The Mississippi River Flows Uphill! (That is, it flows against the force of gravity) ******************************** At any point of time, at any

Transcription

1 The Mississippi River Flows Uphill! (That is, it flows against the force of gravity) At any point of time, at any specific point on the surface of this earth, gravity can be properly represented as a single vector, with one stated value, and one specified direction! At this specific point of time and position and direction, a plane can be positioned so that there is no component or part of the gravity vector lying in the plane. That is, the plane is exactly perpendicular to the direction of the gravity at that point and time. Now at this same point, other than at one of the two poles of this earth, there will normally be a centrifugal force vector. And this centrifugal force will normally not be perpendicular to this plane. Only at or near the equator would such conditions be expected. And thus, in these planes that are perpendicular to the gravity, there will normally be a component of the centrifugal force in these planes. What this means is this: In these planes that are perpendicular to gravity, any water on such a plane will not have any tendency to flow in one direction more than another due to gravity. But there will be some component of the centrifugal force that would cause flow, and would determine the direction of this flow. And let us be sure to understand, that there is a finite angle through which one can tip this plane that was perpendicular to gravity, where the component of gravity begins to increase in the opposite direction to the centrifugal force component. But over this finite amount of tipping, the centrifugal force component will remain the larger value, and it will continue to force a flow even against the small gravity component that would begin to appear. Only until one reaches a specified tipping, will the gravity component actually equal the centrifugal force component, and the tendency of flow will stop. If one had first began to tip the plane in the opposite direction, both gravity and centrifugal force components would be in the same direction, but the centrifugal component remains the stronger until a specified angle is achieved. 1

2 Thus, if a river, such as the Mississippi River, has an angle of flow that is small enough, it could easily be within the range where the flow is due to centrifugal forces, and not due to gravity. It could, in fact, actually be against the direction of the component of gravity in its plane of flow. Now is this hard to understand? No! It is an absolute necessity that on a planet that has a spin, that there exists angles where the flow is only due to the spin, and not due to gravity. Everyone knows the force of gravity at the surface of the earth. Everyone knows the spin of the earth. Everyone knows the radius of the earth. Anyone can calculate the angles where the component of gravity is being over driven by the component of the centrifugal force at each point of latitude that you want to consider. And thus, any river that has such a level of flow or less would be flowing against gravity. Please read the article I wrote on the shape of the earth. If you can really understand the flow of the Mississippi River, then you should be able to start to understand this other article. For those who would like data to work with, use the following: Let us consider the Mississippi River from the ocean to the Falls of St. Anthony, the normal shipping end for this river. This extends from the 45 degree latitude to the 30 degree latitude. This is close to a distance of 1000 miles. The elevation of this Fall is less than a 1000 feet. So the average slope (tangent of the slope) of this river from the given starting point to the ocean is less than: 1) 1000 /(5, ) = The average slope of the earth from the pole to the equator is about: 2) 13.2/6000 =

3 This makes the slope of the Mississippi River to average less than one tenth of the slope of the average for the whole earth. The slop of the earth varies by latitude, it starts out with a zero slop at the pole, becomes a maximum near 45 degrees, and then returns back to zero at the equator. Thus, if its average tangent is , then around the 30 to 45 degree latitude area it must be near this average value or even more than this average value. Thus the flow of this river is well within the range where the centrifugal forces are in control, not gravity. Let us calculate the change in the earth s radius, delta R, that occurs between 45 degrees North Latitude to 30 degrees North Latitude. We will let R be the nominal radius of the earth and L will be the angle of latitude. The following equation can be considered: 3) (Delta R)/(RDelta L) = tan (A) Where A is the angle of the slope of the surface of the earth that is greater than the perfect circle slope at that point. A is very small, and thus its tangent can be reasonable replaced with its value as an angle. The radius for the centrifugal force will be R times Cos(L). This centrifugal force points straight out away from the axis of spin, and so a function, sin(l), is introduced to reflect the centrifugal force that lies within the plane of the surface. This assumption is not perfect, in that we will not make an adjustment for angle A. But angle A is very small compared to L, and we can ignore it for this one relationship. Again, we will let A be the very small angle that the surface of the earth takes to balance the component of the gravity acceleration, gtan(a), with the component of the centrifugal force in this same plane, which will be RCos(L)Sin(L)w^2. Thus: 4) gtan(a) = RCos(L)Sin(L)w^2 3

4 This equation gives us the equality of acceleration forces in the plane. To get the change in R over these 15 degrees we have to integrate the change in R caused by the angle A. Thus: Delta change in radius length, R, equals 5) Delta R = R Tan(A) (delta L) If we put into equation 3) the value of Tan(A), and then integrate over the latitude 45 to 30, we will have what we need to show if the Mississippi River is flowing uphill against gravity. This is done in equation 4. 6) Change in R = /L = 30 /(Rw^2/g)RCos(L)Sin(L)(delta L) /L = 45 7) Change in R = No, I am not going to do this integration. Let us just take a good guess: g = 32.2 feet/sec^2. R = 5,280 ft/mile 3960 miles = ^7 feet Delta L = 15 degrees = 3.14/12 = 0.26 radians w = 3.14/(123600) = ^-5 w^2 = ^-10 Cos(L)Sin(L), over the range of 45 to 30, has a max of.5, and a minimum of.43. Let us use A good guess would then be: ^ ^ ^ /32.2 Or: ^4 feet or 1.67 miles. What one can say is this, over the 1000 miles that the Mississippi River flows from North to South, it sees a drop in its elevation from less than 1000 feet to sea level, and over this distance, the radius of the earth increased by 1.67 miles. This means that this river went against the force of gravity over these 1000 miles, making a climb against gravity of over 7000 feet! 4

5 Now please note that the actual values might really be different. The change in the shape of the earth might truly change the assumptions that were made about the gravity s magnitudes and directions. But no matter what these changes are in gravity, centrifugal forces will not be measurably changed. And the difference in the change in slope of the surface will make it more in line with the5 centrifugal forces, and this effect would actually become greater as the slope of the earth places the vector more in the direction of the surface. Thanks for reading. Gerald L. O Barr 5