A Physics Question

[Bruce King]

This shows more detail
http://saxbyphysics.com/AdvancedPhys...tialEnergy.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.

[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.