we have
[tex]F=G\frac{mM}{r^2}[/tex]
Now
[tex]W=\int_a^hG\frac{mM}{r^2}=GmM\int_a^h \frac{1}{r^2}[/tex]
Which means
[tex]W=GmM((-1/h)-(-1/a))=GmM(1/a-1/h)[/tex]
Where a cannot be zero but maybe it could be R?
Then we have
[tex]W=\frac{GmM}{R}(1-\frac{R}{h})[/tex]
And if we set
[tex]\frac{GM}{R}=g[/tex]
We have
[tex]W=mg(1-\frac{R}{h})[/tex]
But this till does not look like Ep=mgh :D
I get the same problem in the B-case.
Considering Hydrogen for simplicity
[tex]F=\frac{e^2}{kr^2}[/tex]
then
[tex]W=\int_a^h \frac{e^2}{kr^2}=\frac{e^2}{k}\int_a^h \frac{1}{r^2}[/tex]
which gives
[tex]W=\frac{e^2}{k}((-1/h-(-1/a))=\frac{e^2}{k}(1/a-1/h)[/tex]
where once again a cannot be zero.
Can it be the shell radius, then?
There's no real use starting at a point inside the shell radius so let's play with r, this time.
This gives
[tex]W=\frac{e^2}{kr}(1-r/h)[/tex]
Which is about the same result as for the Gravitational Potential Energy above.
Is this totally wrong?
Roger
PS
It is interesting to note that PE increses with distance in both cases. If I'm right, that is :)
I just wish you to help me derive two types of potential energy.
A) Gravitational Potential Energy
B) Electrical Potential Energy
A useful equation is
[tex]W=\int Fdr[/tex]
Which I consider is the work that has to be done to the particle to move it a certain distance from the attractive centre of force (which then becomes it's PE, is my thought).
For A we have
[tex]F=G\frac{mM}{r^2}[/tex]
Now
[tex]W=\int_a^hG\frac{mM}{r^2}=GmM\int_a^h \frac{1}{r^2}[/tex]
Which means
[tex]W=GmM((-1/h)-(-1/a))=GmM(1/a-1/h)[/tex]
Where a cannot be zero but maybe it could be R?
Then we have
[tex]W=\frac{GmM}{R}(1-\frac{R}{h})[/tex]
And if we set
[tex]\frac{GM}{R}=g[/tex]
We have
[tex]W=mg(1-\frac{R}{h})[/tex]
But this till does not look like Ep=mgh :D
I get the same problem in the B-case.
Considering Hydrogen for simplicity
[tex]F=\frac{e^2}{kr^2}[/tex]
then
[tex]W=\int_a^h \frac{e^2}{kr^2}=\frac{e^2}{k}\int_a^h \frac{1}{r^2}[/tex]
which gives
[tex]W=\frac{e^2}{k}((-1/h-(-1/a))=\frac{e^2}{k}(1/a-1/h)[/tex]
where once again a cannot be zero.
Can it be the shell radius, then?
There's no real use starting at a point inside the shell radius so let's play with r, this time.
This gives
[tex]W=\frac{e^2}{kr}(1-r/h)[/tex]
Which is about the same result as for the Gravitational Potential Energy above.
Is this totally wrong?
Roger
PS
It is interesting to note that PE increses with distance in both cases. If I'm right, that is :)