Universal Physics Journal
Article XII: Weight 



 

Author: Ethan Skyler
Publication Date: To be announced
Revision Date: October 29, 2011

 
  Purpose
    What is weight?  What types of forces have a role in producing an object's weight?  What is gravitation's role?  What is acceleration's role?  What is weightlessness?  Is the weight of an object absolute like its mass rating is absolute?  Or does an object's force of weight change with a changing environment?  The purpose of Article XII is to answer these questions, while keeping an eye out for additional truths that may shed new light on our ongoing survey of Universal Physics. 

 

 
 

Article XII
      Let us begin with considering how you determine your body's weight.  Most likely, you will locate a compression scale or balance beam scale, check to see it is set to zero and then step aboard so that the only object your body is touching is the surface of the scale.  Your weight is displayed in lb.force or newton, both being standard units for force.  Force is a push or pull being experienced by an object.  Here the compression scale is the object that is being pushed downward by the contact force of your body's weight against Earth.  This external-to-matter contact force is being transferred down to the scale's surface by the bottom surfaces of your feet.  At the same time and with the same magnitude the scale is being pushed upward by the contact force that results from Earth's weight against your body.  Based upon the description of this weighing event I submit the following definition for weight.   

 

  (2)  Definition:  Weight is the external-to-matter contact force one object freely bears against a second object.

    While an object's weight is often fundamentally caused by the presence of internal-to-matter forces such as gravitation, realize that these internal forces accumulate and thereby stack up through the object's matter in the direction of the second object.  Upon arriving at the second object, the cumulative total of these internal-to-matter forces is presented as the external contact force of the object's weight against the second object. "Freely" means that the object being weighed is free from contact with any object other than the second object.  A second object is required for the first object to present its force of weight against.  If the object has no second object or objects against which to bear with an external contact force, then the object is weightless regardless of whether acceleration is present for the object or absent. 

 

 
  (3)  When Earth is the second object, Type 2 internal-to-matter gravitation forces are the forces mainly responsible for the first object's force of weight against Earth.  When the weighing event takes place against the surface of one of Earth's poles, gravitation is the sole cause of the object's weight.  At every other location on Earth, a small portion of the object's gravitational attraction toward Earth is spent causing centripetal acceleration for the object and therefore is no longer available to contribute to the force of weight the object is freely impressing against Earth's surface.  In other words, the centripetal acceleration present at every non-pole location on Earth causes a reduction in every object's force of weight being presented against Earth.  This centripetal-acceleration-caused reduction in the weight force of an object varies according to the latitude of the object's location. 

 

 
  4)  What is latitude?  Latitude, combined with longitude, forms an angular-based grid for locating the position of any object on, within or above Earth's surface.  To find an object's latitude on Earth, begin by drawing an imaginary straight line from the object to the point on Earth's equatorial plane where it is intersected by Earth's north/south axis of rotation.  The object's latitude in degrees of a circle, minutes of a degree, and seconds of a minute is the minimum angle formed between this straight line and Earth's equatorial plane.  If the latitude is 0 degrees then the object resides on Earth's equator.  If the latitude is 90 degrees N then the object is at the North Pole and 90 degrees S indicates the location is the South Pole. 

 

 
  (5)  I propose an event where we select an object's latitude and then analyze the forces and accelerations present keeping an eye on how they influence the object's force of weight.  Later we will factor in the object's Moon gravitation when the object is closer to the Moon when the Moon is in the upward or overhead direction compared to 12 hours later when the object is farther from the Moon when the Moon is in the downward or underground direction.   I will begin this event by placing the object upon a compression scale at a latitude of 20 degrees N.  Then I will determine the circumference of the "small circle" on Earth that includes every location with a latitude of 20 degrees North.  The plane of this small circle, marked in red in the next drawing,  is north of and parallel to the "great circle" of Earth's equatorial plane.  This small circle marks the path of the object as it rides along on Earth while orbiting Earth's axis. The force responsible for causing centripetal acceleration for the object is directed along the 20 degree latitude plane toward the central point where this plane is intersected by Earth's axis.  In order to find the magnitude of this centripetal acceleration/Action force, which has a role in determining the object's external-to-matter contact force of weight against the scale, we will first need to know the circumference of the 20 deg. Latitude small circle which will allow us to determine the object's orbital velocity.

 

 
           

 

 
  (6)  Constant Values:   
Object mass rating                    =   200 lb.m  (90.7 kg)
Earth mass rating                       =  1.31702804e+25 lb.m
Earth equatorial circumference   =  131,479,775.39 ft   (24,901.47 miles) 
Earth equatorial radius               =    20,925,656.17 ft   (3963.192456 mi)
Earth average radius                  =    20,890,575.79   ft   (3956.548445 mi)
Earth polar radius                      =    20,855,495.41 ft   (3949.904433 mi)
Earth seconds/rotation               =           86,164.1              (23h 56min 04.1 s sidereal time/rotation)
Moon mass  rating                     =    1.61994e+23 lb.m      (Shift the decimal 23 positions to the right for the true value.)
gravitational constant                 =     3.321998855540755e-11  (US units ft, lb.m, and lb.f.  (Shift the decimal 11 positions left for the true value.) 
Moon c/m to object on Earth's near side    =   1,240,245,604 ft       
Moon c/m to Earth c/m                             =   1,261,171,260 ft       (238,858.19315 miles)
Moon c/m to object on Earth's far side      =   1,282,096,916 ft      
barycenter to Earth c/m                           =  15,323,880.3 ft  =  2902.25 miles
barycenter depth below Earth's near side at the equator  =  5,601,776 ft  =  1,060.94 miles
Earth/Moon orbital time about the barycenter                 =  2,360,595 sec  =  27.3217 days

 

 
  (7)  Problem 1:  Determine the circumference of the Latitude 20 deg. N small circle which represents the circular path traveled by the 200 lb.m Earth object.  The formula to solve this problem is Circumference = 2 Pi *  Equatorial Radius * cos(Latitude Angle)

circumference  =  2 * pi * r * cos(latitude angle)
                       =  2 * 3.141593  *  20,925,656.17 ft  *  cos(20 deg)
                       =  6.28318  *  20,925,656.17 ft  *  0.93969262
                       =  123,550,470.26 ft

 

 
  (8)  Problem 2:  Figure the 200 lb.m object's orbital velocity about Earth's axis.

velocity  =  circumference / time
              =  123,550,470.26 ft  /    86,164.1 s
              =  1,434 ft/sec

 

 
  (9) Problem 3:  Determine the radius of this 20 degree latitude small circle.

radius  =  circumference / 2 * pi
           =  123,550,470.26 ft  /  2 * 3.141593
           =  19,663,668.063 ft

 

 
  (10)  Problem 4:  We are ready to determine the inward-directed centripetal acceleration/Action force required to cause the 200 lb.m object to travel the curved path of the Latitude 20 deg. N small circle.  I have included division by 32 so the answer will be converted from absolute Poundal units to more commonly used lb.f units.

force  =  mass * velocity2 / radius / 32
         =   200 lb.m  *  (1,434 ft/sec)2   /  19,663,668.063 ft  /  32 Pdl/lb.f
         =   200 lb.m  *  2,056,356 ft2/sec2  /  19,663,668.063 ft  /  32 Pdl/lb.f
         =   200 lb.m  *  0.10457633 ft/sec/  32 Pdl/lb.f
         =   20.92 Pdl  /  32 Pdl/lb.f
         =   0.65375 lb.f

 

 
  (11)   Now there is a twist in our path to navigate.  We have determined that there exists a 0.65 lb.f  that is causing the action of centripetal acceleration for the 200 lb.m object.  This force is directed along the Latitude 20 deg. N plane which means it is directed at a right angle to Earth's axis of rotation.  The source of this 0.65 lb.f is the object's Earth gravitation force which is being generated within the object in the average direction down along the blue latitude angle line toward Earth's center of matter.  Treating the centripetal a/A force as one component of the object's gravitation force will resolve the issue of their different directions and thereby reveal to us the true portion of the object's gravitation force that is being spent and thereby terminated acting as the cause of the object's centripetal acceleration about Earth's axis.  Hold in your mind the thought that once this true portion acts as the cause of the object's centripetal acceleration, it is no longer available to contribute to the object's contact force of weight against Earth.  There is an adventure unfolding here as we proceed toward the solution. 

 

 
 

             

 
  (13)  Problem 5:  Determine the gravity force portion (side c) being spent on causing Latitude Man's centripetal acceleration about Earth's axis.  See previous Latitude Man force triangle drawing.  We know side a = 0.65375 lb.f and also that angle B = the latitude angle of 20 degrees.  Since this is a right triangle, we are ready to solve for side c being the Gravity Force that is no longer available to act as the cause of Latitude Man's  force of weight.

gravity force  =  a / cos (B)
                    =  0.65375 lb.f  / cos (20 deg.)
                    =  0.65375 lb.f  /  0.93969262
                    =  0.6957 lb.f.

 

 
  (14)  At the latitude of 20 degrees, the gravity force being spent on causing centripetal acceleration for Latitude Man is about 6 % greater than the centripetal acceleration/Action force being caused.  This percentage will increase as we travel to a more northern latitude.  Overall, an object's force of weight at the latitude of 20 degrees is reduced by the ratio of 1/287.5 .  If an object with a "normal weight" of 287.5 lb.f  is placed against Earth at a latitude of 20 degrees, the contact force of its weight will be reduced by 1 lb.f to equal 286.5 lb.f .  Accepting the sudden absence of Sun and Moon gravitation, this reduced weight force will remain the same over a 24 hour period as the object's Earth gravitation will remain constant as will the object's rate of centripetal (inward-directed) acceleration as it rotates about Earth's axis. 

       

 
  (15)  What then is a "normal weight" for an object?   Should such a weight be found, does that mean that all other weights for the object are to be judged as "abnormal" or "apparent"?  At Earth's poles, centripetal acceleration is absent in the downward direction so all of the object's Earth gravitation will be contributing to the object's downward force of weight against Earth.  Also Earth's polar surface at sea level is more than 13 miles closer to Earth's center of matter than is the sea level surface at Earth's equator.  With no centripetal acceleration to take away any of the object's gravitation and being miles closer to Earth's center, any object's force of weight against Earth will be at a maximum at either pole.  So are the poles the location where every object's weight should be judged as "normal"?        

 

 
  (16)  By now I hope you realize that on a rotating Earth the presence and absence of centripetal acceleration is an important factor in determining an object's force of weight.  Other factors are varying distances from Earth's center of matter and an object's gravitation forces caused by the gravitational energy waves it receives from the Moon and the Sun.  Currently, Earth's rotation is responsible for varying rates of centripetal acceleration directed inward toward Earth's axis.  This same rotation causes Earth's surface at the poles to depress 6.5 miles closer than average to Earth's center while at the same time Earth's equatorial matter bulges outward 6.5 miles farther than average from Earth's center.  In search for an average or "normal" weight for an object we could measure its gravitational weight at the latitude 45 deg. N.  Here our object will exist at, or close to, an average distance from Earth's center of matter.  Yet even at this location, if acceleration of one form or another is factored in, then our object's measured force of weight will quickly change.

 

 
  (17)  In truth, all this variability of an object's force of weight means that there is no "normal" weight for an object.  There are several factors at play when one inserts a compression scale between two contacting objects.  For certain there exists more than just the accepted gravity factor expressed in the current, incomplete formula for weight  W = m/g .  In reality, an object's weight is what it is at the moment it is being measured.  The scale displays the true value for a given object at the current location and under the current conditions.  There are no absolute values when it comes to measuring the contact force of weight an object is freely impressing against a second object.  Hence the contact force of an object's weight is relative to the object's environment.  Change the object's environment and its force of weight will likely change to reflect the new conditions.

 
 
  (18)  Let's check Latitude Man's weight at the equator as he orbits Earth's axis.  I will begin by determining his orbital velocity.

Problem 6:  Determine the 200 lb.m object's orbital velocity about Earth's axis.

velocity  =  circumference / time
              =  131,479,775.39 ft  /    86,164.1 s
              =  1525.92 ft/sec

 

 
  (19) Problem 7:  Next we need to calculate the inward-directed centripetal acceleration/Action force required to cause his 200 lb.m to travel the curved path of the Latitude 0 deg. great circle of Earth's equator.

force  =  mass * velocity2 / radius / 32 Pdl/lb.f
          =   200 lb.m  *  (1525.92 ft/sec)2   /  20,925,656.17 ft  /  32 Pdl/lb.f
          =   200 lb.m  *  2,328,431.85 ft2/sec2  /  20,925,656.17 ft  /  32 Pdl/lb.f
          =   200 lb.m  *  0.1112716 ft/sec/  32 Pdl/lb.f
          =   22.254549 Pdl  /  32 Pdl/lb.f
          =   0.6954 lb.f

   

 
  (20)  Problem 8:  Now I need to calculate the force of Earth gravitation for his 200 lb.m when resting against Earth's equatorial surface.  I will avoid rounding since in this case, un-rounded numbers are required to achieve more accurate results.  I am using the Microsoft Windows Calculator in scientific mode.  The values below, expressed in scientific notation, can be copied and pasted into the Microsoft Windows Calculator complete with commas.  For example, the US gravitational constant 3.32,199,885,554,075,5e-11 will paste in without error.

force   = gravitational constant * Earth mass * object mass / Earth equatorial radius2
          = (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2)  *  1.31702804e+25 lb.m  *  200 lb.m  /  (20,925,656.17 ft)2
          = (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2)  *  2634056080000000000000000000 lb.m2  /  437883086145059.0689 ft2
          = (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2)  *  6015432345627.0859133216027686287 lb.m2/ft2
          = 199.8326 lb.f

 

 
  (21)  Here at Earth's equator, both Latitude Man's force of Earth gravity and the centripetal acceleration/Action  force caused by that force of Earth gravity share the same average direction downward toward Earth's center of matter so there is no need to work up a force parallelogram as we did in Problem 5.  All we need to do is subtract the centripetal a/A force of 0.6954 lb.f  from the object's calculated gravitational force which yields a predicted weight of  199.1372 lb.f. 

 

 
  (22)  My friend, James Beyea, posed the following question which prompted the writing of this article.  "Would the weighing of an apple when the Moon is up overhead be any different 12 hours later when the apple is being weighed while the Moon's direction is now underground.?"  Excellent question, Jim.  I will give the following prediction my best effort.

 

 
  (23)  Factoring in the Moon's gravitational effects on our weighing events here on Earth really adds life to such events.   Add Moon gravitation and everything changes.  Now Earth has land, water, and even atmosphere tides.  With Moon gravity restored, weighing an apple over a 24 hour period is no longer an unchanging event.  Consider that when the Moon is overhead, the apple's downward-directed force of Earth gravitation still dominates.  But now there is an upward-directed force of Moon gravitation serving to reduce the contact force the apple is freely impressing against Earth.  Thus the apple's weight is reduced when the Moon is overhead.  But is the apple's weight being reduce by 100% of the apple's force of Moon's gravity?  At first one might think this is the case.  Before we hang the 100% sign on the apple's upward force of Moon gravitation, we need to consider if some portion of this force is being spent causing upward-directed acceleration for the apple.  If this is the case, then that portion will be terminated in that action and therefore will no longer be available to contribute to reducing the apple's force of weight. 

 

 
  (24)    When the Moon is overhead of an apple being weighed against Earth's equatorial surface, is the Moon gravity that is being actively generated within the apple causing the apple to accelerate in the Moon's direction?  The answer is "Yes".  Earth, along with all of its contents including liquids and gases, is accelerating in the Moon's direction.  This acceleration is being caused by each Earth object's internally generated acceleration/Action force of Moon gravitation.  In the Moon-overhead apple's case, downward Earth gravitation is much greater than upward Moon gravitation.  The apple is supported by the scale and Earth's surface below so it is not free to accelerate toward Earth's center.  Meanwhile, the apple is accelerating toward the Moon along with, and at the same average rate as, the rest of solid Earth.  Consider that only when an Earth object resides at the same distance from the Moon's center of matter as is being experienced by Earth's center of matter, will the object's force of Moon gravitation be the correct force to cause the same average rate of acceleration toward the Moon being experienced by the rest of Earth.  Compared to the distance from Earth's center of matter to the Moon's center of matter, the Moon-overhead apple is about 4000 miles closer to the Moon meaning that its internal force of Moon gravitation is greater than average.  The portion of the apple's Moon gravitation force necessary to cause average acceleration for the apple is busy performing and being terminated while accomplishing this single event.  The remaining portion of the apple's upward-directed action force of Moon gravitation is the only portion free to affect the apple's weight by opposing and thereby canceling  a mutual portion of the apple's downward-directed action force of Earth gravitation.  The result is the Moon-overhead apple will freely bear with a reduced weight force against Earth.

 

 
  (25)  Twelve hours later when Earth has rotated 180 degrees positioning the Moon in the underground direction relative to the same apple, the apple's force of Moon gravitation that is being actively generated within the apple's matter is again responsible for the apple's acceleration in the Moon's direction.  Granted, the apple's force of Earth gravitation is this time in the same downward direction but this Earth gravitation force is nevertheless unable to cause acceleration for the apple in the direction of Earth's core since the apple is fully supported by the scale and Earth's surface.  Compared to the distance from Earth's center of matter to the Moon's center of matter, the Moon-underground apple is now about 4000 miles farther from the Moon's center making its force of Moon gravitation less than the previously mentioned average force responsible for the apple's average rate of acceleration when it was previously positioned at the same distance from the Moon's center as is Earth's center.  Here in the Moon-underground position, 100% of the apple's force of Moon gravitation is an acceleration/Action force for the apple.  Yet it is only enough force to cause a below average rate of acceleration in the Moon's direction.  The balance of the acceleration/Action force required to cause the apple to keep up with the rest of Earth comes from the apple's force of Earth gravitation.  In this manner, less of the apple's force of Earth gravitation remains free to contribute to causing the Moon-underground apple to freely bear with its contact force of weight against Earth.  So, despite the fact that the apple's Earth gravitation and Moon gravitation forces are acting in the same downward direction, as with the Moon-overhead apple, the Moon-underground apple also weighs less.   

 

 
  (26)  With the apple weighing less at the Moon-overhead position and also weighing less 12 hours later at the Moon-underground position, I predict it will weigh more at the 6hr and 18 hour positions when the apple's force of Moon gravitation is nearly at a right angle to its force of Earth gravitation.  Also in these two positions on Earth's horizon as viewed from the Moon, the apple's radial distance from the Moon will be very close to the radial distance of Earth's center of matter.  This means that the apple's force of Moon gravity will likely be close to the correct magnitude to cause the required sideways acceleration for the apple in the Moon's direction. 

 

 
  (27)  In the 6 hour and 18 hour positions, due to a small inward component force of Moon gravitation, a squeezing of Earth does occurs with its maximum being for matter located all the way around Earth's circular horizon as determined by an observer on  the Moon.  This squeezing force is a component force that is caused by the misalignment of this matter's direction of Moon gravitation compared to the direction of the matter's Moon-caused acceleration.  While our focus here is on the degree of influence the Earth squeezing force has upon the apple's force of weight at the 6 hour and 18 hour positions, we should keep in mind that this well-known component force must fill a role in causing the formation of Earth's egg-shape and dual high tides.  
 

                            

                                

 
  (28)  Our goal here is to determine if any difference exists between the weights of the Moon-overhead apple and Moon-underground apple.  Better yet, since an apple is so light, I will instead employ the services of Latitude Man with his superior mass rating of 200 lb.m.  This way any difference in measured weight will be larger and hopefully more meaningful.  An important first step is to establish Earth's average acceleration rate due to the Moon's gravitational influence.  In order to make this determination, I first need to establish the balance point between a non-rotating Earth and the normally orbiting Moon.  This balance point, named the barycenter of the Earth/Moon system, may be calculated as if scale models of Earth and the Moon are positioned on a long, scale-length, balance beam where the fulcrum is moved along the beam until balance is achieved.  To locate this barycenter we need to know Earth's mass rating, the Moon's mass rating and the distance between their centers of matter.  The barycenter formula determines the distance from the more massive body's center of matter to the barycenter balance point which in the Earth/Moon system is located deep inside the body of Earth.  This barycenter distance is also the measure of the radius followed by Earth's center of matter as it slowly orbits the barycenter one circuit every 27.3217 days.

Problem 10:
  Determine the distance between Earth's center of matter and the barycenter of the Earth/Moon system.

barycenter distance =  Moon mass  *  Earth-Moon distance  /  (Earth mass  +  Moon mass)             
                                =  1.61994e+23 lb.m  * 1261171260 ft  /  (1.31702804e+25   +   1.61994e+23)
                                =  1.61994e+23 lb.m  * 1261171260 ft  /  13332274400000000000000000 lb.m
                                =   2.0430217709244e+32  /   13332274400000000000000000 lb.m                                          
                                =  15,323,880.3 ft  =  2,902.25 miles  from Earth c.m. to barycenter

 

 
  (29)  Now consider that as Earth's center of matter makes an orbit of the Earth/Moon barycenter, it does so with a radius of orbit of 15,323,880.3 ft or 2,902.25 miles. To determine Earth's average rate of centripetal acceleration directed toward the barycenter and beyond toward the Moon, we need to know the velocity of Earth's axis as it travels along the circumference of its orbital path around the barycenter. 

Problem 11:
 Determine the orbital velocity of Earth's axis about the Earth/Moon barycenter.

orbital velocity  =  orbital circumference / time
                        =   2  *  pi  *  radius / time
                        =   2  *  3.14159  *  15,323,880.3 ft  /  2,360,595 sec
                        =   6.28318  *  6.49153 ft/sec
                        =  40.7875 ft/sec

 (40.7875 ft/sec is about 28 miles/hour which reveals the leisurely speed at which Earth's axis orbits the Earth/Moon barycenter.)

 

 
  (30)  Problem 12:   Determine the average rate of centripetal acceleration of Earth's axis with this acceleration directed along the Earth/Moon centerline toward the barycenter and the Moon beyond. 

Earth average acceleration  =  (velocity) 2  /  radius.
                                          =   (40.7875 ft/sec ) /  15,323,880.3 ft
                                          =  1663.62 ft2/sec/ 15,323,880.3 ft
                                          =   0.000108564 ft/sec2

 

 
  (31) The Earth average acceleration rate from Problem 12 is the same rate of Moon-directed acceleration being experienced by Latitude Man when positioned anywhere on the equator of a full-sized Earth.  To answer the 100% question asked in paragraph 23, we first need to find the magnitude of the acceleration/Action force required to cause Latitude Man, as a small portion of Earth's matter, to orbit the Earth-Moon barycenter.  Then we need to see how this calculated centripetal force compares to Latitude Man's force of Moon gravity when his distance to the Moon's center is equal to the distance from Earth's center to the Moon's center.  If they are the same or at least close to a match then we will have shown that the force predicted to cause Latitude Man's barycenter orbital parameters is equal to Latitude Man's force of Moon gravity that is known to be present.

Problem 13:  Calculate the force required to cause Latitude Man's 200 lb.m , as a small portion of Earth's matter to orbit the barycenter.

force  =  mass * (velocity)2 / radius / g
         =   200 lb.m  *  (40.7875 ft/sec )    /  15,323,880.3 ft   /  32.17 Pdl/lb.f
         =   200 lb.m  *  1663.62 ft2/sec2    /  15,323,880.3 ft  /  32.17 Pdl/lb.f
         =   200 lb.m  *  0.0001.08564 ft/sec/  32.17 Pdl/lb.f
         =   0.0217128 Pdl  /  32.17 Pdl/lb.f
         =   0.00067494 lb.f

 

 
  (32)  Problem 14:  Calculate Latitude Man's Moon gravitational force when he is positioned at the same distance from the Moon as Earth's center of matter.

center force = gravitational constant * Moon mass * object mass / (Moon c.m. to Earth's c.m.)2
                   = (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2)  *  1.61994e+23 lb.m  *  200 lb.m  /  (1,261,171,260 ft )2
                   = (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2)  *  32398800000000000000000000 lb.m2  /  1590552947049987600 ft2
                   = (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2)  *  20369519.95850898045214337431846 lb.m2/ft2
                   = 0.00067668 lb.f

    This is the magnitude of Moon gravitation needed to cause average acceleration for Latitude Man. Logically when Latitude Man is located at the same distance from the Moon as Earth's center of matter, the force of Moon gravitation being generated within his body will be the exact magnitude required to cause a rate of acceleration directed toward the Moon that is a match for the average rate of acceleration being experienced by Earth.  I think the two forces calculated in Problem 13 and Problem 14 are close enough to be called a match.

 

 
  (33)  Barycenter Discussion:  All Moon gravitational forces being generated within Earth are directed toward the Moon's center of matter.  This means that most such forces are not directed toward the Earth/Moon barycenter about which Earth's center-of-matter orbits.  With this being the case, one cannot help but wonder exactly what influence does the barycenter have upon Earth?  The short answer is: "No influence whatsoever".  If all of Earth's Moon gravitational vectors were added, the total vector would be directed toward the Moon along the Earth/Moon centerline.  Since the Earth/Moon barycenter  happens to reside on this centerline, it can be said that this total force vector is the centripetal or center-directed action force responsible for causing Earth's centripetal acceleration about the Earth/Moon barycenter.  In truth this total force vector represents the sum of the myriad of Earth's Moon gravitation forces that are themselves directed toward the Moon and not toward the barycenter for any reason other than for some such forces, the barycenter just happens to be on the line of gravitation.  Earth's Moon gravitational forces would cause short-lived linear acceleration for Earth if the direction to the Moon were to remain constant.  It is the curious mix of Earth's attraction toward the Moon and the orbiting Moon's ever-changing direction that effectively converts a linear acceleration event into a centripetal acceleration event.  Due to these effects on Earth, I find it acceptable to refer to Earth's Moon gravitational forces as centripetal forces responsible for causing Earth's centripetal acceleration that results in Earth's curved path of travel about the underground barycenter.  It is important to keep in mind that Earth is effectively being towed (by the distant orbiting Moon) around in its small-radius orbit of the Earth/Moon barycenter.
    Often the central position, relative to an orbiting object is; 1) occupied by a much more massive, attractive body when the centripetal forces are internal-to-matter forces such as gravitation, or; 2) occupied by some firm attachment point when the centripetal forces are external-to-matter forces such as being provided by a taught string or cable.  In both of these common cases, the centripetal forces being experienced by the orbiting object that are causing the action of centripetal acceleration for the object are caused by and directed toward the central object which resides at the center of orbit.  Yet such is not always the case.  When an automobile rounds a uniform curve at a constant speed, the forces causing the automobile's centripetal acceleration, while directed toward the center of that curve, are not in any way caused by any object that resides at that curve's center point.  Here the curve's center point is only of mathematical importance as it is used to determine the length of the radius of the curve.  The same level of non-influence holds true for the Earth/Moon barycenter.  It is only of mathematical importance just as it is used herein to determine Earth's radius of orbit and Earth's average rate of centripetal acceleration.  Notice how in the following Latitude Man's Moon gravitational force calculations, his distance from the Moon is an important factor while his distance from the Earth/Moon barycenter is irrelevant.

      

 
  (34)  Conditions:  Earth is now back to its normal rate of rotation.  All other Earth-Moon system parameters are back to normal.  Latitude Man is now positioned against Earth's equator on the near side closest to the overhead Moon.

  Problem 15:  Calculate Latitude Man's Moon gravitational force when he is at the 0 hour position on Earth's near side equator with the Moon in the overhead direction.

near side force = gravitational constant * Moon mass * object mass / (Moon c.m. to object on Earth's near side)2
                       = (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2)  *  1.61994e+23 lb.m  *  200 lb.m  /  (1,240,245,604 ft)2
                       = (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2)  *  32398800000000000000000000 lb.m2  /  1538209158241324816 ft2
                       = (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2)  *  21062675.271705184949619183216297 lb.m2/ft2
                       = 0.00069970 lb.f

 

 
  (35)  Here at 0 hour on Earth's near side to the Moon, Latitude Man's force of Moon gravitation is stronger than average.  Yet we know his acceleration directed toward the Moon remains generally average.  An average force is terminated while causing average acceleration.  Once terminated, no portion of this average force can be expected to continue on to act as the cause of any other event.  Only the unterminated remainder of his force of Moon gravity will remain free to act as the cause of some other event such as affecting the contact force of Latitude Man's weight against Earth.  This unterminated portion is 0.00069970 lb.f  -  0.00067668 lb.f  =  0.00002302 lb.f.  This remainder in percent terms is 3.3% of Latitude Man's full force of near side Moon gravity.  To answer the 100% question asked in paragraph 23,  96.7% of Latitude Man's near side force of Moon gravity is being spent causing his acceleration that is directed toward the Moon.  Only the 3.3% remainder is still available to be factored into his weight calculation.  This is a very small force, measuring about 2/100,000th of a pound or 1/6th of a grain.

 

 
  (36)  Next Latitude Man will reach the 6 hour position where his distance to the Moon is equal to that of Earth's center of matter.  His force of Moon gravitation is now at right angle to his vertical force of Earth gravitation.  For this reason, his Moon gravitation will have no effect upon his force of weight against Earth.  Besides, at this central location, his horizontal force of Moon gravity, previously calculated in Problem 14 as  0.00067668 lb.f is just sufficient to cause average acceleration for his body in the horizontal direction.  Thus his force of Moon gravity is not only in the wrong direction to affect his weight but it is also completely terminated while causing his average Moon-directed acceleration.

  

 
  (37)  Finally Latitude Man will reach Earth's far side 12 hour position relative to the Moon.  Here there remains the issue that being located about 8000 miles farther from the Moon, Latitude Man's far side Moon gravity force, which is now downward-directed, is less than the amount of force needed to cause his average acceleration in this same direction toward the Moon. 

  Problem 17:  Calculate Latitude Man's Moon gravitational force when he is at the 12 hour position on Earth's far side equator with the Moon in the underground direction.

far side force  =  gravitational constant * Moon mass * object mass / (Moon c.m. to object on Earth's far side)2
                     = (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2)  *  1.61994e+23 lb.m  *  200 lb.m  /  (1,282,096,916 ft )2
                     = (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2)  *  32398800000000000000000000 lb.m2  /  1643772502016711056 ft2
                     = (3.32,199,885,554,075,5e-11 lb.f*ft2/lb.m2)  *  19710026.75872152084405260733384 lb.m2/ft2
                     = 0.00065477 lb.f

 

 
  (38)  As expected, the far side force is 0.00002191 lb.f  shy of being able to cause average acceleration for our far side Latitude Man.  There is no question that Latitude Man is accelerating at the average rate.  The extra force needed comes from the only other action force that is present in the same direction, Latitude Man's force of Earth gravitation.  This extra force will terminate after being borrowed to aid in Latitude Man's acceleration.  It will no longer be available to aid in causing Latitude Man's contact force of weight against Earth.

 

 
  (39)   It is time to total the forces affecting Latitude Man's contact force of weight against Earth.  Forces present within Latitude Man's matter and directed downward from Latitude Man to Earth will be positive for they serve to increase his weight.  Forces equally present within Latitude Man's matter and directed upward will be negative for they cause or otherwise represent a reduction in his weight.  It is important not to count an action force more than once.  For example, the acceleration/Action force of Earth gravitation causing centripetal acceleration for Latitude Man as he travels about Earth's rotational axis is not a new force separate from his force of Earth gravitation.  It is a portion of his Earth gravitation force that is serving an accelerational role instead of a role affecting his weight.  This portion needs to be subtracted from his total Earth gravitation force.  To accomplish this task using the rules of vector addition, I will instead add his oppositely-directed centrifugal acceleration/Reaction support force that is an internal-to-matter force present within every accelerating component of Latitude Man's matter.  I use the term "normal" as in "normal gravity forces" to indicate the force component that is perpendicular to the surface against which the object's force of weight is being measured.

Problem 18: Total the forces affecting Latitude Man's contact force of weight against Earth's near side at the 0 hour position which is the position that is closest to the overhead Moon.  

Latitude Man's near side weight  =   (normal gravity forces)  +  (normal acceleration/Reaction forces)
                                                 =   (Earth gravity force + Moon gravity force)  +  (Earth axis centrifugal a/R force +   Earth barycenter a/R force)
                                                 =   (199.8326 lb.f   + -0.00069970 lb.f )  +  (-0.6954 lb.f   +  -0.00067494 lb.f)
                                                 =   (199.8319003 lb.f ) +  (-0.69607494 lb.f)
                                                 =    199.13582536 lb.f

    

 
  (40)  Now let's calculate Latitude Man's weight at the 6 hour or 18 hour position where his distance to the Moon is nearly the same as Earth's center of matter.  Here Moon gravity will have no effect on his weight.

Problem 19:  

Latitude Man's center weight   =   (normal gravity forces)  +  (normal acceleration/Reaction forces)
                                              =   (Earth gravity force)  +  (Earth axis centrifugal a/R force)
                                              =  (199.8326 lb.f ) +  (0.00067668 lb.f )
                                              =   199.83327668 lb.f 

                                             

 
  (41)  Now let's calculate Latitude Man's weight at the far side 12 hour position.

Problem 20:  

Latitude Man's far side weight    =   (normal gravity forces)  +  (normal acceleration/Reaction forces)
                                                 =   (Earth gravity force + Moon gravity force)  +  (Earth axis centrifugal a/R force +   Earth barycenter a/R force)
                                                 =  (199.8326 lb.f  +  0.00065477 lb.f )  +  (-0.6954 lb.f   +  -0.00067494  lb.f )
                                                 =  (199.83325477 l b.f )  +  ( -0.69607494 lb.f)
                                                 =   199.13717983 lb.f

 

 
  (42)  Notice there is very little difference in Latitude Man's  weight on Earth's near side verses his weight on Earth's far side.  While still significantly lighter than his Earth center weight, he is 0.00135447 lb.f  heavier on Earth's far side compared to his weight on Earth's near side.  This amounts to 1.35/1000th of a pound or  9.5 grains.  Using a ratio formula, I will convert these results down to apples. 

Problem 21:  Determine how much heavier an apple that weighs 0.5 lb.f on Earth's near side equator weighs when positioned on Earth's far side equator.

apple's far side weight  =  apple's near side weight  *  Latitude Man's far side weight  /  Latitude Man's near side weight
                                    =  0.5 lb.f  *  199.13717983 lb.f   /  199.13582536 lb.f
                                    =  0.5000034  lb.f 
                                                                     
 

 
  (43)  The far side apple will weigh 3.4 millionth of a pound heavier than the 0.5 lb.f weight of the near side apple.  How heavy is  3.4 millionth of a pound object?  Take one grain of barley and with a magnifying glass and a very sharp knife, cut the grain of barley into 50 equal pieces.  Slide just one of those pieces onto the scale of the near side apple and its weight will now be equal to the weight of the far side apple.  A highly sensitive scale would be needed to compare the minute change in the apple's weight from Earth's near to far side.  Of more interest may be a comparison between the apple's near side weight at the 0 hour position and the apple's center weight at the 6 hour or 18 hour position.

Problem 22:  Determine how much heavier an apple that weighs 0.5 lb.f on Earth's near side equator weighs when placed at the 6 hour or 18 hour center position while remaining on Earth's equator.

apple's center weight  =  apple's near side weight  *  Latitude Man's center weight  /  Latitude Man's near side weight
                                 =  0.5 lb.f  *  199.83327668 lb.f   /  199.13582536 lb.f
                                 =  0.5017512  lb.f 

 

 
  (44)  Here at the 6 hour or 18 hour location, the apple's Earth gravitation is not being reduced by Moon gravitation since it is now a sideways force.  Now our apple is 1.75 thousandths of a pound heavier.  This converts to 12.25 grains which should easily submit to measurement.  If only I lived on Earth's equator I would give it a try for myself.  This might make for a good science fair project for a student in Ecuador, Brazil, Columbia, Indonesia, Somalia, Kenya, Uganda, the Congos, Gabon, São Tomé or Príncipe.  Place "Skyler's Apple" on a sensitive digital scale and take time-stamped photos of the scale's differing readings over a 24 hour period with the Moon being overhead at 0 hour.  Attach the photos on a large poster board and explain the forces and their directions plus the accelerations that influence Latitude Man's and Earth's external-to-matter contact forces of weight, each against the other.  Finally, before the science fair ends, send me a few photos of your display and of yourself with your science teacher and fellow science students along with your location and test results.  I will make a place for your photos and test results on UniversalPhysics.Org.

 

 
  (45)  Conclusion I:  Due to the Moon's influence, Latitude Man is lighter than normal on Earth's near side with the Moon overhead which should come as no surprise.  What might be a surprise is that only 3.3 % of Latitude Man's total near side force of Moon gravitation is being spent reducing his weight since the other 96.7% is being terminated while causing his Moon-directed acceleration.  Meanwhile, on Earth's far side, despite the fact that Latitude Man's Moon gravitation is now downward-directed, it comes up short of being able to cause him to accelerate in that same downward direction at Earth's average rate.  As a portion of his Earth gravitation is "borrowed" to complete this accelerational task, less is available to contribute to his force of weight against Earth. Once again he is lighter than average only not as much lighter as he is on Earth's near side.  This is in agreement with our solution to Earth's dual high tidal mounds as presented in Event 5.  Lighter water on Earth's near side mounds up to a higher elevation.

 

 
 

(46)  With weight being the contact force one object freely impresses against a second object, where do you think is the best place on Earth to study your weight?  I suggest you take a compression scale for a ride in an elevator.  I found it best to pick off-hours so elevator traffic is light.  It is hard to keep a straight face when someone boards while you are standing on the scale.  But soon they will join in with this study of your fluctuating weight.  Begin by checking your weight before the elevator starts accelerating toward upper floors.  Gripping the grab bar is not okay but pushing horizontally against the walls is acceptable. 

CAUTION:  If you have any issues with balance and think you might fall from the scale then have a better-balanced friend do the test while you observe.   

 

 
  (47)  Once the elevator door closes, step aboard the scale and read the contact force of your weight against the elevator floor.  This external-to-matter force, present at its maximum at the bottoms of your feet, is sourced from the billions and billions of your body's components of matter.  Each component generates its own internal-to-matter force of Earth gravitation which it adds to the gravitational forces it is receiving from components above while, in turn, it passes on an even greater combined serial transfer force to components below.  In this manner, a myriad of internal gravitational forces in Earth's direction stack up in magnitude to eventually bearing down as the external-to-matter contact force of your feet against the scale.   This same scenario occurs deep within Earth's matter as it's myriad of internal gravitational forces in your direction stack up in magnitude to eventually bear "up" as the external contact force exerted by Earth through the scale against your feet.  This is the full description of this mutual gravitational event accepting either that the test is performed at one of Earth's poles or Earth's rotation has been temporarily suspended. Study the Stacking-of-Forces Effect drawing to see how individual, internal-to-matter gravitational forces stack up to produce the external-to-matter force of weight present between two identical, contacting objects. Equally important notice how each object's maximum serial transfer force is fully present within the other object's matter as it stacks down while providing Type 3 action-force support for the Type 2 gravitational action forces being generated within each of the other object's component's of matter. 

 

 
   
  (48)  Now we could limit our focus to just a portion of the Elevator Man I event, as is commonly done, by referring to Earth's upward external serial transfer force against your feet as a "support" force with no acknowledgement of its gravitational source.  But what advantage is there to describing only half of an event?  Why not describe the entire event as I have done in the Stacking-of-Forces drawing?  This way it can be easily demonstrated that for every force present, there is always present an equal and opposite force.  Surely Isaac Newton would approve of our using the complete event as we continue to verify the truth of his LAW III along with Rule 7a of the Universal Physics Rules for Force and Motion.  

 

 
     

 

 
  (49)  The Elevator Man 1 drawing shows the entire event, with the understanding that it takes place at one of Earth's poles or Earth's rotation is temporarily suspended.  The first drawing shows Elevator Man's and Earth's scale force of weight they each are freely bearing against the other.  The absence of acceleration and relative motion is noted.  The second drawing indicates the action forces present.  There is no third drawing of reaction forces here for there are no reaction forces present.  It may take some getting used to the Universal Physics understanding that an action force is one that is causing the action of acceleration or capable of causing the action of acceleration.  While no acceleration is acknowledged as being present in the Elevator Man 1 event, every force drawn as present is capable of causing acceleration.  All force vectors directed downward relative to Elevator Man are set as positive while upward forces are set as negative.  

 

 
  (50)   
  , (add here a discussion of the first chart.  Include how the internal gravitation vector is drawn beginning at Elevator Man's center of matter.)

 

(51)  Now push the button for a higher floor and watch the scale's display.  For just a second or two your weight will increase perhaps 15 lbs.  Then it will return to the weight you first recorded since you are no longer accelerating.  This non-accelerational weight will

 
 

 

 
 

(52) 

Here discuss how the a/R force is also drawn at Elevator Man's and Earth's centers of matter indicating that it is present at each component of an accelerating object's matter, not just at the interface between the object and its point of contact with a second object as is always claimed to be the case by limited conventional physics.  Also mention the stacking of forces effect which again indicates that the a/A and a/R forces interface with each other at the component level.  This is proof that the a/R force is present throughout the accelerating object's matter again in direct contradiction to the limited conventional view where action and reaction force are said to never affect the same object.  In truth, a/A and a/R always affect the same accelerating object without exception.  Talk about denying the truth of such events right out of the gate.  Is it any wonder that our best "explanations" are so lacking?

 
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Author's Commentary
 

 

 
 
 

Copyright Notice
   
Article XII: "Weight" (C) Copyright 2013 by Ethan Skyler. All Rights Reserved. No portion of Article XII, minus the exceptions noted below, may be copied by any means without the author's written permission and even then only if the author's copyright notice is permanently affixed to each approved copy. Requests for written permission may be directed to Ryan Skyler, Editor, Universal Physics Journal, 9734 Manitou Place NE, Bainbridge Island, WA, USA.

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