Will a Ball placed close to Sun fall into it?

Will a Ball placed close to Sun fall into it?

Like a ball falls on the earth due to gravity, if suppose we are able to place a ball close enough to the sun. Will the ball fall into the sun due to its gravity or will it be pushed away due to its energy?

Short answer is yes. The only way radiation pressure would be a factor was if the ball was in orbit around the Sun, this would mean that the slight pressure it feels would cause it to drift away slowly (assuming its made of the most heat resistant material on Earth!). But in your question you said "placed" near the sun not in orbit around it, so the strength of the sun's gravity would overcome all other forces and pull it in.

The answer depends on the size and mass of the ball. It also depends on its ability to reflect light (albedo $A$), but let's forget that for a moment.

Pressure vs. gravity

Solar pressure decreases with $R^2$ (the inverse square law). At Earth, which is located at a distance of $1,mathrm{AU}$ from the Sun, we receive an irradiance $S_0 = 1361,mathrm{W},mathrm{m}^{-2}$. Since the momentum of a photon of energy $E$ is $p = E/c$, the pressure at a distance $R$ from the Sun is $$ P = frac{S_0}{c(R/mathrm{AU})^2}. $$ If the ball's radius is $r$, this pressure will exert a force $$ F_gamma = pi r^2 P = frac{pi r^2 S_0}{c (R/mathrm{AU})^2}qquad(mathrm{away,from,the,Sun}). $$ Meanwhile, if the ball's mass is $m$, the gravitational force exerted by the Sun on the ball is $$ F_g = frac{G M_odot m}{R^2},qquad(mathrm{toward,the,Sun}) $$ where $G$ is the gravitational constant and $M_odot$ is the mass of the Sun.

Threshold for falling

The threshold for the ball falling into the Sun is found by equating the two oppositely directed forces: $$ frac{pi r^2 S_0}{c (R/mathrm{AU})^2} = frac{G M_odot m}{R^2}. $$ The first thing to notice is that $R$ cancels out; the reason being that flux density and gravity both follow the inverse square law. Second, re-arranging terms we see that the ball will fall if its mass per area is greater than this threshold (if I have calculated correctly): $$ frac{m}{r^2} gtrsim frac{pi S_0}{c G M_odot } mathrm{AU}^2 simeq 2.4 imes10^{-4},mathrm{g},mathrm{cm}^{-2}. $$ Now you can plug in your favorite numbers. You will find that for most macroscopic objects, such as a football, a rock, and even a sand grain, it will fall. On the other hand, small "balls" such as dust grains and atoms, will tend to be pushed away from the Sun. An example of a macroscopic object that won't fall is a solar sail which seeks to maximize area per mass.

For a given density, say $ ho = 2.5,mathrm{g},mathrm{cm}^{-3}$ which is characteristic of rocky materials, you can also calculate the maximum size before it falls toward the Sun: $$ r_mathrm{max} = frac{3}{4pi ho} ,(2.4 imes10^{-4},mathrm{g},mathrm{cm}^{-2}) sim 0.1mathrm{-}1,mumathrm{m}. $$


The above calculations hold of all the radiation is absorbed by the object. If a fraction of it is reflected, the radiation will transfer more momentum to the object, and for a prefectly reflecting object, the radiation force will be roughly twice the amount above (the exact factor depends on the geometry of the object).

Extinction curves

It should be noted, though, that treating small particles as rigid spheres with a geometric cross section becomes imprecise when their size is comparable to wavelength of the light; rather, their absorption/scattering cross section should be treated quantum mechanically. In practice, extinction curves - i.e. the cross section as a function of the wavelength of the light for a given dust particle size distribution - are measured observationally by comparing the light from unobscured star with similar stars behind dust clouds, and subsequently fitted with various functional forms.

Will a Ball placed close to Sun fall into it? - Astronomy

Before we talk about black holes, we need to make a digression into relativity. This is what Einstein is most famous for.

There are two theories of relativity: Special Relativity and General Relativity. These two theories work together, and revolutionized the way that we view the world, almost as much as removing the Earth from the center of the solar system in the 1600s.

Special Relativity has two postulates: (a postulate is something that you assume is true, to see how things turn out. You know if the postulate is false because it leads to a contradiction later on. But you never know for SURE if your postulates are correct. Usually we invent postulates because we have an intuition for how things should turn out, or because they just make sense.)

  1. No reference frame is special. This means that the laws of physics should hold everywhere in the Universe, no matter what speed you are travelling. So, for example, gravity acts on you whether or not you are in your car (thankfully!), and while the value of gravity may be different in different places, such as on the moon, it still works in the same manner.
  2. The speed of light, c, is a constant in all reference frames. This means that if you are travelling in your car at close to the speed of light, and you turn on the headlights, you will measure the light travelling at c. So will somebody watching you drive by! This is completely counter-intuitive. You expect that when you are travelling at some speed, v, and you throw something at a speed t, it will move away at v+t. But light doesn't work that way. Instead, it appears bluer when emitted in the direction you are travelling, and redder when it's emitted in the opposite direction.

We have tested the time dilation consequence, and verified it experimentally by two methods: First, we synchronized highly accurate atomic clocks, and put one of them on a high-velocity aircraft. We flew it around the world a few times, and brought it back to the ground. It was out of sync. with the clock that stayed on the ground the whole time. More importantly, it was out of sync. by exactly the right amount to agree with calculations. Second, we have observed particles called muons in the lab. These particles don't live very long when we create them here on Earth. However, we also find them streaming through our atmosphere at very high velocities, and surviving from the upper atmosphere, where they are created, all the way to the ground. Because they are travelling so fast, they live long enough for us to detect them from the ground!

General Relativity has only one postulate. General relativity says that gravity is indistinguishable from all other accelerations. So if gravity is exerting a force on you, it acts upon you in the same way as if somebody pushes you with the same amount of force. This is neither intuitive, nor counter-intuitive. It's sort of intuitive, because when you are in an elevator that accelerates upward, you feel a force acting on you that seems a lot like gravity, only stronger. On the other hand, what this implies is that the inertial mass (the property that makes it hard for people to push on you and get you moving) is the same as the gravitational mass (the property that the Earth and the Sun and the Moon use to act on you and on each other). If you think about it for a while, you'll say 'HEY! That's WEIRD! Why would a resistance to pushing and a gravitational attraction work in the same way. ' We have no idea why this is. We just know that we observe it to be true.

This postulate has one particularly bizarre consequence.

Near really heavy objects, time is distorted: Consider Ted and Jackie in the accelerating rocket. Jackie is at the back, and Ted is at the front. They have previously synchronized two flashlights to blink at the same rate. Now, while they are accelerating, they blink the flashlights at each other. While the pulses are traveling, Ted is accelerating away from where Jackie was when she sent the pulse. So the pulse has to travel further to get to him. He thinks her time has slowed down. On the other hand, Jackie is accelerating towards where Ted was when he sent his pulse. The pulse doesn't have to travel so far to be intercepted by Jackie. She thinks Ted's time has sped up.

Now, because gravity acts just like any other acceleration, move the rocket to a massive planet. Ted and Jackie will observe exactly the same phenomenon, even though they are not moving. This is called gravitational time dilation.

Yes. It's weird. But it follows directly from the postulate, whether or not we like it.

Will a Ball placed close to Sun fall into it? - Astronomy

The term fall is part of a traditional way of classifying certain sign placements of planets. A planet is said to be in its dignity when it is in the sign it rules (e.g., Mars in Aries, the Sun in Leo). There are also certain placements said to be especially favorable for a planet that are traditionally termed exaltations (to continue with the same examples, Mars in Capricorn, the Sun in Aries). When a planet is placed in the sign opposite its dignity, it is said to be in its detriment (Mars in Libra, the Sun in Aquarius). A planet is in its fall when it is placed in the sign opposite the sign of its exaltation (Mars in Cancer, the Sun in Libra). For example, because the Moon is exalted in Taurus, it is in its fall when placed in the sign Scorpio as the name implies, this is regarded as an unfortunate placement. A planet in its fall is traditionally regarded as being out of harmony with the sign and consequently weakened (in a position of debility).

For the most part, contemporary astrological research has tended to disconfirm that a planet in its traditional fall is weakened. However, it is sometimes the case that planets in fall have unfortunate effects. In the example cited, the Moon, as the planet of receptivity and sensitivity, is not well placed (especially in a natal chart) in Scorpio, a sign noted for possessiveness, obsessiveness, and intense emotions. There are, nevertheless, certain obvious problems with this tradition. The Sun, for example, is exalted in Aries, the sign opposite Libra. This means that the one person out of 12 in the world born with a Libra sun sign has her or his sun in its fall. This particular placement of the Sun, however, is not normally regarded as being unfortunate, making the traditional ascription appear inapplicable, at least in this case. Generally, all the traditional falls should be taken with a grain of salt when found in a natal chart.

The situation is different in horary astrology, where the classical dignities and falls have a definite bearing on the question being asked. In Vedic astrology, a planet that is placed in the sign of its fall is regarded as being unfavorably placed and weak by virtue of this placement. In fact, in contrast to Western astrology, Vedic astrology has elaborate systems for determining the strength of a planet, even assigning numerical values and ranking the strengths of the traditional planets. Sign placement is only one factor in this system, so that, in the final analysis, even a �llen” planet may end up being a strong planet in the chart.


Did u ever watch "Deep Impact" . Or "Armageddon" . I think u ever watched it. So exclaimed,right . The movie story about the asteroids that will hit earth. Could u imagine when an asteroid hit Earth . It will be a big big big disaster,the "dooms day". Because of this ,the dinosaurs was extincted.

Asteroid(planetoid also minor planet) is rock that orbiting the sun. Asteroid placed in the Asteroid belts,where between Mars and Jupiter. There are many asteroid with different size, beginning from 1 km diameter size until 1000km (Ceres,the biggest) . The asteroids amount in our solar are about 1million. Over 8,000 of these have been individually cataloged and named, and have well-determined orbits. Although it is common to depict the asteroid belt as a dense region, asteroids are actually quite well separated, rarely approaching within 1 million km of one another. (A few asteroids have moons of their own: these are certainly the exception to the rule.) All together, the asteroid belt contains about 0.1% of the mass of the Earth. Earth mass is about 5.98 x 10 24 kg. So it is 5.98 x x 10 21 kg

Asteroid Ida with ts satelitte(Dactyl)

In addition to the asteroids in the belt, some asteroids share Jupiter’s orbit. Asteroids in this special group are called Trojan asteroids, and they orbit about 608 ahead or behind Jupiter. Their orbits are stabilized by the combined gravity of Jupiter and the Sun. Over 150 of these are known the largest is about 300km in size.

Astronomers estimate that the asteroid was created the same time as our solar system . They are byproduct. As theory said that planets formed by the joining of the asteroids. Asteroids on the Asteroid belt is the rest of solar system material when was created. They have stable gravitional in there.

An asteroid approching erth
Finally, some asteroids are ‘‘Earth-crossing’’ and are potential impactors. These asteroids come from three different groups—the Apollo, Aten, and Amor asteroids. Most of these are small, less than 40km across, and so they are difficult to find in the sky. About 500 are known. Most of these will strike the Earth some time over the next 20󈞊 million years. Near-misses are common, and are often unpredicted. In 1990, an asteroid came closer to the Earth than the Moon. The asteroid was previously undiscovered, and was not noticed until after it had safely passed the

Asteroids do not emit visible light, they only reflect it. Astronomers determinethe compositions of asteroids by comparing the spectrum of the light reflected by the asteroid and the spectrum of the Sun. Absorption lines that are present in the asteroid’s spectrum, but not in the solar spectrum, must be due to elements or minerals in the asteroid. Asteroids are classified in three major groups: carbonaceous (C), silicate (S), and metallic (M). Most asteroids are C-type asteroids, with very low albedo and no strong absorption lines. The rest are mainly S type, with an absorption feature due to a silicate mineral, olivine.

The amount of light reflected from an asteroid towards the Earth changes as the asteroid tumbles through space. We can use this information to determine how quickly the asteroids rotate. About 500 asteroids have been studied well enough to determine their rotation periods, which are generally between 3 and 30 hours.
Smaller asteroids have irregular shapes. The shape of small asteroids can be determined from analysis of the amount of radiation received over time (light curve).

The Beautiful Universe Astronomy Quiz

T.A. Rector and Hubble Heritage Team What’s that dark outline of a horse’s head? You can find out — and try for a prize — by taking the Lab’s new astronomy quiz.

We have a new astronomy quiz for you, complete with some of the most gorgeous images in the universe.

This astronomy quiz, like previous ones, was created by Christopher De Pree, the director of the Bradley Observatory at Agnes Scott College. You’ll find out right away which answers are correct and what your score is. Then, no matter how you score, you can come back here and try for a prize by answering the extra-credit questions. The person who submits the best answers to the extra-credit questions will win a copy of “The Complete Idiot’s Guide to Astronomy,” by Professor De Pree and Alan Axelrod.

Click here to take the Beautiful Universe Astronomy Quiz. [UPDATE: You can still take the quiz and get your score, but the prize has been awarded.]

The extra-credit questions are inspired by an exhibit opening today in Atlanta: a metropolis-sized scale model of the solar system. The Metro Atlanta Solar System is centered at the Bradley Observatory, where the Sun is represented by a circular granite plaza about 30 feet wide. From there it’s almost half a mile to the 3-inch model of Earth (at the Decatur Public Library), nearly 12 miles to Uranus (at the Hartsfield-Jackson Atlanta International Airport) and more than 18 miles to Neptune (at Sweetwater Creek State Park in Lithia Springs, Ga.)

Now for the extra-credit questions and your chance at a prize:

1. In the scale used by the Atlanta model of the solar system, a distance of 1 kilometer represents 1 astronomical unit, which is the distance from the Sun to the Earth — about 150 million kilometers. The outermost object in the Atlanta model is Neptune, which is shown in a park about 30 kilometers from the Sun. (If you feel Pluto has been unfairly excluded, feel free to suggest a location for it in the model.) Suppose you wanted to add the nearest star, Proxima Centauri, to the model. It is 4.2 light years from the Sun. How far away would it be from the center of the model, and what would be the nearest known landmark?

2. Suppose you wanted to create a scale model showing not just the solar system but the entire Milky Way galaxy — and maybe even more. Could it be done using the same scale as the Atlanta model? How might it be done using a different scale, or using a different kind of model?

You can submit your answers as a comment here. You can check out more beautiful pictures at the source of the ones in our quiz: From Earth to the Universe, a collection of images that have been on exhibit around the United States. The exhibits are part of the International Year of Astronomy celebrating the 400th anniversary of Galileo’s observations with a telescope.

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4.2 light years is 2207520 light minutes, and an AU is 8.2 minutes. So around 269209 miles away. Proxima Centauri would be on the moon for the atlanta model.

2) There would have to be a ton of landmarks for each of the stars in our galaxy, seeing as how you would need hundreds of billions of landmarks if you wanted to count each star. Also since its 100,000 light years it would take a much smaller scale. Maybe they could just dump a couple of truckloads of sand somewhere and cut the scale by 10000.

1) Since a light year = 6.3239e5 au, then 4.2 light years is about 2.656e6 eu. Using the Atlanta scale, that’s 265600 km. Gee, another space mission! Closest landmark is of course Luna (and I don’t mean Luna Park!)

2) Radius of Milky Way is 60000 light years or 14000 times the distance to Proxima Centauri. Obviously, this space mission will take some time to accomplish. Perhaps instead of using a linear scale, use a logarithmic scale. That way would minimize the vast empty spaces of the model, but it still would be visibly impressive.

1. Alpha Centauri is about 165,000 miles (265,000 km) from the center of the model. The closest landmark I can think of is the L1 point of the earth-moon system (200,000 miles)…

2. If the sun is the center of the model, and say we want to depict the far edge of the galactic disk from here….we’re about 28,000 ly from the galactic center and the radius of the disk is about 60,000 ly, so to hit the far edge of the galaxy in this model would be 88,000 ly or 5.6 x 10^9 AU, i.e. 5.6 x 10^9 km in this model….somewhere before the nearest star. I would recommend making the sun the size of a hydrogen atom, which has a radius of 5.3 x 10^-11 m, so that far edge of the galactic disk is only 63 meters away!

This was fun𠄺stronomy rocks!

1. One Light year is approximately 63,240 Astronomical Unit. Making 4.2 Light Years approximately 265,608 Astronomical Units or 265,608 KM using the Atlanta scale. The nearest known landmark would be the moon (albeit the moon is over 100,000KM further away).

2. With a diameter approximated at 100,000 light years across, it would be challenging to build a scale model of the Milky Way large enough to be accurate enough and visible (except a computer model).

1) In the Atlanta model, Proxima Centauri would have to be placed 1,640,419 miles away from the Bradley Observatory. The nearest “landmark” would be the Earth’s Moon but it would be more than a million miles out beyond that. The next nearest “landmark” would be Venus, at least another 40 million miles away.

2) The Milky Way could not be effectively displayed using the Atlanta model scale. If the Bradley Observatory were used for the center of the galaxy, a scale of 1 kilometer for every 2,000 light years would place the edge of the galaxy about 50 kilometers from the Observatory. At that scale, the Earth would have the diameter of an atom. For the Earth to be represented by anything close to a millimeter in diameter, the scale would have to be 1 kilometer for every light-day, in which case the edge of the galaxy would be millions of kilometers away, out between the Moon and Venus. Better to not represent the true size of the earth at all and use the 1 kilometer:2000 light-year scale.

This was a fun exercise, but hopefully my calculations are not too far off. For poor neglected Pluto, I would place the model (very tiny indeed at only 0.54 inches wide!) in one of the towns of Auburn, Acworth, or Buford, GA, each of which are approximately 39.3 km/𠇚U” as the crow flies from the Bradley Observatory. Not sure what the closest landmark of any significance in one of these towns would be, though.
For Proxima Centauri, I had a feeling you would have to go into outer space for the landmark/model since stars are so far away, but was not sure exactly how far. After questionable calculations, I get the appropriate distance from the Bradley Observatory to be 264,900 km/𠇚U”, meaning that the only option I see for closest landmark would be the moon at 384,403 km/𠇚U”. And the model for PC would be about 4.3 feet wide, since its diameter is thought to be only 1/7 that of the Sun (whose granite plaza model is 30 feet wide). Perhaps there is another satellite orbiting around the Earth, even man-made, at a more appropriate distance?

I will be interested to see what folks come up with for #2. Obviously the model would have to be scaled down significantly to fit entirely on Earth. :)

1. Using my trusty envelope back, the Alpha Centauri marker would be 262,000 km away, or about 5/8ths of the way to the mooon, so the nearest landmark would be the moon itself.

2. Using the same scale of 1:150,000, the Milky Way model would stretch from Altanta to Alpha Centauri, more or less. In order to contain the model in near-earth space, the scale would have to be something like 1:15,000,000,000, and the model would stretch from from Atlanta to a point about 65,000 km away. The trouble is that the radius of the sun’s marker would shrink from 15 feet to something on the order of one-fiftieth of a millimeter, way too small to be seen

1. The distance to Proxima Centauri would be about 265,000 km. The nearest “landmark” would be the moon.

2. If the earth-to-sun distance in the model was 1mm (instead of 1km) the diameter of the milky way would be about 6300km, which is roughly the distance between Atlanta and Dublin.

Hvaing posted this, I read the other comments. Amy’s L1 landmark is ingenious. There may be an issue of whether the L1 point counts as a landmark, but this is more a matter of semantics than astronomy. Kudos for an elegant suggestion.

The Milky Way model can’t be linear, because it forces you to choose between having the element sof the model being visible and having them contained within a reasonable volume of space. Pehaps Bob’s suggestion of a log scale is the only way to go, but it would require a trained eye to appreciate, and even then the visual impact would be lost. Perhaps a seies of three models would work, with the Solar System ‘s radius designated by a red line, the Proxima Centauri model with a green line that consistos of 15,000 or so red segments, and the Milky way model a yellow line consisting of 25,000 or so green segments.

1) at the scale of the Atlanta model, Proxima Centauri would have to be located about 264,719 km from the model of the Sun. That’s fairly far off. The closest object upon which the model of PC could be mounted would be the Moon, averaging about 382,000 km from Earth. It might be better to mount the model at the L1 Lagrange point between the Earth and the Moon. It’s not super-stable, but it would be closer to the right place….

2) At the scale of the Atlanta model, the Milky Way (about 100,000 light-years in diameter) would be a disk of 6.3 billion km in diameter. That’s impractical (too hard to make out of granite!). A scale of one meter=1 light-year would result in a 100km wide model, which is something people can (mostly) get their heads around. The downside of such a model is that even objects as seemingly vast as our Solar System need to be rendered extremely small. Back of the envelope calculation comes out to the solar system (all the way out to pluto at about 5.5 light hours) having a scale diameter of .6 mm. That’s pretty tiny in a 100km model.

A better model may be to scale the whole thing up 3820 times, so that the orbit of the moon represents the boundary of the milky way, and the solar system would then be a couple of meters across.

Even better might be to represent the whole thing in a calendar. the calendar could represent the log of an explorer traveling across the Milky Way at light speed, from edge to edge (dodging the black hole at the center). The explorer’s time spent in our solar system, were she to bother visiting, would be about a quarter of a day in a stack of 100,000 1 yr. calendars. Think of all the awesome photos on the back pages! 1.2 million of them, assuming one month per page.

of course, I goofed on #2. The radius of the solar system to pluto is

5.5 light hours, so the diameter is 11. Under the 1 meter/light-year scale, the solar system would be slightly bigger than 1 mm. still pretty small.

In the calendar model, the trip across our solar system would now represent 1/2 day in the 100,000 year trip.

so much for the back of my envelope…next time I’ll use both halves!

1. Since Proxima Centauri is 4.2 light years away, and there are about 63,241 AUs in a light year, Proxima Centauri is approximately 265,612 AUs from the Earth (and 265,613 AUs from the Sun at this distance, the difference is negligible). This means that the scale model of Proxima Centauri would have to be placed 265,613 km away from the Bradley observatory. Since the Earth is only 40,000 km around, you𠆝 have to be creative in finding a way to situate the model near a landmark. For instance, you could go around the Earth six times and then (assuming you went West from Atlanta) plant the model somewhere near Xi𠆚n, China, maybe alongside the Terracotta Army. But how would anyone know you went around the Earth six times before plopping down the model? The other option is to send a rocket two-thirds of the way to the moon (which is 384,403 km from the Earth), and leave the model there. Luckily, production costs would be low since the real Proxima Centauri is 1/7th the diameter of the Sun, the model would only need to be about 4 feet wide. Maybe we could use one of these: //

2. It would be impossible to use the same scale as the Atlanta model for a model of the Milky Way, since the Milky Way is 6,324,100,000 (6+ billion) AUs across. Even shrinking the scale to one foot=one AU would be impossible, since that would require a space with a diameter of 1,197,746 miles [6,324,100,000 ft(AU)/5,280 ft per mile]. Scaling down even further to one inch=one AU wouldn’t solve the problem either, as the model would still need to be more than 99,000 miles across. You could make the scale one mm=one AU and fit the entire Milky Way in a space 6,324 km (about 4,000 mi) across, but then, of course you wouldn’t actually be able to see any of the models, so what’s the point?

Because it’s orbit is so eccentric, the distance a marker should be placed from the center of the solar system model is a bit tricky. At times, Pluto is closer to the sun than Neptune. However, all of the planets have more or less elliptical orbits, so I imagine an average distance from the Sun is used to place the markers. In that case, the marker for Pluto would be approximately 40 km from the Bradley Observatory.

Since one light year is equal to 63,241 au, a marker for Proxima Centauri would need to be placed at a distance of 265,612 km from the observatory. This is more than 10 times the circumfrance of the earth, so the only way to do it would be to place the marker in space. At that distance from the earth, the nearest “landmark” (at least if we put it out there in the right direction) would be the moon. It’s still not particularly close, but the distance from the marker to the moon would be less than half its distance from the earth.

Since we can’t even place a marker for the nearest star on an earth-based model in the scale of the Atlanta model of the solar system, obviously a model for the entire galaxy would either need to be based on a much, much smaller scale or a different type of scale altogether.

If we use a logarithmic scale with the Atlanta model of the solar system as the base unit, then the marker for Proxima Centauri would end up being about 173 km from the observatory, and the model for the entire galaxy would have a radius of a bit more than 300 km. (Using 100,000 light years as the estimated diameter of the galaxy.)

Ooops. After I hit send for my last post, I realized it contains (at least) one error. The distance for the marker for Proxima Centauri is a mere 6 2/3 times the circumference of the earth, not 10 times it. I was mixing miles and kilometers when I made the first estimate.

Gee, getting units of measure mixed up … didn’t we lose one Mars explorer because of such an error?

You can’t find the nearest landmark to Proxima Centauri in the model merely by calculating its distance from Atlanta. You also need to know direction and determine the actual location. And keep in mind that the Earth rotates and the Moon revolves around it. The Earth could very well be the closest landmark most of the time..

I’ve always been fascinated by models of the solar system or universe, starting as a young boy reading a short mystery in which a person was trying to raise money for a universe exhibition in which the earth was an inch across. The clue that it was a scam was that he proposed showing it in gymnasiums around the country!

1. Back to the specific questions, while Pluto’s average distance from the sun is 39.5AU, it’s orbit is very eccentric, and today it’s at roughly 31 AU, having just crossed outside the orbit of Neptune. Marietta Square Park in Marietta GA is a pretty good match for that distance, and with the number of high tech firms with large offices in Marietta it might be possible to get a sponsor for it.

Proxima Centauri, at 4.2 LY or about 266,000 AU, would have to be located in space. Unfortunately it is too far for a geosynchronous orbit and too close to put on the moon. In the interest of accuracy I𠆝 propose instead making a model of Barnard’s Star and putting it on the surface of the moon, as that would be very close to the correct scale. (Barnard’s Star is about 6 LY, or about 380,000 AU, and the moon is at about 384,000 km from earth. We won’t worry about the fact that the distance from Atlanta will vary as the earth rotates.)

2. Any attempt to make a physical model of the Milky Way would run into problems with the scale. If it fits into a building then the individual stars would be extremely small if it spans a greater distance outdoors then the width of the Milky Way means that many stars would have to be located a consierable distance above the ground. I think the best approach would be to create a virtual Milky Way. Not only could people view it from different locations and scales, but as planets are discovered they could be added to the database, allowing zoom in to individual star systems.

didn’t read the comments yet, but my calculation is that,

1. to represent Proxima Centuari, the nearest landmark would be the moon,

300K km from Atlanta. do i need to include calcs? 4*10^13km

265K AUs. if 1AU = 1km, then we’re about to the moon, right?

2. I think you could make a model using a log scale for distance, with 1 meter per order of magnitude. Then P.C. might be only 13.6m from the sun, neputne would be 9.6m away, and earth would be 8m out. Andromeda would be 21m away. if you did the same for scale of objects, then the sun would be 6m in diameter and Andromeda would be 18m in diameter.

now I’ll go back and read how I screwed up!

Proxima Centuri would need to be sited over 200,000 kilometers from Atlanta, at a point in space more than halfway to the moon.

In order to fit that star into a “metropolis-sized” model, one would have to shrink the sun by a factor of about 10,000, down to 100 micrometers wide.

The diameter of the Milky Way is about 15,000 times the distance between the sun and Proxima Centuri. To fit that into the model, the sun would have to shrink proportionately, down to 3 nanometers wide, which would be a challenge to current technology. The planets would just about vanish.

1. Proxima Centauri is 4.2 light-years away. That’s approximately 265,600 AU. At 1 AU = 1 KM, the marker would be about 70% of the distance between the surface of the Earth and the Moon, making the closest landmark Tranquility Base.

Traveling North-South along the Earth’s surface, it would wrap around

6.64 times. Traveling East-West from Atlanta, the Earth’s circumference is 22,189 km, so it would wrap around

Directly E-W, the marker would be approximately 665.67 km (.03 * 22189) from the Observatory, and directly N-S, it would be 14,402.83 km (.36 * 40007.86). Directly West, you end up near Dumas, AR. Straight East, about 40 miles off the coast of Wilmington, NC.

Straight south, you end up off the 𠇌oast” of the Antarctic Peninsula, probably on an ice shelf. Similarly, going north you end up way up by Ellesmere Island in Nunavut, Canada.

Unfortunately, I’m terrible at spherical geometry and don’t have a globe and some string handy, so the points in any other direction are anyone’s guess but mine. I haven’t the slightest clue what the nearest landmarks to any of those locations would be, save the poles.

2. The Milky Way is estimated to be 100,000 ly in diameter, and 1,000 ly thick. We lie (or float, I guess…) somewhere between 24,600 and 26,400 ly from the center. Converted to AU, these numbers are, respectively, 6.32B, 63M, 1.56B, and 1.67B. At the scale of the Atlanta model, using the Sun as the galactic center, we𠆝 be out past Saturn, and the galactic disc would extend into the Kuiper Belt (We could put the landmark on Pluto, just to cheer it up).

If we want to flatten out the Milky Way to make a model similar to Atlanta’s, a scale of 25,000 ly to the km would probably be appropriate. Centering this model at the same observatory would put the Sun at the library they’re currently using for Earth.

1. The distance (in kilometers) would be 265,608. I could only think of L1, also (even though this distance seems a bit short to match that orbital point).

2. To use a model that includes the Milky Way (and I’m just thinking of plotting the center of the galaxy), one could use a centimeter measurement instead of kilometers. This would put a marker for Proxima Centauri about 2,700 km away… let’s say, in Stockton, CA. A marker for the center of the Milky Way would be, well, about 15 million km distant, one-tenth the distance to the Sun.

So let’s use millimeters, instead! Proxima Centauri could be marked in Columbia, SC, and the Milky Way’s marker would could be placed at L2. So let’s call the Herschel Space Observatory the center of the galaxy in this model.

! The Lagrange point is genius!

#1 — Using the Atlanta model, Proxima Centauri would have to be placed about 265,000 km from earth (or from the Bradley Observatory, to be specific). The moon is a obvious candidate, but it’s a bit more than 100,000 km too far away. A better candidate would be IPM-8, a satellite launched in 1973 by NASA to collect data on magnetic fields and solar wind. Though no longer in use, it continues to occupy a nearly circular orbit of earth at a distance of 220,000 km, so I𠆝 nominate this as the nearest “landmark”.

#2 — The Atlanta model won’t work for a model that includes the entire Milky Way — the galaxy would span about 6 billion km in that model. Adjusting the scale of the model to (say) fit the Milky Way within the orbit of the moon ends up making the earth and solar system microscopic — hardly ideal for an educational model.

I𠆝 propose a logarithmic scale or other non-linear scale. I’ll leave it to somebody else to do the math, but we should be able to develop a logarithmic scale that fits the model of the Milky way into (say) the space between here and the orbit of the moon, while still leaving the solar system in a human-perceivable scale on the surface of the earth.

1 – The model for Proxima Centauri, would be about 265,600 km from Atlanta, which would put it well away from the surface of the earth, and past the orbits of even the geostationary satellites. The closest object I could find is actually the moon, whose perigee is 365,000 km. (I tried to find a man made satellite that orbits at around that altitude. The closest I could find was the Ibex probe which has a very eccentric orbit and reaches distances of 320,000 km, but in my opinion it’s eccentric orbit makes it impractical to use for a model).

2 – A model of the galaxy on the same scale would clearly be too large to be useful for the average human to use. (After all, if the model is extra-planetary, that defeats the entire purpose of the model). The standards answer to this is to either make the scale smaller or to ignore the scale entirely. The problem with making the scale smaller so that the model fits in a space that can be perceived is that the objects of interest (like the planets) become too small to be perceived. Ignoring the scale (or making the sizes of the planets not in the same scale as the empty spaces) does a disservice to the viewer and the viewer loses the sense of immense size that exists in the galaxy. The only way I can think of is to make the entire model on a logarithmic scale, where the scale changes regularly as you move away from the sun. In this manner you could make the model small enough to be perceived in its entirety, and still large enough to be able to see things like the terrestrial planets.

Let’s use the speed of light to get a sense of the answers.

It takes about 8:20 minutes or 500 seconds for light from the Sun to reach us one kilometer in the Atlanta park. It takes 30 times longer for that light (what’s left after Earth takes its portion) to get to Neptune, that’s 15,000 seconds or 30 kilometers. One day has exactly 86,400 seconds. It’s nice to know that the whole Solar System always gets day-old light.

It is 5.7 times larger for a day, or 170 kilometers. Are we in Georgia still? Over 1500 times larger still to Proxima Centauri, a quarter million kilometers on the Atlanta scale! This is over half way to the Moon, just under the distance light goes in a second! Does NASA have a commuter line there?

What about the Milky Way, it’s 12,000 times larger! Wow! If we were all composed of photons (brilliant ones, it is hoped), it would take all morning to get all the way across the model this is bigger than our Solar System, but not bigger than the Oort Cloud.

Earth's Orbit

If you really want to understand the L2 orbit, you need to first look at the Earth's orbit. Here is a diagram of the Earth orbiting the Sun (not anywhere close to scale).

Here you have this Earth with only one force acting on it, the gravitational force. The magnitude of this gravitational force is:

  • G is the gravitational constant.
  • MS is the mass of the Sun.
  • mE is the mass of the Earth.
  • rE is the distance from the center of the Earth to the center of the Sun.

And what does this gravitational force do to the Earth? It causes it to accelerate of course. But not the change-in-speed kind of acceleration. The Earth is moving in a circle, so this is called centripetal acceleration. The centripetal acceleration depends both on how fast the object is moving and how large the radius of the circle it is moving in. The direction of this acceleration is towards the center of the circle and it has a magnitude of:

Oh, v is the speed the Earth goes around the Sun and ω is the angular velocity (in radians/sec) of the Earth. Now, since there is only one force acting on the Earth, Newton's second law says: (I hate the term Newton's second law, but people usually know what it is)

The force and the acceleration are both towards the Sun (the center of the Earth's circular motion). So, I write this as a scalar equation by putting in the values for the acceleration and the force.

Let me deal with this in terms of angular velocity (ω) instead of the linear velocity (v). Solving for the orbital radius (since I essentially know ω):

The key here is that for the Earth to be in a circular orbit with a certain angular velocity (in this case about 2π radians per year), the Earth must orbit at a particular radius. What if I wanted to push the Earth into a bigger orbit? In that case, I would have to have a lower angular velocity (it would take longer to orbit the Sun).


Ptolemy's solution was to have planets moving in their own smaller circle, called an epicycle, while the entire epicycle itself revolved around the earth. In the attached drawing, mars (the red circle) moves around the sky according to the combination of the two circular motions.

Copernicus' attributed this apparent retrograde motion (apparent, because the planet didn't really move backwards), as a result of the relative speeds of the earth and the planet being observed from the earth. (See the attached drawing.)

Both models predict the same sort of motion. Ptolemy was able to adjust his model to match actual planetary orbits by altering the size and speed of an epicycle and its path around the earth. Copernicus adjusted the size and speed of a planet's orbit to obtain the same result. It is interesting to note that both can predict the motions of the planets with great accuracy.

Mars travels 1,431,000,000 Km in 687 days or 2,082,969 Km per day, while the Earth travels 942,000,000 Km in 365 days or 2,580,822 Km per day. So the Earth is moving faster! Copernicus correctly reasoned that apparent retrograde motion of a planet is caused by the Earth overtaking and passing a planet as the two revolve around the sun. (Have the children do the math).

Reaching orbit

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When rockets launch our satellites, they put them into orbit in space. There, gravity keeps the satellite on its required orbit – in the same way that gravity keeps the Moon in orbit around Earth.

This happens in a way that is similar to throwing a ball out of the window of a tall tower – to get the ball going, you need to first give it a ‘push’ by throwing it, making the ball fall towards the ground on a curved path. Whilst it is your throw that gives the ball its initial speed, it is gravity alone that keeps the ball moving towards the ground once you let go.

In a similar fashion, a satellite is put into orbit by being placed hundreds or thousands of kilometres above Earth’s surface (as if in a very tall tower) and then being given a ‘push’ by the rocket’s engines to make it start on its orbit.

As shown in the figure, the difference is that throwing something will make it fall on a curved path towards the ground – but a really powerful throw will mean that the ground starts to curve away before your object reaches the ground. Your object will fall ‘towards’ Earth indefinitely, causing it to circle the planet repeatedly. Congratulations! You have reached orbit.

In space, there is no air and therefore no air friction, so gravity lets the satellite orbit around Earth with almost no further assistance. Putting satellites into orbit enables us to use technologies for telecommunication, navigation, weather forecast, and astronomy observations.

Obvious to the Ancients

A little background. Just one example of someone who thought that the earth is a globe was the Greek mathematician Pythagoras (sixth century BC), though we don’t have a record of his reasons. Another example is Aristotle, who lived in the fourth century BC and gave several sound reasons, based on observation, why the earth must be a sphere. Ditto for Ptolemy, who wrote in the early second century AD. Between them, another famous Greek, Eratosthenes, accurately measured the earth’s circumference around 200 BC.

All these sources were known and often referenced in antiquity and throughout the Middle Ages, at least in the West and in the Middle East.

None other than Washington Irving, the famous American author of “Rip Van Winkle,” invented the flat-earth myth in his popular biography of Columbus (1828), where he felt it necessary to embellish the facts to make the story more interesting. Along the way, he took a jab at the church and the supposed errors in the Bible. Others took up his story with gusto to bolster their own attacks on Christianity’s supposed war against science, such as John Draper’s History of the Conflict Between Religion and Science (1874).

Sadly, some Christians in the 1800s chose to adopt the flat-earth myth and push it as truth. The primary instigator in the flat-earth movement was Samuel Rowbotham (1816–1884).

What convinced Rowbotham that the earth was flat? Rowbotham watched a small boat depart along the Bedford Level, a six-mile straight strip of water in England. He could see the boat the entire six miles, though he calculated that if the earth were a globe, the boat ought to have completely disappeared over the earth’s curvature.

Rowbotham then combined this argument with his own hyper-literal interpretations of certain biblical passages. He lectured and even wrote a book promoting his ideas. His work sparked so much interest that by the time of his death in 1884, he had quite a following with flat-earth organizations around the globe.

Other authors wrote their own books in defense of a flat earth, and the movement reached a peak in the late 1800s before it waned. By the late 1900s, few people believed that the earth was flat, and flat earth became a byword for scientific ignorance.

All this has changed in the last decade. Around 2012, videos promoting the notion that the earth is flat began appearing on the internet. The revival of the flat-earth movement seems to have been rekindled by Eric Dubay, a somewhat mysterious American yoga instructor living in Thailand. Apparently, Dubay encountered the century-old writings of Rowbotham and others, reintroducing their ideas to a new generation. While Dubay is not a Christian , some Christians, apparently impressed with the supposed biblical arguments for the flat earth, took up the mantle. YouTube and social media proved to be ideal media for promulgating this belief.


The Galileo Project
Rice University's Galileo Project Web site offers hundreds of pages of detailed information on Galileo, including a timeline of his life and era, a detailed diagram of his family villa, and an extensive bibliography.

Institute and Museum of the History of Science (IMSS)
The Institute and Museum of the History of Science in Florence, Italy features a permanent exhibition of Galileo artifacts. Images of these items, which include telescopes and, surprisingly, the withered middle finger of Galileo's right hand, can be viewed at the Institute's Web site.

Galileo Galilei's Notes on Motion
The IMMS Web site also offers hi-resolution scans of more than 300 handwritten pages of Galileo's notes and diagrams on motion.

Galileo and Einstein
This companion Web site to a University of Virginia course on Galileo and Einstein provides a wide range of information related to Galileo and his place in scientific history. You will find a complete online version of Galileo's Dialogue Concerning Two New Sciences, an overview with diagrams of Galileo's relationship to Copernicus, and the full text of more than 20 related lectures.


Galileo's Daughter: A Historical Memoir of Science, Faith, and Love
by Dava Sobel. New York: Walker, 1999.
Based on 124 letters from Galileo's illegitimate daughter, Maria Celeste, to her father from inside a Tuscan convent, Sobel paints an intimate picture of Galileo's personal and professional life and the times during which he lived.

The Crime of Galileo
by Giorgio de Santillana. Chicago: University of Chicago Press, 1978.
Those interested in reading more about the difficult relationship between Galileo and the Vatican should consult this volume, still considered to be the definitive resource on the subject. Giorgio de Santillana's meticulous prose quotes liberally from official Vatican documents and puts the entire affair in perspective.

The Sleepwalkers
by Arthur Koestler. New York: Arkana, 1990.
Koestler presents the history of cosmology from the Babylonians to Newton and shows how Galileo sat at the center of the scientific revolution that spawned our contemporary worldview.

Seeing and Believing: How the Telescope Opened Our Eyes and Minds to the Heavens
by Richard Panek. New York: Penguin, 1999.
Journalist Richard Panek brings 400 years of the telescope into focus in this slim, highly readable volume. What is the purpose of the telescope? Why did its invention have such a profound effect on science and life as we know it? Find out here.