A note on a straight gravity tunnel through a rotating body
AA note on a straight gravity tunnel through a rotating body
Aleksander Simoniˇc ∗ School of Science, The University of New South Wales (Canberra), ACT, Australia
Abstract
It is well-known that the straight gravity tunnel between any two different positions on a non-rotating Earth, which has uniform density, is traversable, i.e., an object initially at rest will reachits destination through the gravity tunnel in both directions. Moreover, the time taken to fall isalways constant. These facts are no longer true if rotation is allowed. The aim of this note is toderive the necessary and sufficient condition for traversability of straight gravity tunnels througha rotating physical body with spherically symmetric gravitational field. Fall-through times areexpressed in a closed form for linear and constant gravitational fields. In conclusion, these modelsare compared to numerically obtained data using the internal structure of the Earth. a r X i v : . [ phy s i c s . pop - ph ] J a n . INTRODUCTION The concept of the gravity tunnel refers to the travel along frictionless subterranean pas-sages from place A to place B which are positioned on a surface of a massive physical body,e.g., the Earth, with no expenditure of energy for locomotion. Although the correspond-ing vehicle called the gravity train is probably in the domain of science fiction only due toobvious engineering problems, this thought experiment is still an active field of research,which also paves the way for undergraduate physics. A classical result by Cooper assertsthat a gravity train has a simple harmonic motion and always needs a constant amount oftime (roughly 42 minutes) to fall through a chord path (a straight gravity tunnel) betweenany two different positions on a non-rotating Earth, which has uniform density, i.e., lin-ear gravitational field. Recently, several authors have discussed other planet’s interiors andgravitational fields , e.g., constant gravity, the gravitational strength predicted by the Pre-liminary Reference Earth Model (PREM) and polytropes, as well as gravity tunnels withrelativistic and friction effects. The brachistochrone path, i.e., a path which minimizesthe time between any two points, is also studied in some of these papers. Some authorsrefer as gravity tunnels only object’s free fall trajectories since then the gravity train isno longer subject to frictional and inertial forces. Results in this direction are known forrotating homogeneous spheres , rotating homogeneous flattened spheroids and rotatingPREM Earth . However, in this note only straight gravity tunnels are considered. Somereferences to the history of gravity tunnels can be found in Selmke’s note .The most common example of the gravity tunnel is through the center of a non-rotatingEarth. Taking angular frequency ω ⊕ = (cid:112) g ⊕ /R ⊕ where g ⊕ = 9 .
807 ms − is the gravitationalfield strength at the surface of the Earth with the radius R ⊕ = 6 . × m, it followsby simple kinematics that the half-period of oscillation in the linear gravitational field is πω − ⊕ ≈ .
19 min while in the constant gravitational field it is 2 √ ω − ⊕ ≈
38 min. Notethat Cooper’s result about the constant fall-through time T is not true in the constantgravitational field. Moreover, T only depends on the distance of the gravity tunnel from thecenter of the Earth and 2 √ ω − ⊕ ≤ T < πω − ⊕ .Surprisingly, none of the above authors considered the rotating case for straight gravitytunnels. In this note, we would like to emphasize that rotation and the spherically symmetricgravitational field imply traversability of straight gravity tunnels. A traversable gravity2unnel means that the gravity train which initially rests at A will reach its destination B and vice versa. It turns out that there are two necessary and sufficient conditions: the firstone asserts that the gravity energy is large enough while the second one asserts that theabsolute value of A ’s latitude is the same as the absolute value of B ’s latitude. The latteris more important because the first condition is, under reasonable circumstances, alwayssatisfied by constant and linear gravitational fields as well as by the PREM. The gravitytunnels which go through the center of a massive body or are perpendicular to the rotationalaxis are rather specific.Although Simoson addresses this problem by providing several examples in linear grav-itational fields, he does not state this general condition. In the next section we deriveSimoson’s equation of motion, and in Sec. III the general condition. In Sec. IV exact formu-las for fall-through times in linear and constant gravitational fields are provided and theseresults are then compared to the PREM. II. THE EQUATION OF MOTION
We follow Simoson’s paper although our derivation of the equation of motion differsfrom Simoson’s in the sense that we use a non-inertial rotating reference frame instead ofapplying rotational transformation to inertial one. The reader is advised to consult Fig. 1. FIG. 1. The gravity train T , which moves along the straight tunnel from A to B . Assume that our physical body is a rotating ball with radius R and rotational period Q .Take an orthogonal coordinate system ( x, y, z ) ∈ R with the origin O at the center of theball such that the z -axis is the rotational axis. This gives a rise to body’s inertial frame3 = { O ; i , j , k } . Let x = R cos ϕ cos λ , y = R cos ϕ sin λ and z = R sin ϕ be a parametrisationof the surface of this ball with latitude ϕ ∈ [ − π/ , π/
2] and longitude λ ∈ [ − π, π ).Let A and B be two distinct points on the surface of this ball and let A and B becorresponding position vectors. Identify the point A with spherical coordinates ( ϕ , λ ) and B with ( ϕ, λ ). Take the vector u = ( A − B ) / | A − B | and let c = | A + B | be the distancebetween O and M , the midpoint of the segment AB . We have c = R √ (cid:112) ϕ sin ϕ + cos ϕ cos ϕ cos ( λ − λ ) , u = − (cid:112) − ( c/R ) (cos ϕ cos λ − cos ϕ cos λ , cos ϕ sin λ − cos ϕ sin λ , sin ϕ − sin ϕ ) . Points A and B are antipodal points if and only if c = 0. If c (cid:54) = 0, then we take the vector w = ( A + B ) / | A + B | which is w = R c (cos ϕ cos λ + cos ϕ cos λ , cos ϕ sin λ + cos ϕ sin λ , sin ϕ + sin ϕ ) . If c = 0, then we take w = − (sin ϕ cos λ , sin ϕ sin λ , − cos ϕ ) if ϕ / ∈ {− π/ , π/ } , and w = (1 , ,
0) otherwise. Observe that vectors u and w are always the unit vectors and u isparallel to the line AB while w is perpendicular to it. We can take F = { O ; u , w , u × w } as a non-inertial frame which rotates with the angular velocity Ω = (2 π/Q ) k relative to I .Then one can describe the position of the gravity train T in the frame F , which moves alongthe straight tunnel from A to B as a function P ( t ) = c w + s ( t ) u dependent on time t suchthat P (0) = A , i.e., s (0) = | A − B | . By the acceleration transformation formula we have¨ P = − f ( | P | ) | P | P + ( Ω × P ) × Ω + 2 ˙ P × Ω + P × ˙ Ω (1)where f ( r ), 0 ≤ r ≤ R is a spherically symmetric gravitational field of the physical body.The Euler acceleration P × ˙ Ω is obviously zero. After multiplying both sides of Eq. (1) by u , it is easy to see that 2 (cid:16) ˙ P × Ω (cid:17) u = 0 and(( Ω × P ) × Ω ) u = | Ω | Pu − ( ΩP ) ( Ωu ) = (cid:18) πQ (cid:19) (cid:0) s ( t ) (cid:0) − u (cid:1) − cu w (cid:1) . Therefore, the only fictitious acceleration which plays a role here is the centrifugal acceler-ation. Thus the function s ( t ) must satisfy the equation¨ s = (cid:32) − f (cid:0) √ c + s (cid:1) √ c + s + (cid:18) πQ (cid:19) (cid:0) − u (cid:1)(cid:33) s − c (cid:18) πQ (cid:19) u w . (2)In the next section we will write Eq. (2) in a dimensionless form.4 II. THE MAIN RESULT
In order to write Eq. (2) in a dimensionless form with coefficients expressed in sphericalcoordinates, take s ( t ) = Rχ ( τ ), t = Q (2 π ) − τ , F ( χ ) = Q R − (2 π ) − f ( Rχ ) and ¯ c = cR − .Observe that F ( χ ) is a positive function defined on the interval [0 , u and w from the previous section, Eq. (2) becomesd χ d τ = − F (cid:16)(cid:112) χ + ¯ c (cid:17)(cid:112) χ + ¯ c + a χ + b (3)with real numbers a = 1 − (sin ϕ − sin ϕ ) − ¯ c ) , b = sin ( ϕ + ϕ ) sin ( ϕ − ϕ )4 √ − ¯ c . Formula for b is correct also in the case c = 0 since then ϕ = − ϕ and b is zero, as it mustbe due to the last term in Eq. (2).According to the definitions, we have 0 ≤ ¯ c < ≤ a ≤
1. Define ¯ g = ω Q (2 π ) − where ω = (cid:112) g/R and g = f ( R ) is the gravitational field strength at the surface of the body.For the Earth there is ω ⊕ = 1 . × − s − and Q ⊕ = 8 . × s, therefore ¯ g ⊕ = 289 . F are the constant gravitational field F con = ¯ g and the linear gravitational field F lin ( χ ) = ¯ gχ . Here F ⊕ ( χ ) denotes the gravitational fieldpredicted by the PREM.Assume that the gravity train is at rest at A . It is possible to integrate Eq. (3) to obtain (cid:18) d χ d τ (cid:19) = (cid:90) √ χ +¯ c F ( u )d u − a (cid:0) − χ − ¯ c (cid:1) + 2 b (cid:16) χ − √ − ¯ c (cid:17) . (4)The gravity tunnel is traversable if and only if the right-hand side of Eq. (4) is, as a functionof χ , positive for −√ − ¯ c < χ < √ − ¯ c and equals zero for χ = −√ − ¯ c . While thefirst condition ensures that the gravity train reaches B , the second condition means that italso stops there. The latter condition which is equivalent to b = 0 or | ϕ | = | ϕ | is crucial forthe gravity train to be able to move back to A . We thus have two necessary and sufficientconditions for traversability:1. For 0 ≤ χ < √ − ¯ c it follows (cid:90) √ χ +¯ c F ( u )d u > a (cid:0) − χ − ¯ c (cid:1) ;5. | ϕ | = | ϕ | .The first condition is always satisfied in the case of constant and linear gravitational fieldsuntil ¯ g >
1. This is taken for granted in what follows. The PREM model also satisfiesthis condition since F ⊕ ( χ ) is always greater than F lin ( χ ). Therefore, the second conditionis crucial. For the sake of simplicity, we say that the gravity tunnel is horizontal if ϕ = ϕ and vertical if ϕ = − ϕ . Throughout this note, these two types of gravity tunnels are ofinterest to us since we are only interested in traversable gravity tunnels.From Eq. (4) the general expression for the fall-through time is deduced: T = 2 √ ¯ gω (cid:90) c w (cid:18)(cid:0) w − ¯ c (cid:1) (cid:18)(cid:90) w F ( u )d u + a (cid:0) w − (cid:1)(cid:19)(cid:19) − d w. (5)This integral depends on ϕ and ¯ c because ϕ = ± ϕ and a depends only on ϕ and ¯ c . Ifby θ the angle between vectors B (or A ) and w is denoted, then ¯ c = cos θ . Therefore, ¯ c has a similar role as θ in Klotz’s paper. For the horizontal gravity tunnel we have a = 1and T really depends only on ¯ c ∈ [sin | ϕ | , a = 1 − (1 − ¯ c ) − sin ϕ and ϕ must be taken into account where ¯ c ∈ [0 , cos ϕ ]. Of course, T is an even function in variable ϕ . IV. FALL-THROUGH TIMES
In this section, exact formulas for fall-through times in linear and constant gravitationalfields are provided. In general, these equations depend on variables a and ¯ c and on pa-rameters ¯ g and ω . Observe that the limiting case ¯ g → ∞ corresponds to the non-rotatingscenario. This means that we must obtain all non-rotating formulas after taking this limitin equations below. At the end, we compare both models with the PREM. A. Linear gravitational field
If we take F ( χ ) = F lin ( χ ), then Eq. (5) gives T = πω (cid:115) − a/ ¯ g .
6n this case the fall-through time is constant for horizontal gravity tunnels only but changesa little if ¯ g is sufficiently large. For the Earth, this value is 42 .
26 min, slightly more thanfor a non-rotating Earth.
B. Constant gravitational field
If we take F = F con , then it is possible to find an equation for T in terms of completeelliptic integrals. From Eq. (5) it follows T = √ ¯ gω (cid:90) c w d w (cid:112) ( a ( w + 1) − g )( w − w − ¯ c )( w + ¯ c ) . In order to write this integral in a closed form, we need to consider three separate cases: a > c > a > c = 0, and a = 0. Accordingly, we have:1. T = 4¯ cω (cid:112) (1 + ¯ c )(2 − a (1 + ¯ c ) / ¯ g ) (cid:32) Π (cid:32) − ¯ c c , (cid:115) (1 − ¯ c )(2 − a (1 − ¯ c ) / ¯ g )(1 + ¯ c )(2 − a (1 + ¯ c ) / ¯ g ) (cid:33) − K (cid:32)(cid:115) (1 − ¯ c )(2 − a (1 − ¯ c ) / ¯ g )(1 + ¯ c )(2 − a (1 + ¯ c ) / ¯ g ) (cid:33)(cid:33) ; (6)2. T = 2 ω (cid:114) ¯ ga log ¯ g − a ¯ g − (cid:112) a (2¯ g − a ) ; (7)3. T = 2 √ ω √ c (cid:32) (1 + ¯ c ) E (cid:32)(cid:114) − ¯ c c (cid:33) − ¯ c K (cid:32)(cid:114) − ¯ c c (cid:33)(cid:33) , (8)respectively. Here K , E and Π are the first, second and third complete elliptic integrals .Eqs. (7) and (8) can be obtained from Eq. (6) after taking limits ¯ c → a → g → ∞ and thus also presentsthe non-rotating fall-through time. This equation is the same as Klotz’s Eq. A10 exceptthat in his equation a minus must be used between elliptic integrals, the same remark madealso by Isermann who obtained analytic expressions for fall-through times in the case of anon-rotating Earth with piecewise linear approximation of the PREM.7 . The PREM model Klotz used the PREM to get fall-through times depending on θ . Interestingly, thePREM results are much closer to the constant gravitational field model than the lineargravitational field model although the latter is more physically realistic concerning gravityaround the origin. Our results for a rotating Earth are, unsurprisingly, essentially the same,see Fig. 2 and Fig. 3. FIG. 2. Fall-through time T (in minutes) as a function of distance of the horizontal gravity tunnelfrom the Earth’s center per the Earth’s radius ¯ c according to the PREM, constant gravity anduniform density models.FIG. 3. Fall-through time T (in minutes) as a function of distance of the vertical gravity tunnelfrom the Earth’s center per the Earth’s radius ¯ c and departure latitude ϕ (in radians) accordingto the PREM (the blue surface) and constant gravity (the red surface) models. . CONCLUSION The point of this note is that we must have traversable gravity tunnels in order to be ableto speak about fall-through times. While this is irrelevant for non-rotating physical bodies,it is crucial in other cases. In this note, we have investigated the necessary conditions forstraight gravity tunnels to be traversable. One of them greatly narrows the choice of allowedpositions of tunnel’s endpoints on a massive body; the absolute values of endpoints’ latitudesmust be the same. We compare the theoretically obtained fall-through times in linear andconstant gravitational fields for the Earth with numerics of the PREM. The results are closeto the non-rotating times because the Earth rotates relatively slowly. If we want the rotationto have a great impact, then ¯ g must be close to 1. This means that the Earth should rotateclose to the period 1 . ACKNOWLEDGMENTS
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