In some models, the Earth is considered to be a flat cylinder. Here we calculate the escape velocity for such a matter distribution and find that it is similar to the standard sphere. This result is briefly discussed and we show that, for large distances, the force from such a planet would tend to that from a sphere as expected

Flatt, Kieran

Garreffa, Maria

Longman, William

Turnpenney, Sam

English

University of Leicester Open Journals

Journal of Physics Special TopicsA1 1 The Flat Earth’s Escape VelocityK. Flatt, M. Garreffa, W.H.A. Longman, S. TurnpenneyDepartment of Physics and Astronomy, University of Leicester. Leicester, LE1 7RH.Oct 8, 2014.AbstractIn some models, the Earth is considered to be a flat cylinder. Here we calculate the escape velocity forsuch a matter distribution and find that it is similar to the standard sphere. This result is briefly discussedand we show that, for large distances, the force from such a planet would tend to that from a sphere asexpected.Belief that the Earth is flat originates from ancientcivillization, and persists to this day [1] due to a re-vival in the 19th century spearheaded by Samuel Row-botham [2]. This idea is known as the Flat Earth andin this paper we derive an analytic expression for thehypothesised planets escape velocity that is, the min-imum velocity required for a test mass to overcomethe gravitational potential. In the model we propose,the Flat Earth is a disk of height h, radius RE andtotal mass ME , where the latter two parameters arethe same as the radius and mass of the paradigmaticSpherical Earth model. The density ρ is consideredconstant throughout the mass distribution.In a standard escape velocity treatment, the poten-tial energy at a point on the surface of a spherical bodyis calculated: the escape velocity is then that whichgives a high enough kinetic energy to overcome this po-tential. When such a method is tried outside of spheri-cal symmetry, however, it leads to several problems: inparticular it is not possible to solve analytically for thepotential inside the cylinder. The method presentedhere thus considers Newtonian force-at-a-distance inplace of energetics, in order to arrive at an analyticresult.We can consider the forces on a test particle of massM at a point on the z-axis, at a distance λ from a ringof mass MR. Using a cylindrical co-ordinate system,the force can be resolved into z and r components andfrom Fig. 1 it can be seen be clearly seen that com-ponents in the r direction will cancel so that the forceis directed entirely along z. If a distance λ is definedas the total distance of the particle from the ring, asin the geometry of Fig. 1, the force on the particle isthen~F = −GMRMλ2zˆ (1)where the minus sign comes from the fact that theparticle is being drawn towards the centre of the ring.To extend this treatment to a disk we integrate therings over a radius RE . To do this, we write theFig. 1: This diagram shows a point mass at a height z abovethe centre of a ring, radius r, so that the total distancefrom the ring is λ2 = r2 + z2. The gravitational forcefelt from this ring by the test particle will act along thedashed line: it can be seen by resolving this force that onlythe component along the z axis will take effect. In orderto find the total force on an object due to a flat disk, weintegrate a series of rings with radii between 0 and RE . Formore details on this, see the text.mass of the ring as a radius element in terms of thedensity ρ, which we consider constant. This givesMR = 2piρhrdr. We also eliminate λ to give, for a‘force element’d~F = −2piGhMρ rz2 + r2drzˆ. (2)We wish to find the force from the whole Earth byintegrating between r = 0 and r = RE . Firstly, fromthis point on the vector notation is dropped for sim-plicity. However it should be noted that this equationis only valid along the z-axis; and that the force is di-rected along this axis towards the centre. By makingthe substitution q = r/z this can be written asF = −2piGhMρ∫q1 + q2dq, (3)The integral over q is a standard integral that canbe found in standard reference texts [3]. It is given by1The Flat Earth’s Escape Velocity, Oct 8, 2014.∫x1 + x2dx =12ln (1 + x2) + C, (4)with C the arbitrary constant of integration. Sub-stituting this into Eq. 3 and applying the boundaryconditions, the equation for the force on our particlebecomesF = −piGhMρ ln(1 + (REz)2). (5)Before continuing, we should briefly considerwhether this result is reasonable. One way to do thisis to see how this function behaves for large distancesfrom the Flat Earth. In order to this, we can expandthe logarithm in ascending powers of REz . Such an ex-pansion is given byln (1 + x2) ≈ x2 − x42. (6)For the current discusion, x = RE/z. ThusRE/z << 1 for large z and we can ignore the termshigher order than squared and use this approximationfor the long range limit of the force, Eq. 5. If the def-inition of the mass in terms of the density and volumeis included, we findF = −GMEMz2. (7)Thus at long distances the gravitational field of theFlat Earth approaches that of a point mass, as to beexpected. We can then see that the result derived sofar is correct, and can now solve for the escape velocity.The procedure we use for this is fairly straightforward- Newton’s second law can be written in the formF = mvdvdz(8)by application of the chain rule. This is then equatedwith the earlier expression for the force at a point,Eq. 5, and we solve according to a set of boundaryconditions. For this we note that the definition of theescape velocity is the minimum initial velocity for atest mass to be able to travel away from a body withoutits direction being reversed - hence it will have velocityof v = 0 as z → ∞ and a velocity of v = vF at z = 0.We hence have to solve∫ 0vFvdv = −piGhρ∫ 0∞ln(1 + (REz)2)dz. (9)By using the substitution x = RE/z this can berewritten in the form∫ 0vFvdv = piGhρRE∫ 0∞ln (1 + x2)x2dx. (10)Again, the right hand is a standard integral whichgives the result [5]∫ ∞0ln (1 + x2)x2dx = pi. (11)Finally, we can rearrange the results so far to give anexpression for the Flat Earth’s escape velocity. Alsousing the definition of the mass in terms of densityand volume, i.e. ME = piR2Ehρ, we can eliminate thedensity as a parameter. The result of doing this is:vF =√pi√2GMERE=√pivs, (12)where vs =√2GME/RE is the usual result for theescape velocity of a spherical matter distribution - i.e.the Earth [6]. So, interestingly, it can be seen thatchanging the particular shape of the mass distributionsimply introduces an extra factor of√pi into the finalresult. In some respects this is to be expected, as thebehaviour of the gravitational field will simply be thatof a point mass far away from both matter distribu-tions. Numerically, using the standard values for thevarious parameters gives a value of vF = 19.9 kms−1.So, to conclude, we have calculated an expression forthe escape velocity of a cylindrical mass distribution -i.e. the Flat Earth - and found that it is similar inform to that of a spherical distribution. This suggestsan obvious area for further research: what happensfor other mass distributions? For example, it may bepossible to derive expressions for the escape velocityof a Cube Earth or Rod Earth. Could these also beclassified similarly by an extra factor? Sticking to theconceptual difficulties surrounding the Flat Earth, itwould be interesting to further explore claims made bythe Flat Earth Society in the context of mathematicalphysics or experimental tests.References[1] http://theflatearthsociety.org/cms/, accessed:08/10/2014[2] Samuel Rowbotham, Astronomy: Earth Not a Globe,Forgotten Books, 2007[3] http://www.wolframalpha.com/input/?i=int+x%2F%281%2Bx^2%29+dx, accessed: 07/10/2014[4] http://www.wolframalpha.com/input/?i=taylor+expand+log%28x^2%2B1%29, accessed: 07/10/2014[5] http://www.wolframalpha.com/input/?i=integrate+log%28x^2+%2B+1%29%2Fx^2+between+0+and+infinity, accessed: 07/10/2014[6] Tipler and Mosca, Physics for Scientists and Engineers6th Edition, W.H Freeman, 20072

A1_1 The Flat Earth's Escape Velocity

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A1_1 The Flat Earth's Escape Velocity

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