Does the Earth orbiting around the Earth-Moon barycenter cause a measurable centrifugal force?

Does the Earth orbiting around the Earth-Moon barycenter cause a measurable centrifugal force?

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I just realized the Earth is not a stationary object with the moon orbiting around it. As shown in this minimalist animation from Wikipedia, the Earth actually orbits a common barycenter with the moon. If this was a carnival ride, people sitting in the moon and on the far side of the earth would certainly feel the centrifugal force acting on them.

However, this is a very large system in reality, and the Earth also has quite a strong gravitational force felt on the surface anyway. So when it comes to accurately measuring gravity or doing experiments that rely on gravity, how big of a factor is this centrifugal force? Does the apearant force of gravity fluctuate from 9.9 m/s^2 to 9.7 or is it more along the lines of 9.800001 to 9.799999 (assuming the average is exactly 9.8, which is a simplification). Or is there something I'm missing that means the force is non-existent?

That the Earth and Moon orbit about their center of mass from the perspective of an inertial frame of reference is a bit irrelevant. One thing that is quite relevant is that gravitational force is undetectable by a local measuring device. For example, people standing still on the surface of the Earth do not feel gravity. They instead feel the normal force pushing them up, away from the center of the Earth. The gravitational force on astronauts in the International Space Station is about 90% of what they experience on the Earth's surface, but they feel none of that.

Another relevant factor is that the Earth as a whole, along with objects on the surface of the Earth, accelerate toward the Moon (and the Sun, and Jupiter, and Venus, and… ) gravitationally. The gravitational acceleration of those surface-bound objects toward those other bodies is not exactly the same as is that of the Earth as a whole.

The difference between these accelerations results in a force that can be measured. This is the tidal acceleration. An extremely sensitive scale will show that you weigh slightly more when the Moon is on the horizon than when it is directly overhead. For a 61 kg person, this difference in weight between the Moon being on the horizon vs directly overhead is about 10-4 newtons.

Compared to the ~600 newton weight of that 61 kg person, this is a very small effect. This very small effect, along with an even smaller effect from the Sun (roughly half), are however responsible for the tides in the oceans.

The Earth and Moon are in orbit about each other, which means that the centrifugal (outward inertial) force $M_iV_i^2/d_i$ is balanced with the centripetal (real inward force), ie $GM_{oplus}M_{mathrm{Moon}}/(d_oplus+d_{mathrm{Moon}})^2$, where $d_oplus$ is distance from Earth to Earth-Moon center of mass (c.m.) and $d_{mathrm{Moon}}$ is the Moon's c.m. distance. Thus, most of the centrifugal force of the motion about the c.m. is canceled. Since we reside at one Earth radius from the center of the Earth, the cancellation is not exact, and the remainder is exactly what we call the Lunar Tide, i.e. the familiar tidal force of the Moon which causes most of the ocean tides (there is also a component from the orbit about the Sun).

The tidal acceleration is described in Wikipedia page.

Something interesting from Wikipedia:

Scale model of the Earth-Moon system: Sizes and distances are to scale. It represents the mean distance of the orbit and the mean radii of both bodies.

Earth orbits the common barycenter of the Earth-Moon system in about 28 days, the same time the Moon orbits Earth (or more accurately, the common barycenter of the Earth-Moon system which is below Earth surface). It's estimated that the centrifugal force caused by this should be minimal on Earth surface, and only observable by a very large scale (e.g., the tides).

Earth-Moon system

The earth and moon both have ellipsoid orbits (Kepler Laws )) and have the orbits in such a fashion that for the earth’s orbit, the moon forms a focus and for the moon’s orbit the earth forms a focus. This is essentially because both the moon and the earth gravitationally attract each other which are contrary to what I previously believed that the earth had the moon revolving around its center and the earth spinning on its axis.
Since this is the case we will have on the far side of the earth, I.e the side not facing the moon, having tides because of centrifugal force and the lunar side having tides because of gravity of the moon, thus we have gravity of the moon and earth (assuming now that here centrifugal force and gravity is interchangeable, as in case of low velocities and distances) causing tides.
Here I come to pause now to regard to questions which come to my mind
1. Doesn’t this now become a binary system similar to the Pluto charon duplet?
2. Is the moon’s gravity large enough to affect the earth’s rotation? Would the earth have had an ellipsoid rotation had the moon not been there?

Is there an experimental proof that an Earth/Moon barycenter exists?

The root of the matter is in the universal acceptance of Newton's law of universal gravitation which, to simplify, pulls objects toward the center of the earth and that all objects with mass attract each other. This argument may be fundamentally flawed. I and at least one other Mensan in my local group (yes, I am a member) have been pondering whether or not gravitation is actually a 'push' force and not a 'pull' force. Hence my need to find out whether or not it has been proved that an EM barycenter exists in order to prove myself wrong.

The only thing I am seeing that we don't have the technology to detect it and that I should just accept that it is there without proof just like everyone else. Which is bogus.

Frankly, I would have expected a Mensa group to understand how limited and pointless of a request the measurement of the Earth-Moon barycenter is. and of course someone in the group to understand the scientific method itself, and teach it to the rest of you.

If this is representative of Mesa, I'm not impressed.

The Ocean tides are a pretty good detector of that.

miliarcsec or better accuracy of modern ephemerides.

So each time anyone uses ephemeris tables (or software, these days) to point their telescope at a solar system object and find it where it's supposed to be, they provide evidence of the kind required by the OP.
Whether this will be accepted as such by the OP is another matter.

Frankly, I would have expected a Mensa group to include people who understood basic physics and the well known dead-end that is push-gravity.

Frankly, I would have expected a Mensa group to understand how limited and pointless of a request the measurement of the Earth-Moon barycenter is. and of course someone in the group to understand the scientific method itself, and teach it to the rest of you.

If this is representative of Mesa, I'm not impressed.

My problem with our current understanding of gravity started with that blasted simulation of the mass of the Earth warping space/time to create gravity. You know the one. It looks like a bowling ball on a sheet of rubber with a satellite falling into the well that creates an orbit around the bowling ball. I have at least two problems with this:

The first is how this scenario would allow satellites to orbit in something other than an equilateral plane around the mass depending upon their weight when they fell into the well. The heavier the satellite the lower in the well they would orbit. Which does not relate to the real world since all orbits around the earth can only occur when the orbit is centered over the mass of the planet. You cannot have a satellite orbiting above the tropic of Capricorn, for example. Which could happen with the simulation mentioned previously.

Ok, I am going to be pretty blunt, but I think it is necessary here:

So, you have made a number of serious errors here. First of all, the image you are talking about is not a simulation - it is an illustration. Second, it has nothing at all to do with Newtonian gravity - it is an analogue used to popularize general relativity. Third, it is, as most popularizations, an oversimplified image - it has very little to do with the actual description of gravity in GR, it just illustrates curved space. Fourth, you have taken this popularized image way too literally and made your argument based on this.

Each of these errors is a direct killer for any serious argument being made. If you want to have any chance of making a serious argument you need to learn what the theory you are trying to disprove is actually saying, not base your argument on a popularized image from which you have made erroneous extrapolations about what the theory actually says.

The barycentre of any system of particles is not a physical thing. You can't detect it. You can measure where each particle in the system is, measure the mass of every particle in the system and then compute where the barycentre is. For example, if you and your friend drive your cars around, then the barycentre of that two-body system will be roughly half-way between your cars (assuming they are of approximately equal mass). But, no one is going to be able to stand by the side of a motorway and "detect" the barycentre of that system! What would they detect that would tell them that a point on the road is half-way between your cars? Every pair of cars (every pair of objects) has a barycentre. How would they distinguish one "detected" barycentre from another?

In short, the barycentre is a geometrically defined point in space, but there is nothing there to identify it as such. It doesn't exist, because it is purely geometrically defined. There is nothing physical at the barycentre.

I was recently looking at a problem involving the orbit of Mercury round the Sun. All the data is there:

No one is asking you to believe that Mercury orbits the Sun in an ellipse without proof. The data has been maintained for centuries. But, the ellipse precesses to a degree that the Newtonian model cannot explain. The Newtonian model almost exactly predicts the relative position of the planets of the Solar system over time. But, there was always a doubt over the precession of Mercury. In a way you can't ask for more than that. That is the way science works. That is the way physics is done.

And, of course, General Relativity trumps Newtonian gravity by predicting the additional observed precession of Mercury. That is the way science progresses.

Does the Earth orbit the moon?

If this sounds like a joke question, it isn't. I know that the earth orbits the sun and the moon orbits the earth, which means that the moon also orbits the sun. But what I am interested in is do we orbit the moon? I was watching an episode of QI and they mentioned briefly that it is. Although it sounded like it was said on a whim. I could see why the earth would orbit the moon, after all both objects exert a gravitational force on the other. I always thought that it didn't matter because the force of the moon on the earth doesn't accelerate the earth as much. I tried googling the question, but oddly didn't find anything on it (which makes me think I'm either forgetting something rather obvious or didn't search for it well). If it does orbit the moon, do common diagrams exclude showing it because the period of orbit of the earth around the moon is incredibly big in comparison with the earth around the sun? Also, would it look like we're orbiting a point between the earth and the moon rather than the moon itself?

Let's just think about the Earth and Moon. The two bodies orbit a common point, which is the center of mass of the two-body system. This center-point is actually about a quarter of the Earth's radius below the Earth's surface (1700 km below the surface), because Earth is so much more massive. Because the center is inside the Earth, we say that the Moon orbits the Earth, and not the other way around.

Pluto and Charon are different - their barycenter is actually above Pluto's surface. So these two bodies comprise a binary system.

If the center of mass of the Earth and Moon were inside the Moon, then we could say that the Moon orbits the Earth. But it doesn't. EDIT: I swapped the Earth and Moon here. Whoops.

some goes for the sun and every planet. The center of mass is well within the sun we say all the planets orbit the sun. But the sun does wobble slightly due to orbiting the center of mass inside itself.

From what I recall from an online astro 101 course, two bodies orbit a point which represents their total center of mass. So, theoretically yes. But realistically, the sun represents such a larger portion of the sun + earth center of mass than the moon + earth center of mass that it doesn't really make a difference.

But I also know there's some reason the moon is gradually getting further from the earth.

The Earth and Moon both orbit their center of gravity.

1.) Okay, so there's a lot of questions/uncertainties in your question here. This video explains some of the force-acceleration you were hinting at, be nice to Derek since this is one of his earlier videos and a little bit cheesy, but easily my favorite of his since it really clears up this misconception -- was also the first video of his that I saw / introduced me to his work(watch other videos of his too if you're interested).

2.) I have no clue what QI is. I'm sorry.

3.) I think they meant -- and I acknowledge that I could be completely wrong -- is the consequence of different reference points. For example, we're on Earth, so to us, Earth isn't moving, and everything else that we observe from Earth is moving, so we might be compelled to argue that it is the Sun that is moving around the Earth (as they did before 500 years ago), but we now know this not to be the case. So perhaps, and this is where the uncertainty is, they might have meant that from the frame of standing on the Moon, and ignoring the fact the it's tidally locked with Earth (for purposes of this argument), it would appear as if the Earth is moving around the Moon just like when we thought that the Sun was moving around the Earth. But, I cannot tell what they meant exactly by their remark.

4.) I believe you're describing a Lagrange Point, video embedded, but I don't know what that would look like from that reference point.

Tidal locking (also called gravitational locking or captured rotation) occurs when the gravitational gradient makes one side of an astronomical body always face another, an effect known as synchronous rotation. For example, the same side of the Moon always faces the Earth. A tidally locked body takes just as long to rotate around its own axis as it does to revolve around its partner. This causes one hemisphere constantly to face the partner body. Usually, at any given time only the satellite is tidally locked around the larger body, because this is the first stage that occurs and it takes a lot more time to also lock the heavier main body to the partner body. But given enough time, eventually each may be tidally locked to the other. This will occur more quickly if the difference in mass between the two bodies and their physical separation is small which probably helped the fact that this is already the case with Pluto and its moon [Charon](

Image i - Tidal locking results in the Moon rotating about its axis in about the same time it takes to orbit Earth. Except for libration effects, this results in the Moon keeping the same face turned towards Earth, as seen in the figure on the left. (The Moon is shown in polar view, and is not drawn to scale.) If the Moon were not rotating at all, it would alternately show its near and far sides to Earth, while moving around Earth in orbit, as shown in the figure on the right.

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What Really Causes Tides

Solar eclipses are notable for a variety of scientific and social reasons. By blocking the intense sunlight that we normally see, they provide astronomers with information from the much dimmer outer layers of the Sun's atmosphere, the chromosphere and corona. They also enable scientists to see stars when they are almost directly behind the Sun and therefore are normally invisible to us. This was important in 1919 when Arthur Eddington photographed a total solar eclipse and imaged stars whose light paths had been affected by the Sun, thereby providing the first confirmation of Einstein's theory of general relativity.

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The simple answer is that the sun's gravity produces the same acceleration on both the Earth and the Moon. The Sun is pulling both of them along, but they are falling together.

You may imagine two skydiver jumping out of a plane at the same time (and we'd better ignore air resistance). They are subjected to gravitational forces from the Earth that vastly larger than the forces between them, but that doesn't rip them away from each other because they both experience the same acceleration.

Let's do a back-of-the-envelope calculation:

Let $Mapprox 2.0cdot 10^<30> ext$ be the mass of the sun and $mapprox 6.0cdot 10^<24> ext$ the mass of the earth, $Rapprox 1.5cdot 10^<11> ext$ the distance between earth and sun and $rapprox 3.8cdot 10^8 ext$ the distance between earth and moon.

The relative acceleration of the moon respective to earth due to the difference in gravity of the sun can be approximated by $ Delta a = frac - frac <(R+r)^2>= frac left( 1 - frac 1 <(1+frac rR)^2> ight)approx frac<2GMr> $ via Taylor expansion.

The moon's acceleration due to earth's gravity is $ a = frac $ and we end up with $ frac a approx frac<2Mr^3>approx 92 $ Personally, I'd have expected some more powers of ten here, but of course this should still be more than enough to keep the moon from wandering off.

Does the Earth orbiting around the Earth-Moon barycenter cause a measurable centrifugal force? - Astronomy

The Earth, Sun, and Moon

Earth, which is our base from which we look into space, is constantly moving. Understanding this movement is one of the most useful and important things in astronomy.

The earth orbits the sun in an elliptical orbit and the moon orbits the earth with the same kind of orbit. Looking down from the north pole, the earth spins in a counterclockwise direction on an imaginary line called its axis once every day. This accounts for the fact that the sun rises in the east and sets in the west. The earth’s axis is tilted with respect to the plane of its orbit at an angle of about 23.4 degrees. If we position ourselves high above the north pole, we would see that the earth orbits the sun in a counterclockwise motion, coming to the same position among the stars every 365.26 earth days. We would also see that the moon also orbits the earth in a counterclockwise motion. This is illustrated in the following example.

Figure 1: The directions of the orbits of the earth and moon.

The average distance from the earth to the sun, the semimajor axis of its orbit, is 149,597,890 km. This distance was not known until recently and it is called the astronomical unit or AU. The distances of the other planets to the sun are usually measured in astronomical units.

Because of the tilt of the earth, not every place on earth gets light every day. Also, some places have extremely short days.

As the earth revolves around the sun, the place where light shines the brightest changes. This motion gives us the different seasons. For instance, the poles receive less light than does the equator because of the angle that the land around the poles receive the sun’s light. When the north pole is tilted toward the sun, the northern hemisphere is presented to the sun at a greater angle than the southern hemisphere and the northern hemisphere gets warmer. When this happens, the northern hemisphere gets summer while the southern hemisphere gets winter. When the south pole is tilted toward the sun, the two seasons reverse hemispheres. This is illustrated in the following image.

Figure 2: The positions of earth at the different seasons. Counterclockwise from lower left: summer, fall, winter, spring (northern hemisphere).

The earth’s orbit is called the ecliptic. The plane which contains the ecliptic is the reference plane for the positions of most solar system bodies. Viewed from earth, the ecliptic is the apparent motion of the sun among the stars.

The earth’s equator is a circle going around the earth which is on a plane that is perpendicular to the earth’s axis. The equator and the plane on which it lies are illustrated in the following image.

Figure 3: The equatorial plane.

The Equinoxes

This equatorial plane is one of the most important in astronomy because it intersects the plane of the ecliptic and gives us a reference point in space by which we can measure the positions of stars. This plane also divides the earth into halves, the northern half being the northern hemisphere, the other half being the southern hemisphere. The intersection of these planes is a line, which for convenience we will call the line of equinoxes. The real definition of equinox is the point on the celestial sphere which intersects this line, but since the celestial sphere is an imaginary sphere with any size, the equinoxes are really lines. Also, for some purposes and illustrations, it is more convenient to think of the equinoxes as a line extending into space. For other purposes, it is convinient to think of the equinoxes as directions. The two planes are illustrated below.

Figure 4: The vernal equinox from two perspectives.

One half of this line is called the vernal equinox the other half is called the autumnal equinox. At two points in the earth’s orbit this line intersects the sun. These two places mark the start of two of the four seasons, autumn or spring. The autumnal equinox starts autumn around September 23. From earth, this marks the time when the sun looks as if it is crossing the plane of the equator on its way south. The vernal equinox starts spring around March 21. This marks the time when the sun looks as if it is crossing the plane of the equator on its way north. The earth carries the plane of the equator along with it. When the sun looks as if it is on its way north or south, the earth is actually carrying the equatorial plane along so that it crosses the sun.

Perpendicular to this line of equinoxes is a line which contains the solstices. The solstices are points on the ecliptic which start the other two seasons, summer and winter, when they cross the sun. The summer solstice is one half of this line, the winter solstice is the other half of this line. The half of this line that is north of the celestial equator is the summer solstice, the half that is south of the celestial equator is the winter solstice. Currently, the winter solstice starts winter for the northern hemisphere at about the time the earth is closest to the sun. This line is illustrated in the following example.

Figure 5: The summer and winter solstices.

Because of centrifugal force involved when an object spins, the earth is not a perfect sphere, but is somewhat flattened at the poles and bulges out at the equator. The distance from any point on the equator to the center of the earth is longer than the distance from either pole to the center of the earth. This is illustrated in the following image which is exaggerated for clarity. The form caused by this equatorial bulge is called a geoid.

Figure 6: A geoid.

The moon is the earth’s only natural satellite. Its average distance from the earth is 384,403 km. Its revolution period around the earth is the same length and direction as its rotation period, which results in the moon always keeping one side turned toward the earth and the other side turned away from the earth. This type of motion is called synchronous rotation. The side turned away from the earth is called the moon’s dark side, even though it is lit half of the time. The moon’s sidereal period of revolution is about 27.32 days long. This means that a line drawn through the center of the earth and the moon would point to the same star every 27.32 days. Due to slight variations in the orbital velocity of the moon, over a 30 year period, 59% of the moon’s surface is made visible. This is known as libration.

The moon’s orbit is not in the plane of the ecliptic and because of the elliptical nature of the moon’s orbit, it is not always the same distance from the earth. At the two intersections of the moon’s orbit and the plane of the ecliptic are two nodes. These nodes regress along the plane of the ecliptic, making one complete rotation every 18.61 years. See Orbits.

The Effect of the Moon

The moon has a noticeable effect on the earth in the form of tides, but it also affects the motion and orbit of the earth. The moon does not orbit the center of the earth, rather, they both revolve around the center of their masses called the barycenter. This is illustrated in the following animation.

Figure 7: The earth and moon revolving around the barycenter. Notice how the earth moves slightly.

The sun acts on the earth and its moon as one entity with its center at the barycenter. Since the earth revolves around the barycenter, which in turn orbits the sun, the earth follows a wobbly path around the sun. This is illustrated in the following example. To complicate things further, the barycenter is not always in the same place due to the elliptical nature of the moon’s orbit.

Figure 8: The wobble of the earth's orbit.

The sun attracts the moon in such a way that it perturbs its orbit every 31.807 days, this phenomenon is called evection. The moon also changes the position of the earth’s equinoxes. The sun and moon each attract the earth’s equatorial bulge, trying to bring it into alignment with themselves. This torque is counteracted by the rotation of the earth. The combination of these two forces is a slow rotation of the earth’s axis, which in turn results in a slow westward rotation of the equinoxes. Looking down from the north pole, the equinoxes would appear to be rotating in a clockwise motion. The equinoxes and poles complete a rotation every 25,800 years. The equinoxes move at a rate of about 50.27 arc seconds per year. This phenomenon in known as the precession of the equinoxes and is illustrated in the following image.

Figure 9: The precession of the equinoxes. The blue disk is the plane of the ecliptic. The white line is the equinoxes. The green plane is the plane of the ecliptic.

The north pole is currently pointing to a spot near the star Polaris. Because the vernal equinox is the starting point for most star charts, the charts must be made for a certain period. The star charts must be updated periodically to account for this movement of the reference point.

Because of the seasonal changes in the ice, snow, atmospheric distribution, and perhaps because of movements in the material within the earth, the geographic poles constantly change position in relation to the earth’s surface. This phenomenon is known as the Chandler wobble. Scientists have resolved the change into two almost circular components, the first with a radius of about 6 meters and a period of 12 months, the second with a radius of 3-15 meters and a period of about 14 months.

The sun and moon, because of their varying distances and directions in relation to the earth, constantly vary their gravitational attractions on the earth. This makes the poles wander irregularly by about + or - 9 arc seconds from its average, or mean, position. This phenomenon is known as nutation and has a period of about 18.6 years. The primary component of this is from the moon and is known as lunar nutation.

The sun and moon also constantly change the earth’s rate of spin.

Star charts use the mean equinox instead of the true equinox for their zero points. The mean equinox is the position of the equinox corrected for the slight but noticeable changes caused by nutation and the Chandler wobble. The mean equinox is still affected by precession, however, and does change position, but does it at a constant, predictable rate. Scientists requiring up-to-date precision information about the position of the earth can use the International Earth Rotation Service or IERS. This information can be found at the IERS web site at

Because of the elliptical nature of the earth’s orbit and constant changes in the earth’s rate of spin because of the previously mentioned phenomena, the sun, as seen from earth, is moving at a non-uniform rate. This makes it difficult to use the real position of the sun as a reference for time keeping. For these purposes, a point which moves at a constant rate around the earth is used instead of the real position of the sun. This point is called the mean sun and is the basis for mean solar time.

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Answers and Replies

Are you OK with the idea that the barycenter is a point that isn't moving? It's the center, so stationary while the earth and the moon are orbiting around it.

Then consider that there are no external forces acting on the earth-moon system. Therefore, the center of mass of the earth-moon system is not moving.

There can only be one point in the system that is not moving, so if the barycenter and the center of mass are both not moving, they must be one and the same.

Are you OK with the idea that the barycenter is a point that isn't moving? It's the center, so stationary while the earth and the moon are orbiting around it.

Then consider that there are no external forces acting on the earth-moon system. Therefore, the center of mass of the earth-moon system is not moving.

There can only be one point in the system that is not moving, so if the barycenter and the center of mass are both not moving, they must be one and the same.

You both seem to talk about moving points.. relative to what? Or do you mean changing?

Okay Nugatory, I get that the center of mass wouldn't change // move because it has only internal forces and the barycenter is a point that doesn't change assuming no changes happen to the system but why does the system must have only 1 point that is not moving?

I don't think I've answered your question adequately. The two bodies each have a gravitational field, each is attracted to the other. In order for them to NOT collide and become a single body, there has to be angular momentum so that an orbit will be established. Of course the speed of the orbits of both depend on both of their gravities and on their momenta.

Consider a very different yet relevant question. Perhaps you've seen the toy which is composed of 5 steel balls (usually) hanging from a cradle on two strings each. When you lift one ball up on one end, and let it drop, it falls and hits the second ball which hits the third which hits the fourth which hits the 5th which then rises up in an arc coming to nearly the same height as you lifted the first ball, and then drops back and hits the 4th ball and the process is repeated in reverse, back and forth.

If you've never seen it, you should google a video. The amazing thing is how/why does this toy know to give all of the energy to the 5th ball? Why not give half the energy to the 4th and half to the 5th and have two balls rising into the air at the other end? There's plenty of on-line explanations you can find for this.

My explanation is that the system has to conserve not only the energy ½mv² but also the momentum mv that the first ball had right before impact with the 2nd. For 2 balls to move on the other end, you need their initial velocity to match the final velocity of the first ball ½mV² or solving for v, ½mV²= ½(2m)v² = mv². A little algebra gives us the speed (assuming both have the same speed) of v = √(V/2). OK?

Now we also know that momentum is conserved (it is a Law of Physics!) so mV = 2mv, so v = ½V. So what value must V have in order for v to be both ½ of it and √½ of it? There's only one solution. V=0. Which isn't really a solution, since it's telling us that the 1st ball must hit the 2nd at V=0. In other words, the only feasible solution is that the mass rising up from the other end is equal to the mass delivering the first impact.

One ball comes in, one ball leaves.In the case of a 2 body orbit, they will start with some velocity along the axis between them, and also have off-axis components. If the only velocity they have is on that axis, then they will either hit (and you've now got a different problem) or they will escape each other's gravity (and no orbit will happen). Because of the conservation laws (energy, angular momentum, linear momentum) they will eventually form a system of two circular orbits around their CM. (Ultimately, such a system isn't stable, and will eventually decay, but these second order effects are ignored here).

Since the system's angular momentum is conserved, it turns out that there must be a point about which the two objects rotate (the CM). The 2-body problem can be viewed as if gravitational attraction were a rod connecting the two objects. If you can imagine this "bar bell" (or baton) to be spinning, even if the two "bells" are of different sizes, can you perhaps intuit that the point (their axis of rotation) they are spinning around lies somewhere on the rod between them? (By spinning, I mean end-over-end.). So they orbit around a point, the point is on the line connecting them, and it must be "stationary". I still don't think I've fully addressed your question. I will shut up now, and defer to others.

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437. Second Moon for Earth?

I have a question, which might be more fictional than anything else, but I thought I'd go ahead and ask anyway.

For a while now, I've been very interested in our Solar System. One question keeps returning in my mind, which is the following: Earth somehow receives a second moon, and for the fun of it let's get Ganymede into this email, and what would happen?

What if, 99.999999% unlikely, Ganymede somehow breaks loose from its orbit of Jupiter and somehow escapes the gravitational pull, strays towards us and somehow slows down and reaches our planet Earth. (the somehows in this line are lack of my knowledge on how this should happen.. (I'm not a Hollywood actor who suddenly knows the answer.. hah!)

What if Earth somehow is able to grab Ganymede within its gravitational pull and Ganymede becomes our second moon. We now have two, similar sized, moons orbiting our planet. And for the heck of it, let's say that Ganymede and Luna would oppose each other and both would have a certain distance from Earth so they won't collide. What would happen to this change of gravity within our tiny Earth-Luna-Ganymede system? And more important, what would happen to our own planet, having two moons pulling on Earth?

Would this be a simple 1+1=2 and whatever we experience now would just be doubled? Would it keep our Moon from moving away any further or actually push it away? Would there be drastic climate changes that could harm us?


Having more than one moon is nothing special--Earth is the only planet in the solar system which stopped at one moon. On the other hand, Earth's moon is special in having an unusually large mass compared to the planet, more than 1%.

I do not think capturing Ganymede is a possibility. For this to happen, a huge amount of energy needs to be given to it--first, to tear it away from Jupiter, second, to bring it close to Earth, and third, to allow Earth to capture it. An asteroid currently too distant to be seen might just possibly whiz close to Earth, avoid hitting it (that would be a real miracle!) but get captured by some strange combination of motions with the Moon.

It would not be easy. The "Messenger" mission currently on its way to Mercury is trying to enter an orbit around that planet, but to take way its extra speed it needs swing several times in specific encounters with Earth, Venus or Mercury, and even then, it will only have a long, elliptical orbit.

Your question has been asked before--look up

The motion of two large moons around the Earth could be complicated. The "Three body problem" in celestial mechanics has few regular solutions, and is hard to predict for the indefinite future (although Earth, Sun and Moon also form such a system). My guess is that the moons will settle into some resonance--say, three orbital periods of nearer one would equal two of the more distant one, like the motion of Pluto (stargaze/Skuiper.htm).

438. Distant companion of our Sun and 2012

The reason for my question comes from my fascination with the 2012 end-of-times scenario to which so many people have succumbed. Much of it is hogwash but as with any crackpot theory there exists the minuscule possibility of truth. I think there is more credibility in the theory that earth could have a brown dwarf twin with it's own small planetary system than the theory of a planetoid body having a huge elliptical orbit such that it only comes round once every 3600 yrs.

Thanks and have a great day!


I certainly do not believe that the Maya astronomers--a stone age culture with no telescopes--observed astronomical processes which modern instruments fail to detect. Many people have written about it, for years, and some of the correspondence is linked from the end of

I expect many more letters after November, when a film "2012" is due to be released--also when "Sky and Telescope" will feature a discussion of 2012. Can't help it.

A large astronomical object with a period of 3600 years seems too close--it would probably affect planetary motions in a detectable way. Anyway, known distant small planets are very small. The idea that the Sun may have a dim distant companion star is old: see
So far there exists no real evidence

439. Distance between two points on a sphere

I am professional java software engineer.. yesterday I read about you on internet so decided to talk with you. these days I am exploring the google map services. last week I created an application which take any location name on earth and provide me the correct latitude and longitude as found on Google Earth for example

Lahore Lati: 33.7181510 Long: 73.0605470
Islamabad Lati: 31.5450500 Long: 74.3406830

I want to get the distance b/w these two cities

I take the haversine farmula to get the result from latitude and longitude. [continues with a computer code]


Then a unit vector along the radius from the Earth's center to a point has components

(Ux, Uy, Uz) = (sin θ cosφ, sin θ sinφ, cos θ)
For any given point, all three components can be derived from the latitude and longitude.

Let (Vx, Vy, Vz) and (Wx, Wy, Wz) be unit vectors to the two points whose distance you seek. Then the angle A between them has a cosine equal to the scalar product (where * denotes multiplication):

The distance D between the points is simply RA with A in radians. If a quarter of the circumference of Earth is 10,000 kilometers (original definition of the meter) and A is in degrees, then
D = 10000 (A/90)

If D is much smaller than R (as in your example) you probably get reasonable precision by assuming the Earth is locally flat and using the theorem of Pythagoras in rectangular coordinates. Each degree of latitude then has length 10000/90 kilometers and each degree of longitude length (10000/90) cos L, with L the average longitude of the points.

440. Getting sucked in by Gravity

Also, basically the same question may be asked about the black hole in the center of the Milky Way. Blacks holes take in anything if I remember right, not even light can't escape. So it should be slowly sucking the Milky Way into it, and in time earth would be sucked in, though that would be a very long way away. Is that possibly right?


None of these will happen, because gravity conserves energy. Let me give an example. The Sun is also surrounded by a huge cloud of slow small icy objects, left over from the time when the solar system formed. Like planets, they move in ellipses (Kepler's first law, a result of the laws of motion).

Some of them do not have enough sideways motion and very slowly fall towards the Sun, in an ellipse that is very narrow and stretched. As they fall, they gain speed, and as they get near the Sun, sunlight evaporates their outer layers, revealing them at comets. However, unless they are aimed exactly at the Sun (very few are) they miss it, swing around and fly away again, losing the speed and kinetic energy they have gained.

The Earth also moves in a nearly stable ellipse--the size of which depends on the energy, which is practically conserved. It is expected to keep its distance from the Sun (if not, the year would get shorter, something sure to be noted). The situation is similar for an object orbiting a black hole, even the one at the center of our galaxy, In the galaxy, energy would be conserved, too (from all we know), even though the motion may not obey Kepler's laws, because of invisible dark matter, which we still do not understand. About that , see

441. The Work of Nikola Tesla

I have been enjoying your website but noticed that you haven't mentioned the work of Nikola Tesla at all. Did he not contribute anything to our understanding of the nature of electricity?


Tesla was essentially an engineer who played an important role in introducing alternating currents to general use. See

He contributed nothing important to the understanding the nature of electricity itself, the way Volta, Ampere, Faraday, Maxwell, Hertz and Thompson did (to mention a few).

Later in life, he developed all sorts of ideas for extracting electric power from the atmosphere, and that gave him a reputation of sorts in the area of fringe science. See

442. Firing a cannon straight up

Recently several teacher colleagues and I were on a camping trip, and while sitting around the fire found ourselves in a strange debate. The question was posed: If you fired a cannon straight up, would it be possible for the cannonball to return and land back into the barrel? Our debate focused on the effects of wind resistance, etc., as you might suspect, but if such an impediment was removed, would it be possible? Or would the rotation of the earth cause a change in its trajectory on the way back down? Any help would go a long way in preventing such an extreme discussion from recurring. Thanks very much,


Few ideas are so crazy that no scientist would look into them, and your experiment was done in 1634 , by Marin Mersenne and his friend Petit.

The result was unexpected: the cannonball was never found! Mersenne wrote about it to Descartes, his friend and correspondent, and Descartes suggested the cannonball escaped the Earth's gravity.

I attach an article telling the story (also showing a picture) which appeared in 1954 on p. 76, vol. 22, of the American Journal of Physics . It also cites some older references.

May the Force be with you!

443. Does one see half the sky--or more, or less?

Hello, I am currently going into grade 12 and an interesting question came up between me and some friends at a campfire while staring at the stars. This question could definitely be phrased better for your understanding but this is what it is. If you were on the ocean surface "perfectly flat, largest horizon distance" how much of space is visible from horizon to horizon? Some of us suggested far less than 50%, exactly 50%, and or greater than 50% .

We then created theories and diagrams supporting each percentage. Anyways by the end of the night we concluded that we are technically on the surface or top of a ball looking to either side is an angle SLIGHTLY greater than 180 degrees and as this angle extends out to areas as far as other galaxies we are seeing a larger and larger amount "as the cone's rounded edge circumference increases" If you could please revise this statement or say if it is incorrect. A diagram would be much appreciated!


If you lie on the ground and neglect the atmosphere, you see half the sky, 50%. If you stand up you see a bit more, if you climb a building or mountain or look out from an airplane, the higher you go the more you see. You can calculate the extra angle by deriving the distance D to the horizon

Then the tangent of that angle is D/R, with R the Earth radius. The angle does not change with distance of the viewed object.

But in addition there exists another factor, the refraction of light in the atmosphere , which also extends your view. See

Because of refraction, the setting Sun looks a bit flattened. The bottom of the Sun may already be below the horizon (in the "extra" visible part of the sky) but you still see it because of refraction, while the top of the Sun, more distant from the horizon, is refracted by a smaller angle. See table in the first of the letters above.

444. Why are nights dark?


Darkness is the absence of light--it is not a color, but the lack of any light. Almost all our light comes from the Sun, including sunlight reflected by the Moon and planets. Only the light of stars does not come from the Sun--they are very far-away suns, but I guess you do not count their light as a factor. So with the Sun and Moon facing the other side of Earth, the sky has no source of bright light.

People once wondered, "if the universe is infinite, there should be a star in any direction we look" ( "Olbers' Paradox" ). True, but the universe is not infinite, and in any case it began at a "big bang" about 13.7 billion years ago. So we cannot expect to get any light from distances more that 13.7 billion light years

445. Origin of the Solar System

While I'm at it: Does our sun rotate in any particular direction around the center of the Milky Way galaxy? Say, as viewed from Polaris?
If we were looking down on the Milky Way as a great disk, what angle would the Sun's pole and planetary disk subtend? Is that the right word?


No, the main reason astronomers discount solar origin of the planetary system is the conservation of angular momentum. All but a few percent of the angular momentum of the solar system reside in the planets: it would be hard to see how this would happen if the planets came from the Sun.

446. Flying to other planets


If your spaceship plans is to imitate the flight to Mars described on "Stargazers," the data for Hohmann ellipse transfer orbits for all major planets are tabulated on p. F-149 to F-153 of my 1982 CRC Handbook of Chemistry and Physics. That handbook is relatively inexpensive (especially if you buy an earlier year's edition!) and should be in most science libraries, though I am not sure that tabulation is still being printed.

However, a mission from Earth to a planet more distant than Mars will almost certainly NOT use a Hohmann ellipse. To do so requires an increase in velocity, and it is usually much more economical to obtain that velocity by a gravity-assist encounter with the Moon and/or another planet, as described in
Unfortunately, deriving such orbit requires very precise calculations, and often a "mid-course correction" during the flight (or more than one), because at launch it is very hard to achieve the precise velocity needed.

447. The heat of the Sun's Corona

I recently found out that the Sun's corona can be hotter than the surface of the Sun. Is this correct - it doesn't seem to make any sense. If so, why? My only guess could be that particles in space are farther between, and therefore heat up more easily, so the actual temperature of the corona wouldn't really "feel" hotter. I'm probably wrong, though.

By the way, I have learned a lot from your website and it has helped me greatly in answering high school students' questions. I might have some more questions in the near future, if that's ok!


It is not related to the separation of particles in space--when you see spectral lines from iron atoms that are missing 13 electrons or so, you cannot but conclude that those atoms are hit by something very powerful (or hot). The X-rays from the corona confirm the heat, and the existence of the solar wind depends on it.

You may send me questions now and then, but please only as last resort!

448. Crossing our galaxy's equator


You ask a difficult question. Presumably, the solar system is orbiting the center of the galaxy, and there should be no change in its path just by passing the plane of symmetry, which I read somewhere is 15 light years away (I quote that reference without endorsing it). However, we do not know the attracting mass, which presumably includes not only the black hole at the center (section S-7A of "Stargazers") but also the much bigger mass of all stars closer to the center than us.

However, we can estimate the velocity of the solar system's rotation around the center (look up " solar apex ") by the average Doppler shift of distant stars. From that, assuming Newton's laws hold and that the attraction is contributed by the stars we see (plus the centeral black hole) the average force of gravity can be calculated, As confirmation, Kepler's third law should be valid for objects rotating around the center of our galaxy, or any other.

The trouble is, Kepler's third law is NOT obeyed for objects rotating around the center of our galaxy (or any other). The common interpretation is that galaxies have greater mass than is visible, that they contain "dark matter" as well. (A minority opinion proposes that the cause is departure from Newton's inverse-squares law at great distances). See
This discrepancy is currently a hot subject in astronomy. Stay tuned.

449. Newton's 3rd Law--In Statics and Dynamics

The topic is Newton's 3rd law . I demonstrated the law with various obvious examples (boat, skates, rocket etc.), but when I talked about static systems (e.g. - I press with my finger on the table), I was asked where the opposite force comes from . If a truck stands on the road that means that the road is "pushing" the truck back upwards. Where is this force coming from? If the road has so much force to start with, can't we "harvest" it? Does that mean that static objects has a huge amount of force within them? Thank you very much for you help.


When forces are encountered in nature, it is usually in one of two situations. Either the force encounters an equal and opposite force, and no motion occurs (brick on a table). This is not really Newton's 3rd law: Newton's laws involve DYNAMICS, whereas this is a case of STATICS.

The force may be quite large (chimney standing on its foundation), but it produces no motion, and therefore does not perform mechanical work. It therefore doesn't require energy, and can't provide any.

Or else , the force can produce motion (brick held at arm's length and let go). To be sure, two bodies are involved here too (the brick and the Earth which pulls it down). The difference is that motion is possible, energy is released (or converted), and the brick is accelerated.

By the 3rd law, the Earth moves too, accelerating towards the brick, so that the common center of mass (brick plus Earth) remains the same. Of course, considering the size of Earth, that motion is too small to observe.

I hope that answers your students. I hope they enjoy the Q&A.

Request for clarification


    "When two compact objects ("point masses" in phystalk) act on each other, they accelerate in opposite directions, and the ratio of their accelerations is always the same. "

This only makes sense in dynamics. The way the 3rd law is usually formulated, though ("every action has an equal and opposite reaction"), is ambiguous and may be viewed as covering statics as well.

450. Barycenter of Earth-Moon system


I do not know who first estimated the mass ratio based on observations, but suspect this was before Charles-Eugene Delaunay (1816-1872), a French astronomer who undertook to calculate the motion of the Moon. He worked well before computers, so he had to manipulate lengthy algebraic equations, and his calculation fills 2 large books (1860, 1867). A modern repetition by computers took only a few minutes and reputedly found no more than 3 minor errors.

451. What if Earth rotated in only 10 hours?


The effect is easiest to calculate at the equator, where it is also largest. The centrifugal force there reduces gravity by less than 1%. If Earth like Jupiter rotated in 10 hours and its shape did not change, the reduction might be 5%, but you would probably not notice it in casual observation. See
But that is not all. The Earth would also get deformed into a more flattened spheroid , again, like Jupiter. This might reduce gravity at the equator even more.

Considering the mass of the Earth, the forces generated by rapid rotation are enormous. George Gamov in "Biography of the Earth" promotes the Gerstenkorn theory that the Earth originally rotated much faster, and broke up to create the Moon--I vaguely recall a rotation period like 4 hours (there certainly exists a limit of this order and any faster rotation would break up the Earth). The tides raised by the Moon then would gradually slow down our rotation, as they still do (very slowly). Today most astronomers believe that Earth and Moon evolved separately at the time when the smaller bodies of the solar nebula joined up to form planets.

Does the Earth orbiting around the Earth-Moon barycenter cause a measurable centrifugal force? - Astronomy

I'm a journalist, working for a tabloid here in South Africa. I answer a Q&A page, based on general knowledge for this market. One of the questions I received seems so simple, yet nobody seems to be able to answer it. The question was, "How does the moon stay suspended in the air? Why does it not fall out of the sky if it is not hanging from anything?" Now I know that its because of gravity, but I don't know enough of the details of it to answer the question. Can you help?

Your question has two answers: an elementary one and a rather subtle one. The elementary answer stems from Newton's laws of gravity. A complete answer, however, was only worked out by Einstein at the start of the last century.

A the simple answer to "why does the Moon stay suspended in the air?" is this: There is a gravitational force between the Moon and the Earth, that tries to pull the Moon toward the latter. This constant tug on the Moon as it moves around the Earth is called a "centripetal" force. This force is balanced by the "centrifugal" force, that pulls on the Earth and keeps the moon in motion. For a technical description of the (subtle) difference between centripetal and centrifugal forces check out this page. It is the balance between the centripetal and centrifugal forces that keep the Moon in orbit around the Earth.

One can be more persistent than that, however, and this is where the subtlety comes about. Why do the centripetal and centrifugal forces exactly balance each other? Because otherwise the Moon would come crashing into Earth. But why does the Moon not come crashing into Earth? Because the centrifugal force exactly balances the centripetal force. The reasoning here is circular: the simple explanation above provides a way of understanding how the Moon stays in orbit around the Earth, but not exactly why. The why was only supplied by Einstein in the early 1900s.

A more thorough answer to "Why does the Moon stay suspended in the air?" is the following. We think that Einstein's Theory of General Relativity explains the properties of space and time in the Universe. Within that theory, objects with mass curve spacetime in their vicinity, and this curvature influences the motions of other objects. The greater the mass and density of the object, the larger the curvature of spacetime that results. So, the Moon orbits the Earth because the Earth curves spacetime in the vicinity of the Moon. Though the Moon itself curves space as well (since it has mass), the curvature in the vicinity of the Moon is dominated by the Earth, which "tells" the Moon to orbit the Earth, given the Moon's current position and motion in the sky. It is this interplay between mass and curvature that causes the gravitational and the centripetal forces in the first place, and thus why the simple explanation holds.

The difference between the two answers is very subtle, but boils down to this:

First case: - Why does the Moon orbit the Earth? It just does. And you can understand how it does by analyzing the forces on the Moon caused by its orbit and finding the forces pushing in and out are equal.

Second case: - Why does the Moon orbit the Earth? Because the Earth distorts spacetime in the vicinity of the Moon, and causes it to orbit the Earth the way it does and the balance of forces to come out the way it does.

This page was last updated on July 18, 2015.

About the Author

Kristine Spekkens

Kristine studies the dynamics of galaxies and what they can teach us about dark matter in the universe. She got her Ph.D from Cornell in August 2005, was a Jansky post-doctoral fellow at Rutgers University from 2005-2008, and is now a faculty member at the Royal Military College of Canada and at Queen's University.

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