Purpose to compare and contrast geostationary flat earther claim that object on a round spinning earth when throwing vertically and not horizontally should not land in the same place it started relative to the thrower because the earth is moving. If the differential equations provide a certain solution than the object on a round spinning earth would land in the same place it started relative to the throwing observer due to the initial horizontal velocity of the observer caused by the rotation of the earth being identical to the initial horizontal velocity of the ball caused by the rotation of the earth from a different reference frame in which the earth is rotating and the thrower is initially moving horizontally but not initially vertically in Cartesian coordinates at the initial starting time and the thrown object is moving horizontally at the same velocity as the observer and moving vertically at it's maximum positive vertical velocity which is it's initial vertical velocity identical to a positive value of the initial speed the thrower throws the object from the throwers frame of reference. The two reference frames share the same origin, but one reference frame is rotating relative to the other at a constant rotational velocity and air friction is ignored. If the result shows that the object would land where the throwing observer initially threw it from in the observers frame of reference this would fail to refute the claim of a rotating spherical earth without proving a flat geostationary earth, however if the result is different by a large enough margin this would refute a spinning spherical earth without proving a flat geostationary earth. Provided of course I set up the differential equations correctly. The refutation or lack thereof is presuming the standard equation for calculating the force vector from gravity using the universal constant for gravitation is correct if the earth is a rotating sphere which does not necessarily logically follow but is none the less used because it is the current standard physics model which most rotating globe shaped earthers claim.
R=Radius=Radius of sphere earth
Mobject = Mass of object
MSphere = Mass of sphere earth
Xinitial = 0 for both thrower and thrown object
Yinitial = R for both thrower and object
THETAinitial=0 radians
Vxinitial = Vx0
Vx0 = Radius*dtheta/dtime
dV/dtime= -1*G*Msphere*Mobject/(Mobject*[|X|^2+|Y|^2])
dVx/dtime = -|dV/dtime|*X / (|X|^2+|Y|^2]) ^ 0.5
dVy/dtime = -|dV/dtime|* Y / (|X|^2+|Y|^2]) ^ 0.5
dVx/dtime =-1*G*Msphere*Mobject*X/(Mobject*[|X|^2+|Y|^2]^1.5)
dVy/dtime =-1*G*Msphere*Mobject*Y/(Mobject*[|X|^2+|Y|^2]^1.5)
Vfinalobject and TIMEfinal occurs when |Xfinalobject|^2+|Yfinalobject|^2 = |R|^2 for a X value other than zero
Xfinalthrower = R*sin(Thetafinal)=R*[(timefinal-timeinitial)*(dtheta/dtime)]
And
Yfinalthrower = R-R*cos[(timefinal-timeinitial)*(dtheta/dtime)]
If
Xfinalobject=Xfinalthrower and Yfinalobject=Yfinalthrower
Or in other words the thrower and the thrown object both exist at
(R, Thetafinal) = (R, (dtheta/dtime)*(Timefinal-Timeinitial))
Then
An observer throwing an object vertically standing at Xinitial, Yinitial at TIMEinitial would observe the object landing where he stands at Xfinal, Yfinal, TIMEfinal which would be equivalent to R, THETAfinal
Which would mean from the throwers point of view throwing an object vertically (and not horizontally) while standing on a rotating sphere would result in the object landing in the same place it is thrown from failing to refute a rotating spherical earth.
Else if
XfinalObject does not equal Xfinalthrower or if Yfinal object does not equal Yfinalthrower
Then
if the difference is large enough this would not fail to refute a rotating spherical earth
Now for a simplification by rounding which might not hold correctly you can use
dVx/dtime =-1*g*|X|/([|X|^2+|Y|^2]^0.5)
dVy/dtime =-1*g*|Y|/([|X|^2+|Y|^2]^0.5)
If you wish to round further yet the following can be used however it will result in a different position between the thrower's final position and the landed thrown object 's final position none the less if the difference is small enough a strong case can be made for failure to refute a rounded globe earth.
Of course this rounding results in the typical equations for a thrown projectile from a flat surface onto the same flat surface without the direction or magnitude of forces from gravity changing possibly because physicists assume a flat earth model when simplifying projectile problems although they actually believe in a round earth and this is not necessarily contradictory because the difference over small distances is presumed negligible. Some of these rounded points might not actually exist on the Sphere but could be very close to points on the Sphere.
only do rounding for small enough values of (tfinal-tinitial)
dVx/dtime = 0
dVy/dtime = - 1 * g
Xfinalthrower= Radius*sin[(dtheta/dtime)*(Timefinal-Timeinitial)]
Yfinalthrower = Radius*cos[(dtheta/dtime)*(Timefinal-Timeinitial)]
Vy0 = Vyinitial
Vx0 = Radius*dtheta/dtime
Rounded Timefinal-timeinitial=2*|Vy0/g|
Rounded Xfinalobject = Rounded (Timefinal-timeinitial) * Vx0
Rounded Xfinalobject = 2*|Vy0/g|* Vx0
Rounded Xfinalobject = 2*|Vy0/g|* Radius*dtheta/dtime
rounded Xfinalthrower= Radius*sin[(dtheta/dtime)*(2*|Vy0/g|)]
Xfinalthrower / Xfinalobject =
= (sin[(dtheta/dtime)*(2*|Vy0/g|)])/[(dtheta/dtime)*(2*|Vy0/g|)]
Limit as theta approaches 0+ of (sin[theta]) / theta is 1-
Limit as Vy0 approaches 0+ of Rounded Xfinalthrower/RoundedXfinalobject is 1-
For small values of Vy0
Rounded Xfinalthrower is approximately equal to rounded Xfinalobject
Conclusion
When throwing objects vertically for small enough vertical heights with a small enough duration of time between throwing and landing a rotating earth should have approximately the same results as a stationary earth and a flat earth approximately the same results as a globe shaped earth. There is a failure to refute the round earth claim with rounded simplified equations, failure to refute a round rotating earth does not refute a flat geostationary earth. Perhaps if the differential equations were used without rounding the object might not be calculated to land in the same place but typically physicists round so much that a flat earth and round earth produce the same results with projectile motion for objects thrown over short distances for a short period of time.
Copyright Carl Janssen 2020 March 26
