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**Play the video below** to watch a rock released into orbit over the Dirty Devil river (in Utah, USA), from the highway 95 bridge.

QuickTime offers the best 'scrubbing' control (the ability to move forward/backwards smoothly, frame by frame), so if your browser has a QuickTime plug-in installed, this web page will try to use it. Otherwise, it'll try using HTML5 with a controller I wrote that permits frame control. Finally, it'll resort to YouTube. If, despite all that, none of it works for you, here are direct links to the video files: rockDrop_f30.mov or YouTube.

Fig 1. Rock drop into the Dirty Devil river, Utah, from Highway 95. 120 frames/sec (1/4 real time).

Soon after that rock drop, over dinner that evening, a question arose: **How much would Earth's rotation cause the rock to deflect from a strictly vertical drop?**

Guesses varied from millimeters to centimeters.

After dinner, while relaxing on a warm slab of slickrock, watching the sun set, we tried making an estimate of the deflection from vertical. The bridge is higher than the river, so it must need to travel further and thus faster to make one rotation of Earth. As the rock drops, it has and maintains the higher 'horizontal' velocity of the bridge, and therefore ought to 'overshoot' the spot directly below. The Earth rotates eastward (sun rises in the east), so the rock ought to land east of vertical.

The rock was in free-fall for about 4 seconds, so using high school physics, the height was probably about (1/2)(gravityAcceleration)(time)^{2} = (5)(4)^{2} = 80 meters. So the difference in radial distance (bridge elevation minus river) is about 80 meters. We want the difference in distance travelled by the bridge versus the river along their daily circlings of Earth, which would be 2π (bridge - river) = 2π (80) ~= 500 meters / day. The rock drop took 4 seconds, so the deflection would be the difference in bridge/river travel during that 4 seconds, ie., 4 / (24 * 60 * 60) * 500 m ~= 7/3 cm ~= 2.3 cm. That'd be at the equator. At the latitude of the bridge (~38°), we need to multiple by cos(38°), which is somewhere between cos(45°) ~= 0.7 and cos(30°) ~= 0.86, say 0.8, which gives 2.3 *0.8 ~= 1.8 cm deflection. So, for an 80 m drop, at latitude 38°, we guessed the deflection would be about 2 cm, eastward.

Objections were raised; the above assumes gravity is constant during the fall, both in direction and magnitude, but we know it is neither; as the rock travels eastward, it would be increasingly pulled (accelerated) back toward Earth's center, and as it fell, gravity would strengthen. Both of those would reduce the deflection from vertical. In effect, when the rock was released, it'd be going into orbit (!), albeit brief. But making an estimate of the deflection using that approach seemed like more than we'd be able to do in our heads, so with the sun setting (ie., earth rotating), we left it at that.

When we got home, I looked into it further. I expected it'd be a fun trip down memory lane (high school physics), but more brain cells were exercised and stretched than anticipated. With the help of online course notes and web pages (referenced below), below is some of what I found out. I focus on what I thought was interesting and relevant to the rock dropping questions above.

(This page uses javascript to insert results of many of the calculations, so to see values or calculations, you can look at the javascript.)

Fig 1. Bridge and falling rock as viewed from a non-rotating reference. Stop the animation if it gets tiresome.

Figure 1 shows the view from a non-rotating inertial reference frame. From that viewpoint, the bridge and the rock before its release are moving eastward with Earth's rotation.

When the rock is released, it is no longer being held to the Earth and, under the influence of Earth's gravity, begins orbiting Earth. The bridge and river continue rotating with Earth's surface. The rock's orbit soon intersects with the Earth's surface (the river, in this case) and, alas, is stuck in the mud, again locked with the Earth's rotating reference frame, which, to the external observer, are all moving eastward along a circle of latitude.

Here's some information about the bridge and Earth:

Parameter | Value | Source |
---|---|---|

Latitude of bridge | ° N | Google Earth |

Elevation of bridge | meters | Google Earth |

Height of bridge | meters | Estimated, using a string^{*} |

Elevation of river | meters | Calculated: Bridge elev minus height |

Equiv radius of Earth (sea level)^{**} | km | Wikipedia |

Mass of Earth | kg | Wikipedia |

Rotation period of Earth (sidereal day) | seconds | Wikipedia |

Gravitational constant | m^{3}/kg s^{2}
| Wikipedia |

* A rock was lowered to the river on a twisted nylon kite string. The length of string measured 70.21 meters without tension; with the tension of supporting a 350g weight, the string stretches ~0.7% for a length of 70.68 meters.

** Hides Earth's oblateness. If Earth were a sphere of equivalent volume, this would be its approximate radius.

The Earth completes one rotation (360° = 2π radians) in a day, which with respect to the stars is seconds. Thus Earth's angular velocity ω is:

ω = 2π / = radians/s

Due to that rotation of Earth, the bridge and everything co-rotating (most notably, the atmosphere) are moving eastward with a tangential velocity, at the bridge's latitude of ~38°, given by:

v_{tangential} = ω [r_{bridge} cos(θ_{bridge})] = m/s = km/h

Fig 2. As viewed from an inertial frame. The bridge travels the arc A to A', which is slightly longer than the arc B to B' travelled by the river, left-to-right as Earth rotates. Lines AB and A'B' go to Earth's center; they are 'vertical'.

(This is a repeat of what was discussed near the top of this web page, but with more accuracy. It turns out to be a simplistic approach, with shortcomings.)

The bridge is further from Earth's center than the river and therefore has a slightly higher tangential velocity around Earth's axis than the river. At the bridge, the rock shares that higher tangential velocity, and during the drop, it retains it (Newton's first law). This velocity difference suggests that when the rock hits the river, it will have travelled further east than the river (figure 2), and therefore not land directly below the point of release; it will overshoot.

Difference in latitudinal arc distance travelled by bridge & river |
= ω [(r_{bridge} - r_{river}) cos(latitude)] t_{travel} = cm |

Note: Methods discussed below yield more accurate results

Objections: While falling and travelling eastward, the direction of the gravity experienced by the rock is changing as the rock moves eastward; it has an increasing westward component. This suggests that the distance above overstates the rock's deflection. Let's try another approach, in the next section.

If we can calculate the rock's orbital path, then we can determine the deflection from vertical by finding its orbital position at the time of its intersection with the river and comparing it with vertical.

Fig 3. Rock's orbit is the narrow dashed loop. Horizontal circles are latitude lines (38° and the equator). Red line is a great circle tangential to both the 38° latitude circle and to the rock's orbital path. Created using GeoGrebra.

With help from MIT/Edx course notes (Dourmashkin, 2013) on celestial mechanics and orbits, the following parameters were obtained for the rock's orbit:

Parameter | Value |
---|---|

Apogee | km |

Perigee | km |

Period | minutes |

Minor axis | km |

Major axis | km |

Eccentricity | (long and narrow) |

So the rock is in an orbit that is about 6400 km long and 400 km wide, taking about 30 minutes to complete one revolution. Figure 3 shows th rock's orbit, the narrow dashed loop, in true scale, if all Earth's mass were packed into something smaller than the orbit's closest approach (perigee). The apogee (greatest distance from Earth's center) is at the elevation of the bridge, and the perigee is about 7 km on the other side of Earth's center. Like a comet diving toward and looping the Sun, the rock would like to loop around Earth's mass. Of course it only gets as far as its intersection with the river.

Fig 4. As viewed from an inertial frame. Angle of Earth rotation (φ), and angle of orbit (θ).

The plane of the orbit is nearly the same as that of a great circle tangential to the latitude circle and passing through Earth's center, but not identical because of Earth's oblateness.

Based on the orbit equations, the rock should reach the elevation of the river seconds after release. Referring to figure 4, the impact is slightly further than the latitudinal arc between B (where the river was at the time of release) and B' (where the river is at the time of impact), but hasn't travelled as far as the bridge (arc A to A' ~= B to C).

Deflection from vertical suggested by orbital path = cm

The direction of this deflection is along the orbital path. As illustrated in figure 3, the orbit is tangential to the line of latitude, so most of the deflection is eastward with a tiny fraction toward the equator (south, in this case).

The textbook approximation of the deflection due to rotation is given by:

Deflection = | ω cos(latitude) (8h^{3}/g)^{1/2} |

3 |

Two sources I found discuss how that is derived in a ways that I particularly enjoyed: West (2013) and Persson (2014) (though I believe his figure 4 may be misleading; see Marginalia below, regarding orbits within Earth).

The approximation assumes constant gravity. Using height h = m and the value for g at the bridge (see appendix B):

Deflection = cm

That's pretty close to the presumably-more-accurate (because it has fewer approximations) deflection based upon orbit (above), providing some confidence in both approaches and the calculations.

*Assuming a bridge height of m*, estimates of the rock's deflection from strictly vertical during its fall ('orbit') to the river (in the absence of air friction) are:

Source | Estimate | Comment |
---|---|---|

Difference in tangential velocities | cm East | Overly simplistic model |

Orbit method (intersection with Earth) | cm East (with a bit of South) | (probably most accurate) |

Textbook method | cm East | Makes use of approximations |

Air friction would slow the rock's descent, providing slightly more time for the rock to travel tangentially, suggesting that the actual deflection would be slightly greater (perhaps by a few percentage points) than the amounts predicted above.

Fig A. On a rotating sphere, gravity (equals gravitational minus centripetal acceleration vectors) has an equatorially-directed component. The red line is an equipotential surface for gravity, the tan line (coincident with the sphere's surface) is an equipotential surface for gravitation.

As many can attest from childhood use of playground carousels, on a rotating platform one must hang on (apply a centripetal force) to avoid being thrown off (ie., to avoid travelling in a straight line and thus leaving the rotating platform).

Gravity acceleration is the vector sum of gravitation acceleration minus centripetal acceleration (figure A):

Gravity acceleration = gravitational acceleration - centripetal acceleration

The gravitation vector (orange in fig A) due to attraction between two masses:

F_{gravitational} = |
Gm_{1}m_{2} |

r^{2} |

is not affected by whether the masses attracting each other are rotating (excluding possible relativistic effects: frame dragging).

The centripetal acceleration (purple in fig A) is dependent upon the angular velocity (ω) and distance from the axis of rotation, and thus varies with latitude:

F_{centripetal} = ω^{2}r

where r is the axial distance, eg., distance to Earth center * cos(latitude).

The gravity vector (red in fig A) is the vector sum F_{gravitational} minus F_{centripetal}. For the rotating sphere of figure A, the equipotential gravity surface is not aligned with the surface of the sphere, thus the gravity vector (red) is not perpendicular to the sphere's surface (blue). The gravity vector has a tangentially-oriented component, which would urge objects on the surface to move toward the equator. Thus water, for example, on that sphere's surface would move toward the equator, flattening the sphere -- as happened on Earth. Water (and rocks) at the Earth's equator are approximately 21 km higher (from Earth's center) than at the poles.

The resulting shape is a slightly flattened ball; an oblate ellipsoid (or oblate spheroid).

Fig B. On a rotating ellipsoid, gravity (equals gravitational minus centripetal acceleration vectors) can be perpendicular to the surface. The red line (coincident with the ellipsoid's surface) is an equipotential surface for gravity; the tan lines are equipotential surfaces for gravitation. Based upon figure 2a in Hvoždara and Kohút (2012).

The result is not an infinitesimally-thin spinning disc because there are negative feedbacks: As mass moves outward, rotation slows (via conservation of angular momentum), and as the shape changes, the gravitation vector changes direction. There can be equilibrium, as shown in figure B.

In figure B, orange lines illustrate some gravitational equipotential lines for an ellipsoid (which depend on shape but not whether the body is rotating) (Hvoždara and Kohút, 2012). As before (fig A), the gravitation vector is perpendicular to its equipotential surface. But in this case, the mass surface (blue) is aligned with the equipotential gravity (equals gravitational minus centripetal) surface, and thus objects on the surface feel only a force perpendicular to the surface; there is no force component tangential to the ellipsoidal surface, so a rotating ellipsoid shape can be stable. The gravitation vector points slight toward the equator of center.

For a rotating sphere, the mismatch between the equipotential surfaces of gravitation versus gravity (fig A) would cause it to flatten (if it can flow). Similarly, if a rotating ellipsoid were to slow or stop rotating, the same mismatch but in the opposite direction (fig B) would cause it to become more spherical (if it can flow). Earth's substance can flow, so Earth has an ellipsoidal shape.

Earth is nearly spherical; the equatorial radius is 6378.1 km, the polar radius is 6356.9 km (Wikipedia), giving an eccentricity of 0.0816 (for comparison, a sphere has an eccentricity of 0). A body would have to spin much more rapidly than Earth (or have less gravitation) to be as oblate as in figure B.

The Moon probably also influences Earth's shape; due to the viscosity of water and rocks, there's probably a static component to the Moon-induced oscillations.

As discussed in Appendix A, gravity is the vector sum of gravitation minus centripetal acceleration. Using the equations above, the table below gives a feel for the magnitudes of these vectors at the elevation of the bridge:

At the bridge: | |

Latitude | ° N |

Distance of bridge (elev. m) from Earth's center | m |

|Gravitation acceleration| | m/s^{2} |

|Centripetal acceleration| | m/s^{2} |

|Centripetal acceleration parallel to gravitation vector| | m/s^{2} |

|Centripetal acceleration perpendicular to gravitation vector| | m/s^{2} |

|Gravity parallel to gravitation vector| | m/s^{2} |

|Gravity| | m/s^{2} |

Angle between gravity and gravitation vectors | ° |

So at the bridge where the rock was dropped, there's only a 0.1° difference between the direction of the gravitation and gravity vectors. The direction of the gravity vector determines what is vertical (eg., how a plumb line would hang), and the gravitation vector determines the path of the rock's orbit. When the rock is released, it no longer being accelerated (by connection to the Earth) along a line of latitude; after release, the centripetal acceleration vector is then aligned (but opposite) with the gravitation vector. The Earth is almost spherical, so the gravitation vector points close to Earth's center and the rock's orbit would be close to the plane of a great circle.

Clever and interesting use of dimensional analysis is illustrated at Bohren (2004) by considering influences upon objects in free-fall, including the error introduced by assuming g (gravity) is constant over a fall, estimating the deflection due to Earth's rotation, the effect of relativity, and the effect of air friction. There's a comment on that article (Rojas, 2011), not accepted by the publishing journal but interesting.

What would happen if the rock were able to 'tunnel through Earth'? An internet search on 'hole through center of earth' returns many web pages (eg., this, this, and this). The short answer, ignoring Earth's rotation and friction, is that the object would oscillate from one end of the tunnel to the other, with a period of about 84 minutes (the same period an object would have in an at-surface great circle orbit).

That's because while gravitational force *outside* a sphere is given by the familiar:

F_{gravitational, outside Earth} = |
Gm_{1}m_{2} |

r^{2} |

*within* a sphere of uniform density, the gravitational forces of material outside a given radius cancel out, leaving only the force from the sub-sphere closer to the center. The mass within a sub-sphere of radius r is 4π (Earth density) (Earth radius)^{3} / 3, which when substituted into the gravitational force equation simplifies to:

Fig C. Initial path of a pendulum when released with non-zero tangential velocity. If the dropped rock could pass frictionlessly within Earth, this is the precessing, quasi-elliptical orbit it might follow, in the plane of the great circle in figure 3. Based upon a screen capture from Zabunov (undated).

F_{gravitational, inside Earth} = kr

where 'k' is a constant, which, for motion in one dimension, produces simple harmonic oscillation, like a weight on a linear spring, or a pendulum swinging in a plane.

What about an object possessing initial tangential velocity, such as our dropped rock? In that case, I believe the applicable idealized analog is a pendulum swinging in two dimensions (rather than just in a plane) (a spherical pendulum). Compared with a pendulum swinging in a plane, a pendulum swinging in two dimensions can have a more complex path (Gray, 1918).

In the idealized and imaginary case of a uniform density spherical Earth offering no friction to a rock moving within it, I think the rock's orbit would follow a precessing quasi-elliptical path, such as illustrated roughly to scale in figure C. It would remain within the plane of the great circle tangential to the latitude of the bridge (red circle in figure 3), because, at the moment of release, the great circle and latitude tangent lines are identical and thus there would be no initial velocity component outside of the great circle plane.

My guess for the dimensions of the quasi-ellipse, based upon the tangential kinetic energy of the rock when it is released, is an ellipse centered upon Earth's center with its major axis spanning Earth's diameter, and a minor axis of approximately km — wider than that ( km) of the orbital path found above, where Earth's mass was assumed to be packed inside the rock's orbit, eg., as a point.

Fig D. Gravity vs radius. Earth's radius is about 6371 km, so this is all within Earth. From Preliminary reference Earth model.

Within an inhomogeneous, oblate spheroid like Earth, the path would be more complicated (eg., not planar) than the idealized approximation in figure C.

Figure D shows how the strength of gravity is thought to vary with depth (charting values from this table, which is from Preliminary reference Earth model), based upon seismic data.

Of course once *outside* Earth, starting where the line in fig D ends, the strength of gravity declines proportional to 1/r^{2}.

Bohren, Craig F (2004). "Dimensional analysis, falling bodies, and the fine art of not solving differential equations". *Am. J. Phys.* 72, 534537. http://dx.doi.org/10.1119/1.1574042. Retrieved 2014-Dec-10 from https://www.amherst.edu/system/files/bohren_notsolving_AJP000534_0.pdf.

Dourmashkin, Peter (2013). "Classical Mechanics 8.01 MIT/Edx edition: Chapter 25 Celestial Mechanics". Retrieved 2014-Dec-10 from http://web.mit.edu/8.01t/www/materials/modules/chapter25.pdf.

Gray, A (1918). A Treatise on Gyrostatics and Rotational Motion: Theory and Applications. New York: Dover, 1918.

Hvoždara, Milan and Kohút, Igor (2012). Gravity field due to a homogeneous oblate spheroid: Simple solution form and numerical calculations. Contributions to Geophysics and Geodesy, 41(4), pp. 279-327. Retrieved 2014-Dec-10 from doi:10.2478/v10126-011-0013-0.

Persson, Anders O (2014). "Proving that the Earth rotates by measuring the deflection of objects dropped in a deep mine; The French-German mathematical contest between Pierre Simon de Laplace and Friedrich Gauß 1803". Retrieved 2014-Dec-10 from https://www.bibnum.education.fr/sites/default/files/137-analyse-laplace-en.pdf.

Rojas, Sergio (2011). The math must be right: Comment on "Dimensional analysis, falling bodies, and the fine art of not solving differential equations" by C. F. Bohren [Am. J. Phys. 72, 534-537 (2004)]. Retrieved 2014-Dec-10 from http://arxiv.org/pdf/1102.1120v1.pdf.

Strong, Kimberly (2005). "Introduction to Earth Observations PHY499S: Satellite Orbits". Course notes, section 2. Retrieved 2014-Dec-10 from http://www.atmosp.physics.utoronto.ca/people/strong/phy499/section2_05.pdf.

West, Brian (2013). "Mechanics in Non-Inertial Frames". Lecture notes, chapter 11. Retrieved 2014-Dec-10 from http://legacy.wlu.ca/page.php?grp_id=2590&p=13566.

Zabunov, Svetoslav S (undated). "3D Pendulum Simulation". Accessed 2014-Dec-10.

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