A Physics Question

[Tom Bender]

Just to keep the old grey matter exercised I am rereading my college textbooks. It is a humbling experience. Did I ever understand this?

So here's my question.

Per the text the gravitational potential energy of an object is;

Energy = -G (Mobject Mearth)/radius

Since G and M and M are constants,

E = C/r

But if we consider the molecule at the center of the earth it must have infinite energy. Clearly it does not.

Please try to keep to a reasonable level with your debunking of my thought process.

Thanks,
Tom

[roger wiegand]

I think you need to define a few more symbols/variables for us for the question to make sense. What are G, M, E, C, and r?

[Prashun Patel]

The potential energy gets bigger with the height right? So wouldn’t the /r be a *r?

[Steve Wurster]

The gravitational potential energy of an object is:

mass * g * height

where g is the force of gravity, and height is the height of the object above your chosen zero point.

For most scenarios, the surface of the earth is the chosen zero point, therefore an item resting on the ground has zero gravitational potential energy. As you raise the item above the ground it gains gravitational potential energy.

For your scenario, a molecule at the center of the earth has a height of zero; therefore it has no gravitational potential energy.

[Jim Matthews]

I thought the dinosaurs in the Antarctic caldera caught anything dropped down the hole and sent something up to replace it where the dot products canceled.

I believe the exit point is somewhere in New Jersey.

[Zachary Hoyt]

All I know about physics is that if I drop something heavy enough on my foot, it hurts. I admire you for ever having tried to understand things like what you posted, let alone trying to understand it twice.

[John Stankus]

If you are at the center of the earth you have to take into account the gravitational pull of all the matter pulling from each direction. You can no longer assume the earth is a point mass. If you are in a hollow at the center of mass of the earth the gravitational potential will go to zero because all the mass is surrounding you pulling in opposite directions

A way to think about this is to have two identical earths at some distance apart. If you are on a line between their center of masses, half the distance they are separated, then the gravitational pull of earth 1 is of equal size but opposite direction of that of earth 2. So the gravitational potential goes to zero.

For the case at the center of the earth you have add up all the masses and account for the direction they are pulling on you. Do able if you know the distribution of the masses (and can do a little bit of calculus)

John

[Steve Wurster]

I thought the dinosaurs in the Antarctic caldera caught anything dropped down the hole and sent something up to replace it where the dot products canceled.

I believe the exit point is somewhere in New Jersey.

You're thinking of a different kind of portal. https://en.wikipedia.org/wiki/Being_John_Malkovich

[John Stankus]

F=Gm1m2/r2

F is force due to gravity
G is gravitational constants
the m’s are the masses
r is the separation

Don’t think about m1 and m2. Being your total mass and the earths total mass, but rather adding up the forces between every pair of individual masses.

At a large distance apart we can approximate that the separation is about the same and can use the difference in the centers of mass and the total masses.

[Prashun Patel]

Is it that your equation is for gravitational force not potential energy? It makes intuitive sense to me that if the distance between 2 object centers is zero, they are occupying the same space and probably have a darn high attractive force .

[Bill Dufour]

r is the seperation. so if a particle is inside the middle of the earth the seperation is zero. anything times zero is zero. So at the exact center of the earth there is no gravitational force.
Bill D

reading the equation again I see you are dividing by r, not multiplying. dividing by zero gives "undefined" as an answer. So the answer is not zero it is "undefined"

[Doug Garson]

I think you need to reread your textbook, don't think the formula is correct. As Steve posted gravitational potential energy is mass x g x height and as John pointed out if the formula is for gravitational force the r should be squared. Now what I find is the tricky part is determining the height. If you have a mass six feet above a four foot high table is h six or ten? Does it matter if the table is strong enough to support the mass when dropped from the six feet? If h is ten feet what if the floor under the table can't support the mass when dropped and the basement floor is eight feet below the main floor? And how about if there is a cave under the basement floor that leads to the center if the earth?

[David L Morse]

Just to keep the old grey matter exercised I am rereading my college textbooks. It is a humbling experience. Did I ever understand this?

So here's my question.

Per the text the gravitational potential energy of an object is;

Energy = -G (Mobject Mearth)/radius

Since G and M and M are constants,

E = C/r

But if we consider the molecule at the center of the earth it must have infinite energy. Clearly it does not.

Please try to keep to a reasonable level with your debunking of my thought process.

Thanks,
Tom

The bold text is the problem. As other posters have said, once you are inside the Earth "Mearth" (your notation) is no longer a constant, rather, it varies with the cube of r.

Your equation for potential energy is not it's definition. It's definition is the integral of force with distance. (hard to review physics without reviewing calculus too;)) The equation you stated is the result of that integration under a particular set of conditions. To calculate the total potential the integral is broken into two parts; one for outside the Earth and the other for inside. The "inside the Earth" integral will converge to a finite value at zero.

[Tom Bender]

Ok, at first read I ignored Roger's post about defining the terms. On rereading I realized that G is not a constant. As the object approaches the center of the earth, G (the acceleration due to gravity) decreases to zero (John's point) so we have 0/0. This is a reasonable result.

Prashan and Steve,there is a minus sign in the equation so energy does increase with height.

Thanks all for helping me to think this thru.

Tom

[Michael Weber]

Thank heavens we got that cleared up:rolleyes:

[Bruce King]

This shows more detail
http://saxbyphysics.com/AdvancedPhysicsLab/Mechanics/Gravity/UniversalGravitationalPotentialEnergy.pdf

[Stephen Tashiro]

Some vocabulary technicalities:


On rereading I realized that G is not a constant.
According to current theory, G is constant. The force between two "point" masses m_1, m_2 that are a distance r apart is G m_1 m_2/ r_squared. Whether you include a minus sign in front of this formula depends on which mass you consider the force acting upon and which direction you call positive.


(The formula E = G m_1 m_2/r for potential energy is different than the formula for the force of attraction F = G m_1 m_2/ r_squared.. The formula e = m g h for potential energy is an approximation valid in distances near the surface of the earth, where (percentage wise) the value of r doesn't change much with changes in height. The constant g is a different constant than G )



As the object approaches the center of the earth, G (the acceleration due to gravity) decreases to zero (John's point) so we have 0/0.


A better name for "G" is "the gravitational constant". The term "acceleration due to gravity" is used for the constant denoted by a lower case g that is used in the formula e = m g h.

It isn't G that changes, but yes, if we imagine a point object inside the earth and the earth as set of many point masses the net force on the object now involves some points in the earth that are pulling the object generally away from the earth's center and other points in the earth that are pulling the object toward the earth's center. So the total force on the object is no longer given by the formula that applies to only 2 point masses. (We can no longer consider the earth to be a point.) Acceleration on a mass is proportional to the net force on it, so the acceleration of the point mass is affected by its being inside the earth. Yes, the "acceleration due to gravity" changes if that phrase refers to the acceleration of the object. But it isn't the constant G that changes - and the constant g (which is also called "the acceleration due to gravity" ) isn't relevant since we are no longer talking about an object near the surface of the earth.

[David L Morse]

Some vocabulary technicalities:


According to current theory, G is constant. The force between two "point" masses m_1, m_2 that are a distance r apart is G m_1 m_2/ r_squared. Whether you include a minus sign in front of this formula depends on which mass you consider the force acting upon and which direction you call positive.


(The formula E = G m_1 m_2/r for potential energy is different than the formula for the force of attraction F = G m_1 m_2/ r_squared.. The formula e = m g h for potential energy is an approximation valid in distances near the surface of the earth, where (percentage wise) the value of r doesn't change much with changes in height. The constant g is a different constant than G )



A better name for "G" is "the gravitational constant". The term "acceleration due to gravity" is used for the constant denoted by a lower case g that is used in the formula e = m g h.

It isn't G that changes, but yes, if we imagine a point object inside the earth and the earth as set of many point masses the net force on the object now involves some points in the earth that are pulling the object generally away from the earth's center and other points in the earth that are pulling the object toward the earth's center. So the total force on the object is no longer given by the formula that applies to only 2 point masses. (We can no longer consider the earth to be a point.) Acceleration on a mass is proportional to the net force on it, so the acceleration of the point mass is affected by its being inside the earth. Yes, the "acceleration due to gravity" changes if that phrase refers to the acceleration of the object. But it isn't the constant G that changes - and the constant g (which is also called "the acceleration due to gravity" ) isn't relevant since we are no longer talking about an object near the surface of the earth.

Well, yes, G is constant. However, g is not a constant. It's different on the Moon, and different at various altitudes on Earth. The acceleration due to gravity is simply the ratio between the force on a test mass and it's mass. So at the center of a massive sphere g is zero. It increases linearly with distance from the center up to it's surface. Past the surface in then varies with the inverse square of the distance from the center of mass, just like the force equation you stated.

[Doug Garson]

Well, yes, G is constant. However, g is not a constant. It's different on the Moon, and different at various altitudes on Earth. The acceleration due to gravity is simply the ratio between the force on a test mass and it's mass. So at the center of a massive sphere g is zero. It increases linearly with distance from the center up to it's surface. Past the surface in then varies with the inverse square of the distance from the center of mass, just like the force equation you stated.
Sounds good but what do you define as the surface? The diameter of earth is 12,742 km, I assume that's an average since the earth is neither perfectly round or uniform. So if a mass is 12,742.5 km from the center of the earth it could be 0.5km above the surface or 8.35km below the surface of the top of Everest. (Everest is 8.85 km high).

[David L Morse]

Sounds good but what do you define as the surface? The diameter of earth is 12,742 km, I assume that's an average since the earth is neither perfectly round or uniform. So if a mass is 12,742.5 km from the center of the earth it could be 0.5km above the surface or 8.35km below the surface of the top of Everest. (Everest is 8.85 km high).

That's why I specified a sphere. A sphere is an ideal geometric solid. It has a well defined surface. It doesn't have an Everest.

[Perry Hilbert Jr]

This whole thread makes me glad I was a Political Science major.

[Doug Garson]

Yeah, politics makes much more sense than science. :rolleyes:

[Jim Matthews]

The "inside the Earth" integral will converge to a finite value at zero.

The dinosaurs are going to be miffed to hear that.

[Rick Potter]

I never did get past a b c and x. Flunked out of Trig twice, but I do understand ZZZZZZZZZZZZZZZZZZZZZZZZZ.


Sorry, couldn't resist.