Asked by science
How do you calculate gravitational potential energy?
Answers
Answered by
PsyDAG
From Google:
PE_{\text G} = mg \Delta h
PE_G = potential energy due to gravity
g = acceleration due to gravity
m = mass
\Delta h = distance above a surface (such as the ground)
PE_{\text G} = mg \Delta h
PE_G = potential energy due to gravity
g = acceleration due to gravity
m = mass
\Delta h = distance above a surface (such as the ground)
Answered by
Damon
Say I have a rock of mass m kilograms in my hand.
The weight of the rock pushing down is m g or about 9.81 m Newtons.
If it is not accelerating the force up from my hand is then m g Newtons.
Now
If I lift that rock up a distance z, the work I do with my hand is F z = m g z
If there are no losses to friction or radiation or heat or anything energy is conserved so that work I put in, m g z, is an increase of potential energy of the rock of amount m g z Newtons
or
increase in PE = m g z
Note, important ---- all I calculated with that m g z was increase in potential energy from moving up distance z
There was no starting z = 0, the original height of my hand defined
Therefore to say my potential energy is U = m g z
I must say where z = 0 and U = 0. Often one chooses the level of the earth for z =U = 0, but you can pick any other height for U = 0
This is true of any other potential, always it is relative to some defined zero point.
The weight of the rock pushing down is m g or about 9.81 m Newtons.
If it is not accelerating the force up from my hand is then m g Newtons.
Now
If I lift that rock up a distance z, the work I do with my hand is F z = m g z
If there are no losses to friction or radiation or heat or anything energy is conserved so that work I put in, m g z, is an increase of potential energy of the rock of amount m g z Newtons
or
increase in PE = m g z
Note, important ---- all I calculated with that m g z was increase in potential energy from moving up distance z
There was no starting z = 0, the original height of my hand defined
Therefore to say my potential energy is U = m g z
I must say where z = 0 and U = 0. Often one chooses the level of the earth for z =U = 0, but you can pick any other height for U = 0
This is true of any other potential, always it is relative to some defined zero point.
Answered by
Damon
ah well, who am I to criticize Google ?
Answered by
Damon
Now if you are not just working on the surface of earth but in the much larger universe you must use the more general form of gravitational effect
F = G m M /r^2
m is that mass you are moving
M is the planet or star or whatever
r is the distance between the center of gravity of your rock and that star or whatever.
G is Newton's universal gravitational constant
[ Note if near earth surface r is about constant earth radius and m is earth mass F = [G M/r^2] m where G M/r^2 = g pretty much constant 9.81 m/s^2]
now change in PE = [(GMm)/r^2] dr = (GMm) integral dr/r^2
= GMm/ [1/R1 - 1/R2]
that is
G M m [ R2 - R1] /R1 R2
if R1 is close to R2 like near surface of earth that is
(GmM/R^2 )*( change in height ) or the same old m g z
which is the same old m g * change in height
F = G m M /r^2
m is that mass you are moving
M is the planet or star or whatever
r is the distance between the center of gravity of your rock and that star or whatever.
G is Newton's universal gravitational constant
[ Note if near earth surface r is about constant earth radius and m is earth mass F = [G M/r^2] m where G M/r^2 = g pretty much constant 9.81 m/s^2]
now change in PE = [(GMm)/r^2] dr = (GMm) integral dr/r^2
= GMm/ [1/R1 - 1/R2]
that is
G M m [ R2 - R1] /R1 R2
if R1 is close to R2 like near surface of earth that is
(GmM/R^2 )*( change in height ) or the same old m g z
which is the same old m g * change in height