Physics with Samir Thursday
04-05-2018, 11:33 AM
#9
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[corrected]
After some more thought, I have haphazardly gone, no pun intended, flying past a key point that, while not devastatingly inaccurate in terms of calculated values to which I don’t usually care since numbers are for engineers, does misapply the physical fundamentals of this scenario to which I do, in fact, care.
My initial condition, which was flawed, was that the gravitational force, g, due to Earth’s mass was constant for the duration of the journey through our tunnel.
It is not.
Gravitational acceleration, g, is constant, for most intents and purposes, at the surface of the Earth and decreases as the inverse square of the distance as you move away from the Earth. HOWEVER, below the surface of the Earth, the gravitational acceleration becomes a linear function of the distance to the center of mass or radius. Therefore, g = g(r).
Here is how.
The gravitational force between two objects, g, goes as:
(1) g = (G*M_1*M_2)/r^2
Here, G is the gravitational constant, M is the mass of the two objects and r is the distance separating them. If this were applied to an object with a mass much smaller than that of the Earth, the mass of that object can be disregarded and the force of gravity at the surface of Earth is then defined as:
(2) g_earth = (G*M_earth)/R_earth^2
For points below the surface of the Earth, since now not all of the Earth’s mass is imparting a force toward the center of mass--some of that mass now being above pulling in the opposite direction--the effective gravitational force is a function of the distance to the center of mass, just as before, but, unlike before, this is no longer the radius of the Earth as though the object were on the surface. Additionally, the mass of the Earth is no longer constant but goes as a function of the effective radius, r. The effective gravitational force is then defined as:
(3) g_effective = (G*M(r))/r^2
The effective Earth mass, M(r), imparting a force on the object is now limited to what is between the object and the center of mass or the volume contained within a sphere of radius r, where the density of the Earth, ρearth is the defined as the total mass of the Earth divided by the total volume, V, assuming a uniform distribution of mass.
(4) ρ_earth = M_earth/((4/3)*π*R_earth^3)
(5) M(r) = V_effective*ρ_earth = (4/3)*π*r^3*M_earth/((4/3)*π*R_earth^3) = M_earth* r^3/R_earth^3
Rearranging (2) to isolate G and plugging (5) back into (3), we get:
(6) g_effective = (g_earth/R_earth )*r
This means that, below the surface of the Earth, the gravitational force exerted on an object decreases linearly with the radius as you approach the center of mass—much like a spring.
The force on a spring goes as follows:
(7) F_spring = -kr
Looking at equations (6) and (7), the similarity exists in that, for our sphere of uniform distribution with a hole tunneled through it, the “spring” constant, k, is just the gravitational force at the surface divided by the radius of the Earth.
Modeling our traveler oscillating back and forth as a weighted spring, the period of oscillation, T, for a spring constant, k, is given as:
(8) T = 2π√(m/k)
Expanding the constant coefficient of (6) and substituting that into (8), we find that:
(9) T = 2π√(R_earth/g)
Plugging and chugging numbers, we can calculate the period of oscillation for one roundtrip to be about 5066 seconds or roughly 84 minutes. A one-way trip from one side of the rock to the other would then be about 42 minutes. This not terribly far off from the 4560 seconds or roughly 76 minutes I originally calculated assuming the gravitational force to be constant, but it was, nonetheless, incorrect.
It is interesting to note that this correction resulted in a shorter time than previously calculated so even though the gravitational force below the surface of the Earth is decreasing with distance to the center and conservative--meaning the object dropped into the hole would slow down and stop at the other end--the net force it is still greater than if it were taken to be constant.
After some more thought, I have haphazardly gone, no pun intended, flying past a key point that, while not devastatingly inaccurate in terms of calculated values to which I don’t usually care since numbers are for engineers, does misapply the physical fundamentals of this scenario to which I do, in fact, care.
My initial condition, which was flawed, was that the gravitational force, g, due to Earth’s mass was constant for the duration of the journey through our tunnel.
It is not.
Gravitational acceleration, g, is constant, for most intents and purposes, at the surface of the Earth and decreases as the inverse square of the distance as you move away from the Earth. HOWEVER, below the surface of the Earth, the gravitational acceleration becomes a linear function of the distance to the center of mass or radius. Therefore, g = g(r).
Here is how.
The gravitational force between two objects, g, goes as:
(1) g = (G*M_1*M_2)/r^2
Here, G is the gravitational constant, M is the mass of the two objects and r is the distance separating them. If this were applied to an object with a mass much smaller than that of the Earth, the mass of that object can be disregarded and the force of gravity at the surface of Earth is then defined as:
(2) g_earth = (G*M_earth)/R_earth^2
For points below the surface of the Earth, since now not all of the Earth’s mass is imparting a force toward the center of mass--some of that mass now being above pulling in the opposite direction--the effective gravitational force is a function of the distance to the center of mass, just as before, but, unlike before, this is no longer the radius of the Earth as though the object were on the surface. Additionally, the mass of the Earth is no longer constant but goes as a function of the effective radius, r. The effective gravitational force is then defined as:
(3) g_effective = (G*M(r))/r^2
The effective Earth mass, M(r), imparting a force on the object is now limited to what is between the object and the center of mass or the volume contained within a sphere of radius r, where the density of the Earth, ρearth is the defined as the total mass of the Earth divided by the total volume, V, assuming a uniform distribution of mass.
(4) ρ_earth = M_earth/((4/3)*π*R_earth^3)
(5) M(r) = V_effective*ρ_earth = (4/3)*π*r^3*M_earth/((4/3)*π*R_earth^3) = M_earth* r^3/R_earth^3
Rearranging (2) to isolate G and plugging (5) back into (3), we get:
(6) g_effective = (g_earth/R_earth )*r
This means that, below the surface of the Earth, the gravitational force exerted on an object decreases linearly with the radius as you approach the center of mass—much like a spring.
The force on a spring goes as follows:
(7) F_spring = -kr
Looking at equations (6) and (7), the similarity exists in that, for our sphere of uniform distribution with a hole tunneled through it, the “spring” constant, k, is just the gravitational force at the surface divided by the radius of the Earth.
Modeling our traveler oscillating back and forth as a weighted spring, the period of oscillation, T, for a spring constant, k, is given as:
(8) T = 2π√(m/k)
Expanding the constant coefficient of (6) and substituting that into (8), we find that:
(9) T = 2π√(R_earth/g)
Plugging and chugging numbers, we can calculate the period of oscillation for one roundtrip to be about 5066 seconds or roughly 84 minutes. A one-way trip from one side of the rock to the other would then be about 42 minutes. This not terribly far off from the 4560 seconds or roughly 76 minutes I originally calculated assuming the gravitational force to be constant, but it was, nonetheless, incorrect.
It is interesting to note that this correction resulted in a shorter time than previously calculated so even though the gravitational force below the surface of the Earth is decreasing with distance to the center and conservative--meaning the object dropped into the hole would slow down and stop at the other end--the net force it is still greater than if it were taken to be constant.
Last edited by Rev. Rob Large; 04-05-2018 at 04:53 PM.
