Undergrad Ball free fall 100m at equator, where is the landing spot?

[Vanadium 50]

If you can't be accurate, at least be precise. 0.0338777555225012 implies the Earth taking 86400 seconds to spin. In fact, it;s 86,164.1.

 

[Baluncore]

Thanks for the commentary.

I should have used an average sidereal day of 86164.0905 SI seconds. But with a warming climate, I expect that will be increasing to about 86,165.0

Finding a disused equatorial vacuum-well, for an air resistance free experiment to verify the numbers, is going to be somewhat difficult.

I would like to drop the ball, at the same time as hanging a chain next to a plumb line, to see the differences.

 

[jbriggs444]

If you can't be accurate, at least be price. 0.0338777555225012 implies the Earth taking 86400 seconds to spin. In fact, it;s 86,164.1.

Let me check that calculation again. I believe that I used 86,164.1.

my $pi = 3.141592653589;
my $omega = 2 * $pi / 86164.1;
my $r_e = 6371000;
my $centripetal_acceleration = $omega ** 2 * $r_e;
print "Centripetal acceleration = $centripetal_acceleration\n";
Centripetal acceleration = 0.0338777555225012

Edit: Wikipedia gives a better radius figure of 6,378,137.0 meters.The answer I got when I googled for "equatorial radius of the earth" turns out to have been the average radius, not the equatorial radius.

Centripetal acceleration = 0.0339157064785778

 

[DaveC426913]

Y'all fail for not following directions.

There is no need to give an answer accurate to the nanometer.

 

[Trikenstein]

Hello gentlemen,

Was busy since the last 2 days. I am quite intimidated by the solutions you suggested. Need sometime to digest. Here is the answer of the submitter mentioned in the original post.

The submitter introduced by a short visual proof that dead center cannot be a solution. Because a Foucault_pendulum would sit still without swinging if there is no lateral displacement. There is one of such pendulum in the museum of science in Washington DC.

Here is the submitted solution. The base concept is just the conservation of angular momentum. The submitter did mention the Coriolis effects.

Sorry I don't know how to write math symbols. If there is any flaw in the reasoning, I would appreciate your patience to clarify.

// The duration of the free fall, from mvt equation with acceleration:
displacement H = 1/2 * acceleration * t^2 + V0 * t + d0

// Taken referential at the top center with initial V0 = 0 and d0 = 0
Duration free fall T = sqrt(2H/acceleration) = 4.5 seconds

Angular speed of the Earth is ω = 2*Pi/24/60/60 = 7.27 e-5 radian/second

// The linear speed of the top/bottom of the well:
Vt = Rt * ω  (Rt = radius from Earth to the top well in meter)
Vb = Rb * ω  (Rb = radius from Earth to the bottom well)

// The displacement of the those points during the freefall duration
Dt = Vt * T = Vt * 4.5
Db = Vb * T = Vb * 4.5

// Difference in linear displacement
Dt - Db = (Vt - Vb) * T = (Rt - Rb) * ω * T

// Rt - Rb = 100m = height of the well
Dt - Db = 100 * ω * T = 100 * 7.27 e-5 * 4.5 = 0.032 meter = 32mm

// Why off-center on the East side?
Because Earth rotation is West to East.

 

[Trikenstein]

As a second approximation, we must realize that during the fall, the direction of the acceleration of gravity is changing. We can decompose the resulting acceleration into a vertical component and a horizontal component

Can you please elaborate as why the acceleration of gravity should be changing during the free fall?

I suppose you gave the explanation. Would appreciate if you can make a drawing to better explain to the un-initiated.

Yes. Intuitively and simplistically, one could think of the ball falling through the interval, affected by a force of gravity that remains always in the original "vertical" direction. Or of the ball falling through the interval affected by a force of gravity that remains always in the final "vertical" direction.

 

[Baluncore]

The submitter introduced by a short visual proof that dead center cannot be a solution. Because a Foucault_pendulum would sit still without swinging if there is no lateral displacement.

All Foucault pendulums must be pushed to start them swinging. Foucault pendulums on display are started again each day, or have an electric circuit that maintains the swing. If you start a Foucault pendulum swinging at the equator, it continues to swing in the initial plane without rotation.

A surveyor's plumb line is defined as vertical. Once the plumb line is stationary, it will not swing, the plumb line string will mark the central vertical axis of a well. If you stop a Foucault pendulum swinging, it becomes a plumb line.

The code you posted is the first approximation. It overestimates the Easterly impact point by about 10 mm.

Imagine the ball was not subjected to gravity. It would continue to fly along the tangent to the top of the well. The ball would rise and move further away from the Earth. After a while, the Earth would be left far behind the ball. The direction to the centre of the Earth is no longer perpendicular to the ball's original line of flight. It is that change in direction that causes the significant second term, which reduces the distance from the central axis of the well.

 

[DaveC426913]

Can you please elaborate as why the acceleration of gravity should be changing during the free fall?

He didn't say the acceleration of gravity is changing; he said the direction of acceleration of gravity is changing.

This happens to any object that is moving around the Earth - whether orbiting it or just via inertia (as is the case with the dropped object in the well).

 

[jbriggs444]

Can you please elaborate as why the acceleration of gravity should be changing during the free fall?

I suppose you gave the explanation. Would appreciate if you can make a drawing to better explain to the un-initiated.

We should start by picking a reference frame that we will work in. We will use a non-rotating inertial frame. [In a rotating frame fixed to the Earth the acceleration of gravity would be fixed in direction. So we must be talking about a non-rotating frame instead]

We picture a large circular Earth. It is rotating clockwise (from our point of view). So we must be looking at it from above the south pole.

We have a an object (small bold-faced circle) that is rotating along with the the earth, staying above a fixed point on the ground. It drops and lands at the position indicated by the small non-bold circle.

We are not, for this portion of the analysis, concerned about whether or not the falling ball keeps up with a point on the Earth's surface, moves ahead of that point or falls behind. We are just trying to figure out what effect, if any, the Earth's gravity might have on the object's horizontal velocity.

The initial acceleration of gravity is vertically downward according to our non-rotating drawing. The final acceleration of gravity is diagonally a little bit left of straight vertically downward according to our non-rotating drawing.

If we use the fixed non-rotating frame of reference we said that we were going to use, this means that the acceleration of the falling object is changing direction as the fall proceeds. We should not expect the falling object's horizontal velocity as measured against this non-rotating frame to remain unchanged.

The claim that I made and supported with calculations is that the resulting effect is in the leftward direction and has a magnitude of 1/3 of the deflection that is calculated with the simplistic first approximation. So the true deflection is only about 2/3 of the simplistic result.

 

[Trikenstein]

It now makes all senses. Thanks gentlemen and maybe gentleladies for your expert explanation. In particular

- @DaveC426913 for rectifying my misundestanding about the direction change of the acceleration
- @Baluncore for the extra explanation about the Foucault pendulum.
- @jbriggs444 for the clear drawing and explanation

Seriously, I wish I could buy you all a beer.

 

[Trikenstein]

Gentlemen, the challenge didn't end there. In our company chat channel, some respondents found Answer2 (32mm East) not intuitive. Then someone upped the ante with a variation of the challenge:

CHALLENGE 2: If the well is a tunnel traversing the Earth. What is the minimum diameter to guarantee that the ball would fall through to the other opening without touching the wall:

2.1) When the tunnel is perpendicular to the axis of rotation of the Earth
2.2) When the axis of the tunnel coincides exactly with that of the Earth
2.3) When the axis of the tunnel is parallel to but distant from Earth axis

Bonus: describe the dynamic of the trajectory. Stable, unstable, shape, etc.

The answers we got were more guessing than rigorous. Summarized here for you to laugh a bit. Before enlighten us with correct answers.

- Attempted answer 2.1: Following the initial mistake of neglecting the change of direction of the acceleration. The value was: tunnel radius = 525km, diameter 1050km. 525km = 1140 secs (freefall to Earth center) * 460 m/s (linear speed at surface)

- Attempted answer 2.2: The ball will fall straight through. Trajectory inline with the axis. Then would swing back and forth on that line for eternity. If the atmosphere or cosmic dust are neglected.

- Attempted answer 2.3: the trajectory will be pulled a bit toward the center. Because there are more mass distributed at the center compared to farway from center. The ball will still travel back and forth in the tunnel for eternity. But with an unstable trajectory. ie, possibly never takes the same path.

 

[vanhees71]

If you take into account both the Coriolis and the (residual) centrifugal force the free-fall solution for motion close to the Earth's surface, where for an observer at rest on the surface of the Earth the effective gravitational force is described by $-m g \vec{e}_3$, you get when letting the body fall from a height $h$ and initially at rest
$$\vec{x}_1(t)=\frac{(4 h+g t^2)(\omega t)^2}{8} \cos \vartheta \sin \vartheta +\mathcal{O}(\omega^3),$$
$$\vec{x}_2(t)=\frac{g \omega t^3}{3} \sin \vartheta +\mathcal{O}(\omega^3),$$
$$\vec{x}_3(t)=h-\frac{g}{2}t^2 + \frac{\omega^2 t^2}{8}(4h+g t^2) \sin^2 \vartheta + \mathcal{O}(\omega^3).$$
Here $\vec{e}_1$ points to south and $\vec{e}_2$ to east and $\vec{e}_3$ perpendicular to the Earth's surface; $\vartheta=\pi/2-\lambda$ is the "co-latitude", i.e., $\vartheta=\pi/2$ at the equator. Further $\omega=2 \pi/1 \; \text{d}$.

Note that the deflection to the south of order $\omega^2$ differs from the usual textbook results, because the centrifugal force has been taken into account in this calculation. The deflection to the east is the same, because it's of order $\mathcal{O}(\omega)$ with the contribution of order $\mathcal{O}(\omega^2)$ vanishing.

For the calculation, see Sect. 2.9.3 of (in German)

https://itp.uni-frankfurt.de/~hees/publ/theo1-l3.pdf

 

[jbriggs444]

2.1) When the tunnel is perpendicular to the axis of rotation of the Earth

What model are we expected to use for the density of the earth? The wikipedia entry suggests three possibilities for the portion of the graph which is sub-surface. The constant density model is definitely the most mathematically tractable. However, it is also the least accurate.

With the constant density model, one nifty result is that the time for the traversal of a straight frictionless tunnel will be the same as for a circular orbit. 90 minutes, give or take.

This holds regardless of whether the tunnel goes through the center of the Earth or is just a short perfectly straight tunnel crossing under the English channel. The shallowness of the angle exactly makes up for the shortness of the tunnel.

 

[jbriggs444]

Attempted answer 2.1: Following the initial mistake of neglecting the change of direction of the acceleration. The value was: tunnel radius = 525km, diameter 1050km. 525km = 1140 secs (freefall to Earth center) * 460 m/s (linear speed at surface)

This attempt makes the mistaken assumption that the acceleration of gravity is uniform at 9.8 m/s throughout the interior of the earth.

$\sqrt{\frac{2g}{h}} \approx 1133 \text{ seconds}$

There is some amazing reading to be done about the Binet equation. It is somewhat tough slogging if one is just trying to skim enough to make a post to Physics Forums.

Another article cuts to the chase and invokes Bertrands's Theorem...

Bertrand’s Theorem states that the linear oscillator, and the inverse-square law (Kepler problem), are the only two-body central forces that have single-valued, stable, closed orbits of the coupled radial and angular motion.

The article goes on to explore the case of an attractive central force that is directly proportional to displacement (a so-called "harmonic force") and gives a solution in polar coordinates.

The resulting closed orbit has an ellipitical shape. However, instead of the center of the Earth being at one of the foci of the ellipse as is the case for an inverse square force, the center of the Earth will be at the center of the ellipse instead.

I have not used the equation in the article to derive the orbital half-period which would be relevant to the rotated position of the Earth when a free-falling object first returns to the surface 180 degrees away from where it originally fell.

 

[vanhees71]

Sorry, I didn't get that it's a long tunnel through the Earth. Then my solution doesn't hold of course.

 

[jbriggs444]

Attempted answer 2.3: the trajectory will be pulled a bit toward the center. Because there are more mass distributed at the center compared to farway from center. The ball will still travel back and forth in the tunnel for eternity. But with an unstable trajectory. ie, possibly never takes the same path.

As has already been pointed out, a bound orbit under an attractive linear central force will be a stable ellipse. However, it will not be geosynchronous.

From the point of view of a rotating earth, the orbit will appear to precess, always emerging at the same line of latitude as it started, always anti-spinward by a fixed increment of longitude from the previous point of emergence.

If the orbital period and the Earth's sidereal rotation period are not in a rational ratio, this means that over infinite time there will be a dense, countable mesh of emergence points all around the lines of latitude where the object continues to emerge.

This is not to say that every point on those lines of latitude will be an emergence point. There are uncountably many such points. So "most" of them will not be emergence points. But for any point on the line(s) and any tolerance you choose, no matter how tight, there will be an emergence point within that tolerance of the selected point. That is what "dense" means in the mathematical vernacular.