How to calculate a planet's apparent size when the planet is viewed from a moon in orbit around it?

How to calculate a planet's apparent size when the planet is viewed from a moon in orbit around it?

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Specifically in this case, how large would Jupiter be seen in the Io's sky in comparison to the moon in the Earth's sky? How to calculate this?

Also, if somebody knows any online software which lets you input moons and planets to see how their sky looks like to also fill in the answer, it's appreciated.

By a straightforward bit of trigonometry, if the distance from the observer to the (centre of) the planet is x km, then the radius, r km, of the planet subtends $arcsin(r/x)$, and so the angular size of the planet is twice this: $2arcsin(r/x)$

For Io x = 420,000 km, and the radius of Jupiter is r = 70,000 km, so the angular size is $2arcsin(1/6)=19^circ$. For comparison, the moon from the Earth has an angular size of about half a degree.

Stellarium can be set up to view from other worlds of the solar system

How to calculate a planet's apparent size when the planet is viewed from a moon in orbit around it? - Astronomy


KEEP SAFE! W A R N I N G ! It is never NEVER safe to look directly at the real Sun with the naked eye! Moreover, looking at it—even for an instant— through a telescope, binoculars, camera or similar instrument without adequate safeguards can cause permanent blindness! NEVER DO IT! To learn how you can safely "observe" the Sun, consult the pro- fessionals at your local planetarium or observatory.

Sun Distances: Perihelion = 0.9833 AU,
Average Distance = 1.0000 AU, Aphelion = 1.0167 AU

It is an intriguing "cosmic coincidence" that the Sun has a diameter about 400 times that of the Moon and is about 400 times farther than the Moon from Earth! This makes the two most prominent celestial bodies in our sky look very similar in size from our planet. However, the apparent sizes of the Sun and Moon are not constant —the Sun appearing to grow and shrink due to Earth's elliptical orbit, the Moon due to its own elliptical yet far more complex orbit . The views accessible at the top reveal how great these changes can be, while the adjacent side views of Earth's and the Moon's orbits show the current distances of 1.) our planet from the Sun, and 2.) the Moon from Earth. Light-times too are given, and the view of the lunar orbit lets you send a photon to the Moon and back, showing that even communication with our nearest neighbor is not instantaneous!

Moon Distances (in Earth equatorial diameters):
Min Perigee = 27.9, Average Distance = 30.1, Max Apogee = 31.9

Together these views highlight how the current solar and lunar apparent sizes depend on the current distances of the Sun and the Moon from Earth, and—as we'll soon see—they can be particularly useful in predicting the charac- teristics of an upcoming eclipse! You'll find that the Moon's apparent size varies quite a bit more than the Sun's. This is because on average the Moon's orbit around Earth is over 3 times more eccentric than Earth's orbit around the Sun! The side views of Earth's and the Moon's orbits confirm this, showing very
clearly that—in proportion to their orbits' overall sizes—the "variation" in the Earth-
Moon distance is significantly greater than the "variation" in the Sun-Earth distance.

Shown to the right, on average the Sun looks
just slightly larger than the Moon, their average
apparent diameters 0.533° vs 0.518° respectively.

As you can verify at the top of this page, if both bodies are compared at their maxi-
mum apparent diameters, the Moon looks larger, whereas at their minimum apparent
diameters the Sun looks larger. Of course, the various comparative Moon views avail-
able at the top are simplified for clarity. So neither the current lunar phase nor the current lunar libration are displayed. Nonetheless, these comparative views can tell us a lot. For example, over time the variations in the "current" views confirm that the Moon's apparent size changes over 12 times faster than the Sun's! This should come as no surprise, since the Moon orbits Earth about once a month, while Earth orbits the Sun only once a year.

Periods: (Years)
Sidereal: 365.2564 days (rel. to the stars)
Tropical: 365.2422 days (rel. to equinox)
Julian: 365.25 days (astronomical computation)
Mean: 1 AU (essentially defines measurement)
Mean: 149,597,870.7 km ( defined in 2012)
Dates of: early April & early October
Perihelion: 147,098,300 km (0.9833 AU)
Date of: betw. Jan 2 and Jan 6
Aphelion: 152,097,400 km (1.0167 AU)
Date of: betw. Jul 3 and Jul 7
Eccentricity: 0.0167
Periods: (Months)
Sidereal: 27.32166 days (rel. to the stars)
Synodic: 29.53059 days (rel. to Sun, Phases)
Anomalistic: 27.55455 days (rel. to Perigee)
Distance: (Earth diameters are "equatorial")
Mean: 384,400 km (30.13 Earth dia.)
Mean: 363,300 km
Min: 356,371 km (27.9 Earth dia.)
Mean: 405,500 km
Max: 406,720 km (31.9 Earth dia.)
Eccentricity: 0.0549 (mean) 0.0255 - 0.0775

As you will discover, about every 27 1/2 days—on average—the Moon's apparent diameter grows larger than the Sun's, at times as much as 6 1/2 percent larger! However, the Moon's orbit around Earth is quite complex, far more complex than Earth's orbit around the Sun! While Earth's "variations" in perihelion and aphelion are small (less than 0.02%) and have minor effect, the Moon's "variations" in perigee and apogee are not insignificant. Lunar perigees vary by over 3.5%! So, on occasion the Moon can pass through perigee yet still appear smaller than the Sun—though only from part of Earth! Because Earth is a sphere, parts of its surface are closer to the Moon than others. So, even at the most distant lunar perigees, on the side of Earth facing the Moon is a bowl-shaped region from which the Moon looks just slightly larger than the Sun. But from the "ring" around this "bowl", the Moon still looks smaller. This occurred on 2019 Dec 18, when the lunar perigee's distance from Earth was over 370,255 km, near the maximum possible for any lunar orbit! These "more-distant" lunar perigees, where the Moon—for a full orbit—looks smaller than the Sun from part of Earth, recur about every 13 to 15 months. Even so, it is the opposite situation, when perigee distances shrink and the Moon is "closer" than usual to Earth, that can have consequential effects. At such times the Moon's gravitational pull on Earth increases by an amount that can be significant, and—as the graph below shows—it can result in Extreme Perigean Tides ! Such higher-than-normal tides can substan- tially exacerbate shore erosion, especially if they coincide with strong storms! (Note: the term Supermoon has come into use relatively recently to describe a Full Moon which occurs near an extreme lunar perigee. While this term as yet has no rigorous astronomical definition, it is becoming popular in weather reports.)

With the graph above, which plots the lunar distance from 2020 through 2022, you can obtain a good sense of whether the Moon should appear larger or smaller than average at a particular time. In addition, you can use this Lunar Perigee and Apogee Calculator to study how the Moon's least and greatest distances from Earth vary from month to month. The excellent AstroPixels site provides a 100-year Perigee & Apogee List that can help with this too. (That site has a 100-year Perihelion & Aphelion List as well.) With these resources you can also determine how close this year's lunar perigees and apogees approach their historical extremes. The values of the extreme minimum perigee and extreme maximum apogee of the Moon—shown in the interactive Sun-Moon diagram at the top of the page—are those computed by noted astro-mathmatician Jean Meeus for the years from CE 1500 - 2500. However, as the graph above reveals, in just three months lunar perigees can vary by well over 10,000 kilometers! This demonstrates just how inconstant the Moon's orbit—and especially its perigees—can be, a clear reminder
that in any given month the Moon may not approach anything near its historical minimum distance!

Now, if you like to measure things like screen images, then the "Average Sizes of Sun & Moon" at the top of this page may intrigue you. The Moon's average apparent size is not "halfway" between its maximum and minimum sizes, but a bit smaller! This is because the Moon's average distance of 384,400 km is not "halfway" between its extreme perigee and apogee distances! Near the top, the side view of the Moon's orbit also confirms this. The reason can be seen in the graph above. It is apparent that successive lunar perigees vary considerably more than lunar apogees , so on average the Moon spends more time father from Earth than "halfway" between its extremes of perigee and apogee! The average lunar apparent size reflects this.

Knowing the current relative apparent sizes of the Sun and the Moon can be quite helpful during Eclipse Seasons . How large the Sun and Moon appear can suggest if an upcoming solar eclipse may be Total, Annular or Hybrid.
It is in early January when the Sun and Moon "can" appear most unequal in size to us. This is when the Sun is "nearest" (yes, "nearest"!) to Earth and thus looks its largest in our sky. Near this time, if the Moon is its "farthest" from Earth and thus looks its smallest, it cannot "cover up" as much of the Sun if a solar eclipse occurs. Also, because the Moon is at apogee, it is moving slower than at any other point in its orbit. The result of all this is that the longest Annular Solar Eclipses—at rare times exceeding 12 minutes—occur near a year's beginning or end. In contrast, the Sun is its "farthest" from Earth and smallest in our sky in early July. Near this time, if the Moon is its "closest" to Earth and thus looks its largest, it can hide the Sun to the greatest degree if a solar eclipse occurs. So the longest Total Solar Eclipses—at rare times almost 7½ minutes long—occur near mid-year! Of course, as the figures at the top of this page show, the "maximum possible" size differences between the Moon and Sun are less in July for Total Solar Eclipses (0.559° - 0.524° = 0.035°) than in January for Annular ones (0.542° - 0.490° = 0.052°). This is one reason why the longest Total Solar Eclipses are shorter than the longest Annular ones. Another is that, when it covers the Sun most greatly, the Moon is at or near perigee, so its higher speed shortens the Total Eclipses it produces. Hover over the image directly below for a visual comparison of these circumstances.

Near the top of the page we saw how the side view of Earth's orbit shows the current position of our planet between perihelion and aphelion. In a related way the map of the Inner Solar System below shows Earth's current position in orbit, giving you another way to see how close our planet is to these orbital extremes. To make this easier to visu- alize, the grid has been rotated to align it with the Line of Apsides of Earth's orbit, i.e. the line joining its perihelion and aphelion. Given the map below and the diagrams above, you can even check one against the others and see if they agree! (They should!) Meanwhile, don't forget to also check out the Current Apparent Sizes of the Planets .

Finally, as all of the above deals with the "apparent" sizes of the Sun and the Moon, it seems that—for comparison —an image of Earth, the Sun, and the Moon together in their relative "absolute" sizes is appropriate here as well. Look closely. The Moon—its diameter 1/400th that of the Sun (remember?)—is hard to see. An "edge-on" view of the lunar orbit is also included, shown with its average radius of 384,400 km. And in case you may be wondering, yes, those are the actual sunspots visible from Earth in the SDO video of the Sun that day!

© 2007- by Gary M. Winter. All rights reserved.

Interested in political cartoons and humor?
Check out The HIPPLOMATS™ .

1 Answer 1

The inconsistency comes from assuming the planet has a greater-than-infinitesimal size while leaving the star as a point source.

Usually in transit diagrams we think of the star as a disk of radius $R_mathrm$ emitting parallel rays perpendicular to its surface. The $pi R_mathrm^2$ area of the star is partially blocked by the $pi R_mathrm^2$ area of the planet. This is because, to first order, all $pi R_mathrm^2$ of the star is emitting the same amount of light in our direction (each point on the surface emits isotropically in all $2pi$ steradians outward).

If you want to look at the energy intercepted by the planet, things get more complicated. You found the total reduction in luminosity (turned into a brightness via the Earth-star distance) if we were to capture starlight from all $4pi$ steradians around the star. But of course there is no reduction in the brightness seen by a small detector when the planet is not blocking the line of sight to the star, and at the same time the reduction is greater when the planet is actually transiting.

Take a look at this eclipse diagram from Wikipedia:

Imagine drawing a large sphere around the star. If I define $L_mathrm = 4pi r_mathrm^2 ell_mathrm$, then the $L_mathrm - L_mathrm$ you calculated is the change in the total power reaching this sphere between the cases of not having a planet and having a planet. But you can't take that reduction in power and distribute it evenly over the sphere, since during a transit we are in either the penumbra or antumbra, and the amount of light blocked is viewing angle-dependent.

The question you answered correctly is "how much power and power per unit area escapes the star in all directions given there is a planet in orbit?" The problem is during the transit we have a special line of sight. You can't take total power (equivalently, power per unit area averaged over $4pi r^2$) and propagate it out to our location.

Astrolog's Planetary Moons Report

Astrolog offers a unique planetary moons report, which will show all positions and aspects involving planetary moons. Display it for the active chart with the "Setting / Planetary Moons / Moons Chart" menu command. The location of each moon is indicated in both standard geocentric coordinates, and planet centered coordinates.

  • Lead/Follow: The "Lead" column indicates how far each moon is leading or following the planet it orbits, ranging from +100% (fully leading) to -100% (fully following). Interpretation: A moon leading its planet indicates emerging qualities or where the planet is heading towards energetically. A moon following its planet indicates a quality being left behind, or past talents that one can rely upon or use.
  • Close/Distant: The "Close" column indicates how far each moon is relative to the planet it orbits, ranging from +100% (as close as possible, and potentially in front of the planet) to -100% (fully distant, and potentially behind the planet). Interpretation: A moon closer to the viewer than its planet indicates something that's obvious or strong. A moon more distant from the viewer indicates something that's subtle, weak, or expressing internally.
  • Above/Below: The planet-centric latitude indicates how much the moon is above or below the planet it orbits, in which positive means above and negative means below. Interpretation: A moon above its planet indicates an aspirational or more spiritual guiding quality. A moon below its planet indicates something supportive or more densely material.

Astrolog's planetary moons report will also show aspects involving planetary moons. The following six types of aspects are listed in additional sections:

  1. Planet/moon overlaps: This shows how close moons are to being conjunct with the planet they orbit, as viewed from Earth (or whatever the current central body is). Because moons are very close to their planet when viewed from other planets, these conjunctions are only displayed if they’re precise, i.e. if moon’s disk overlaps (or almost overlaps) the planet disk’s position in longitude. If the moon overlaps the planet disk in latitude too, then there’s a transit or occultation taking place between them, which will also be indicated. Interpretation: Overlaps indicate emphasis on the the moon (and the part of the planet it represents) as the planet affects Earth. Eclipses between a moon and its planet are one of the strongest influences!
  2. Moon/moon overlaps: This shows how close different moons orbiting the same planet are to being conjunct with each other, as viewed from Earth. Like section #1, these conjunctions are only displayed if they’re precise, i.e. if the moons’ disks overlap each other in longitude, or almost overlap within a number of moon disk widths of each other. Eclipses visible from Earth will also be indicated, which means the two moons’ disks are aligned in vertical latitude too and actually overlap. Interpretation: Like a conjunction aspect, these overlaps indicate fusion between the parts of the planet, and indicate emphasis and union as they affect Earth.
  3. Moon/planet aspects (planet centered): This shows aspects between moons and other planets, viewed from the planet the moon orbits. If a moon is Conjunct or Opposite the Sun, then it will be a New Moon or Full Moon as seen from that planet. In such a case, the moon will also be listed as planet-centric “Close” plus or minus 100% above. Similarly, a half moon (moon Square Sun) will happen exactly when the moon is leading or trailing the planet in its orbit, and the moon will also be planet-centric “Lead” +/- 100%. An aspect of moon Conjunct or Opposite Earth will happen when the moon is closest to or most distant from Earth, which means the moon will be also geocentric “Close” +/- 100%. Interpretation: Aspects touching moons stimulate those parts of the planet, which influences the energy of the planet itself, when the planet radiates its own energy toward Earth.
  4. Moon/moon aspects (same planet centered): This section shows aspects between moons orbiting the same planet, viewed relative to the planet they orbit. It’s possible for conjunctions to be eclipses or disk overlaps between the two moons, and if so they’ll be indicated. Interpretation: These aspects indicate connections between different parts of the planet, which affects the energy of the planet itself, which is then radiated toward Earth.
  5. Moon/moon aspects (other planet centered): This is identical to section #4, but shows conjunctions between moons orbiting different planets. In other words, this treats it as if the two moons were orbiting the same planet, and checks for aspects relative to the virtual shared center point. Eclipses in this section are still possible, however they will be “virtual eclipses”, which means they would only be visible if the moons actually did orbit the same body. Interpretation: These aspects are subtle, and indicates sympathetic connections between different parts of different planets, which can still be insightful.
  6. Moon/planet aspects (overlayed): This section shows aspects between planet-centric moons and geocentric planets. In other words, similar to section #5 this treats it as if the moon were orbiting Earth instead, and checks for aspects relative to the virtual shared center point. Moon phases and eclipses will be “virtual”, which means they would only be apparent if the moons actually did orbit Earth. Interpretation: These aspects show how planets interact with each other, and more specifically which parts of planets (indicated by the moon) interact with other planets.

The six types of aspects above can be divided into three general categories: (A) Geocentric aspects as seen from Earth or some other body, which generally requires very narrow orbs (or rather orbs defined in terms of planet and moon disk widths instead of degrees) since moons orbit very closely to their planet when viewed from other planets. (B) Planet centered aspects as seen from the planet itself that the moon orbits, which can affect the planet's energy and therefore the energy the planet sends to Earth. (C) Overlayed aspects that compare objects seen from two different bodies, which can still have a sympathetic connection even though the aspect is "virtual" and is acting as if the two objects orbited the same point.

Astrolog showing Saturn and its rings and moons, viewed from its moon Iapetus.

8 Answers 8

In general, you cannot treat arbitrary mass distributions as if they were concentrated at the center of mass for the sake of calculating the gravitational force, as your example shows. You can, however, do this for spherically symmetric mass distributions if you're outside that mass distribution (even if the radial mass distribution is not uniform) by the shell theorem 1 . So for a point outside the (roughly) spherically symmetric earth your daughter's calculation is correct, but for a more general setup, one would need to do it more explicitly.

1. The same theorem also includes the interesting fact that if we have a hollow shell, the gravity inside that shell caused by the shell's mass is zero, while outside of it, it acts the same as if all its mass was concentrated in the center.

So, you're right that it can matter.

In this case it doesn't, because of a marvelous cancellation which may seem less marvelous after you see the “Gaussian pillbox” approaches of a classical electromagnetism course.

In this cancellation, we discover that if you spread out mass evenly across a spherical shell, first off, the gravitational force inside that shell is zero no matter where you are inside of it. (At least, due to that shell. Other masses could be inside or outside of the shell and could be exerting forces on you of course.)

The “Gaussian pillbox” arguments say essentially that “all of the field lines describing the force have to go out to infinity, because we do not have negative mass, and they can't go into the sphere in any easy spherically-symmetric way without flowing into a negative mass at the center of the sphere. So because we don't have a negative mass they must just all flow outward in a spherically symmetric fashion.”

That is the other part of the argument, it turns out that the gravitational force of the sphere on anything outside of the sphere, is mathematically exactly the same as if you crunched all of the mass into a single point at the center of the sphere. Other mass distributions don't happen to do this, but the spherical shell does.

In turn, any mass distribution that's kind of like an onion—nested concentric spherical layers of homogeneous density, can be built out of these spherical shells and have basically the same property. The field inside the onion is the field from all of the layers underneath you, as if they were compressed to a point mass at the center the layers above you cancel themselves out. The field outside the onion is the field that you would get if you compressed all of the onion to a point mass at the center.

In particular a homogeneous sphere has a force inside of it that goes like $r^3/r^2propto r$ as the volume of “the layers beneath you” goes like $r^3$ and the force laws goes like $r$ , so if you could tunnel through the Earth, and it was approximately homogeneous and not rotating, you would experience simple harmonic motion between the point you stepped in and its antipode, same as the spring force.

These sorts of field line arguments only work for these $1/r^2$ force laws that you happen to see in both gravity and electromagnetism. Change the fouce law and you don't get this nice behavior. But it's really a wonderful cancellation and you would not have expected such a complicated integral to have such a simple result if you hadn't known to look at the force lines.

As others have said, she is correct.

In general, you can’t just substitute a shaped object by two or three objects with the same centre of gravity. That's why your suggestion fails. The gravitational effect of earth on an object is obtained by adding up the vast number of tiny gravitational effects of every last particle that takes up earth, towards every last particle that makes up that object.

Adding up a near-infinite number of near-zero items, is the realm of calculus (integration), and so in a real-world situation, you'd have to integrate the gravitational force, over the volume of the bodies that attract each other.

However in simple homework, and for many many real-world situations, we can simplify this by using two mathematical results.

  1. In many cases, we can represent a relatively small or distant object, by a point object - it’s not precise but it’s virtually the same answer. So the gravitational force of earth pulling on a mountain, or sun pulling on earth, we can often replace by a point mass at the centre of gravity for simplicity. That's usual in homework-land, and also in many real situations.
  2. The shell theorem, which says that any time you have a spherically symmetrical "shell" of matter (think the "skin of a tennis ball"!) and a force that works according to an inverse square law (like gravity does), then when you integrate the total gravitational force from it, you find remarkably, it has no net effect on any point inside it, and has the same net effect as the same point mass at its centre, on any point outside it. Since many objects in the universe are spherically symmetrical, we can treat objects like the earth, moon, and sun, as well as balls, like point masses at their centres. But that's only because they are spherically symmetrical it’s not generally true. So it will not work for your two 1 kg objects 6371 x2 km apart. They do not act the same as a 2 kg object at its centre.

These two results simplify many problems, and explain why your daughter is right.

Consider the same problem where your weight is a 1kg weight 250 km above 'earth', but replace earth with two 1kg weights spaced 6371*2 km apart, which are inline with the object floating above it. Note that the new earth has the exact same center of gravity as the old earth.

The Earth is not a pair of 1kg masses, nor does it resemble this model in any way. If your calculations had resulted in a similar effect I would have been shocked. Two masses separated by distance will produce a very different result than a single mass, even if the center of mass of each system is the same.

The reason we can calculate the Earth's gravity as if it were a point source is that the math works the same for point masses, spherical shells, spheres of uniform density and layered spheres. As long as the mass distribution is spherically uniform (no big concentrations of mass in the body) then the math works out the same.

One way to demonstrate this is to simulate a shell with a cloud of smaller point masses distributed uniformly (or as uniformly as possible) on the surface of the shell. The higher the number of point masses the closer the result resembles the result for a single point mass. We can extend this to a layered collection of such shells, and as long as the total mass remains the same the answer is the same.

Of course in the real world the Earth is neither a uniform sphere nor a layered sphere, since the Real World is not very tidy. In reality the gravitational field of the Earth is a complex structure because of non-uniform mass distribution and so on. You just have to look at the Geoid map to see it.

Fortunately for the sanity of physics students the world over, we don't insist that they calculate the actual values for an object at a particular spatial location relative to Earth's Geoid. Not until long after they've already understood the basics anyway. We're not monsters.

Your daughter is correct. Let's not confuse her.

The mistake you are making is common enough.

While it is true that the center of gravity of the earth is in the middle of earth, I would imagine that the effective center of gravity is actually closer.

Your model of two objects of different mass is not the same as single object of the same total mass in the middle of the two.

Your incorrect caluclation is the same as :

You don't have to understand much math to see that these are completely different in form and will not in general be equal.

In effect what you did was prove that the model you made (two parts) is not the same as the single object of the same mass.

Another way of arguing that the radius of the Earth doesn't matter is to use Gaussian surfaces. These are generally introduced in the context of electrical charges, but they also apply to gravity. The gravitational force on a sphere around the Earth depends only on the mass of the Earth, not the distribution of mass of the Earth. In the hypothetical you mention with the mass distributed into a barbell shape, the distribution of gravitational force on the Gaussian surface will be different, but the total flux will be the same. If the distribution of the mass is spherically symmetric, then the distribution of flux on the Gaussian surface will also be spherically symmetric.

Actually your daughter is correct. For Newton's gravitational force you can consider the earth as a point mass having the whole mass (although you will encounter problems where e.g. explicitly the inner gravitational potential of the earth is calculated, but for these calculations of an object in the earth's orbit the assumption is fine).

If the two one-kilogram point masses (the ones you use in the new calculation) would each be point masses containing half the mass of the Earth, their center of mass lies exactly on the middle of the line connecting them. Their "effective center of gravity" (the point on which you can concentrate the two Half-Earths) lies somewhere on that line too but it doesn't coincide (except for two situations) with their center of mass. That is because the effective center of gravity is dependent on where you put the 1-kilogram point mass relative to the two Half-Earth masses. If the 1-kilogram mass is aligned with the Half-Earth masses, then the effective center of gravity will lie closer to the Half-Earth mass that is closer to the 1-kilogram mass. When you rotate the 1-kilogram mass around the center of mass of the Half-Earths (keeping the distance to the center of mass to it fixed, while the three masses lie in one plane), the effective center of gravity will move to and fro wrt the center of mass. They only coincide two times in one whole rotation of the 1-kilogram mass (when the 1-kilogram mass has equal distances to both Half-Earth masses).
You can say that the effective center of gravity is a relative point (only to be defined if, in this case, a third mass is present) while the center of mass is an absolute point. If the mass distribution is spherically asymmetric, as for the two Half-Earths point masses, the effective center of gravity will change if you change the position of the 1-kilogram mass. Only for a spherically symmetric mass distribution (like the point mass emerging if you make the two Half-Earths coincide at one point or the mass distribution of the real Earth), the center of mass and the center of effective gravity will always coincide, no matter where you place the 1-kilogram mass (even if you place it inside the mass distribution).

So, even though the half of the Earth closest to the 1-kilogram mass contributes more to the gravity "experienced" by the 1-kilogram mass, the effective center of gravity will reside exactly in the middle of the Earth (in contrast to the effective center of gravity of two Half-Earth point masses if they are aligned with the 1-kilo mass).
An even better way to see this is by considering the contribution of the two halve pieces of the Earth that are at equal distance to the 1-kilogram mass (perpendicular to the halves close and far from the mass). In that case, the effective center of gravity lies clearly in the middle of the Earth (like was the case two times when the 1-kilogram mass rotated around the two Half-Earth point masses, namely when that mass had the same distance to both Half-Earth point masses). Because there can only be one effective center of gravity (for a given position of the 1-kilogram mass), the center of effective gravity for the halves of the Earth that are close and far from the 1-kilo mass must lie in the middle of the Earth too (so not closer to the Earth's surface).

So you are making the mistake to assume that the center of effective gravity doesn't lie at the center of the Earth. That is, your daughter is right.

The Brightness of Venus

Anyone, even the most casual observer, looking at the evening sky in the last month will have noticed the brilliant white planet Venus shining in the west. Often known as the Evening Star, Venus is the third brightest natural object in the sky after the Sun and the Moon. In this post I’ll talk about why Venus is so bright.

Because Venus is a planet it doesn’t emit any visible light of its own like a star does. All planets shine by reflecting starlight from the star they orbit, which in the case of Venus is the Sun. The brightness of a planet is determined by a combination of three different factors.

Factor one is the planet’s distance from the Sun. This is because the intensity of sunlight falling on a planet diminishes as the square of its distance from the Sun. This is the well-known ‘inverse square law’, which many of you will have studied in high school science lessons.

The Intensity, which is the amount of radiated power falling on a unit area, falls as the square of the distance

Clearly the more radiation falling on a planet the brighter it is, all other factors being equal. If we compare Venus to Mars, then Venus is on average 2.1 times closer to the Sun than Mars. So, it receives 4.4 (which is 2.1 squared) times as much sunlight per unit area.

Factor two is the proportion of sunlight hitting the planet which is reflected back into space. This is known as the albedo (strictly speaking the Bond albedo) and has a value between zero and one. An albedo of zero means that the planet reflects no sunlight back. Such a planet would be totally black and thus invisible. Clearly no such bodies exist, but a hypothetical planet covered in soot would have an albedo of only 0.04, meaning that only 4% of the sunlight hitting it would be reflected back. At the other end of the scale an albedo of one means that all the sunlight hitting it is reflected back. Although no bodies exist with an albedo of one. A planet completely covered in fresh snow would have an albedo of 0.9. The albedo of Venus at 0.77 is higher than any other planet in the Solar System. For comparison, Mars has an albedo of only 0.25.

Factor three is how large the illuminated part of the planet appears in the sky. This depends on:

  • the diameter of the planet – factor 3A,
  • its distance from Earth – factor 3B and
  • its phase i.e. the percentage of its sunlit face which is visible from Earth – factor 3C.

Both the distance from Earth and the phase are continually changing as the planet and the Earth move around the Sun in their respective orbits.

Examples of phase for the Moon

The way that these three factors interplay to make Venus the brightest object in the sky is best illustrated if we take the examples of the three brightest planets Venus, Mars and Jupiter.

Data from Williams (2018 a, b, c)

*An astronomical unit (AU) is the average distance between the Earth and the Sun and is equal to 149 597 871 km.

** Because the planets travel in elliptical, rather than circular orbits, their distance of closest approach to Earth and thus their maximum brightness achieved varies from orbit to orbit. The ranges of the distances from the Earth for Venus, Mars and Jupiter are given below.

In the main table the areas are given in arc seconds squared (arcsec sq.). When viewed from Earth, planets are very small and appear to the naked eye as points of light because they are too small for the human eye to resolve them into discs. Astronomers measure the apparent size of small objects in the sky in arcseconds. An arc second is 1/3600 of a degree (or roughly 1/1800 of the diameter of the Moon). It is the size that an object 2 cm in diameter, such as a US 1 cent or British 1 pence coin, would appear from a distance of 4 km.

The relative sizes and phases of Venus, Mars and Jupiter when they are at their brightest. Venus is shown in the crescent phase, because that is its phase at it brightest

Planets which are outside the Earth’s orbit, such as Mars and Jupiter, are always at their brightest when they are at their full phase (i.e. 100% illuminated) and are at their closest to Earth. This is known as the opposition and is discussed in detail in a previous post Venus orbits inside the Earth’s orbit and when Venus is closest to Earth its sunlit side faces away from the Earth and the planet is at its lowest brightness. Venus is at its maximum brightness when its phase is around 26%, shown as the two points labelled A in the diagram below.

And finally…

I hope you have enjoyed reading this post and have plenty of clear skies to observe Venus in these difficult times, when many of my readers are having to remain at home due to corona virus. If you want to read any of my previous posts on Venus, please click on

Notes on Magnitude

When discussing the brightness of objects in the sky, astronomers use a scale called magnitude, where the lower the magnitude the brighter the object. The scale was originally invented by the ancient Greek astronomers who classified all the stars visible to the naked eye into six magnitudes. The brightest stars were given a magnitude of 1, and the faintest a magnitude of 6.

Values in the magnitude scale were standardised by nineteenth century astronomers to make each decrease in magnitude value by 1 an increase in brightness of 2.512. The range of values was also extended as well, to cater for the brightest stars and most planets which are brighter than magnitude 1 and stars fainter than magnitude six.

In the standardised scale for example

  • a bright star having magnitude 1 is 9 times brighter than a star of magnitude 4. This is because 2.512 x 2.512 x 2.512 = 15.9.
  • a star having magnitude 1 is 100 times brighter than a star of magnitude 6. This is because 2.512 x 2.512 x 2.512 x 2.512 x 2.512 = 100

The brightest natural objects in the sky are (obviously) the Sun, which has a magnitude of -26.7, followed by the Moon, which has a magnitude of -12.7 at a typical full Moon. Third comes Venus, with a magnitude of around -4.5 at the two points in its orbit when it is brightest. The magnitude of the faintest star that can be seen by someone with good eyesight in a rural location, once their eyes have fully adapted to the dark, is normally taken to be around 6.5 to 7.0.

There is now a video on the Explaining Science YouTube Channel which describes Venus’s orbit and phases. To view it, please click on the link below

The 'Morning Star'

Today (Nov. 3), Venus is the dazzling "Morning Star," rising more than 3.5 hours before sunup all month (more than 2 hours before the first light of dawn). The planet's altitude a half hour before sunrise declines only a little for mid-northern viewers from about 37 degrees to 32 degrees over the course of the month. Venus was at its greatest elongation (47 degrees west of the sun) on Oct. 26, but its dichotomy or apparent half phase as seen in a telescope, is probably happening now. As November marches on, Venus becomes more gibbous while also shrinking in apparent size. Venus is receding into the distance ahead of Earth as it outraces us in its faster orbit around the sun. This morning, take note of the much dimmer (by a full six magnitudes) yellow-orange Marssitting just 0.7 degrees to the upper left of Venus.

Venus in the Ptolemaic and Copernican Systems

The planets in our solar system and their moons are not independent sources of visible light like stars. They appear as bright objects in the sky, because they are illuminated by the Sun. Only the side of a planet or moon facing the Sun is bright. It is usually possible to see only part of the bright side of a planet or moon at any given time. For example, as our Moon moves through its roughly 29 day orbit, its appearance changes as shown in Figure 1. The balance of light and dark is called the phase of the Moon. The other planets and their moons are distant enough from us that they appear to the naked eye to be comparable in size to stars, but it is possible to observe their phases with the aid of a telescope.

Figure 1 : The phases of the moon. The cycle of phases is divided into quarters. In the first quarter, only a bright crescent is visible. In the second quarter, the bright portion of the Moon grows (waxes) from a half to a full circle. The bright portion then decreases in size (wanes) through the latter two quarters. The completely dark phase is called the new moon .

The phase of a planet or moon depends on its position relative both to the Sun and the observer. Hence, any model of the solar system can be used to make predictions regarding the phases of the planets and moons. From our perspective on Earth, Mercury and Venus are inner planets . That is, they are closer to the Sun than we are. The Ptolemaic and Copernican models of the solar system make different and incompatible predictions regarding the phases of the inner planets viewed from Earth. Hence, observations of the phases of the inner planets lead us to reject one or both of these models. In 1613, in his Letters on Sunspots [Dr57], Galileo Galilei reported that Venus displays a full set of phases similar to those of the Moon. He later presented these observations as evidence in his Dialogue on the Two Chief World Systems [Dr67].

The Animations

The animations discussed below show the motions of the Sun, Earth, and Venus according to the Ptolemaic and Copernican models. They also show Venus from the vantage point of the Earth, displaying the phases predicted by each model. Snapshots of one of the animations appear in Figure 2. Based on your own observations of these animations, you should be able to make your own judgment as to whether Galileo's observation of the phases of Venus favored one model over the other.

Figure 2 : Snapshots of an animation showing the Sun, Earth, and Venus from "above" and from Earth.

Installing and Running the Software

The animations below with file names ending with the suffix .py are written in the Python programming language with the Visual Python 3D graphics library. Both the Python language and the Visual Python package are open-source software - they are available for free. You can download both from the Visual Python Web site :

To save local copies of the the animations, right-click on each of the following animation links

If you have installed Python and Visual Python, double-clicking on your local copy of an animation program will run it. You can control the speed and direction of the animations and exit the programs with the following keyboard commands:

Disclaimers, Cautions, and other Details

  • The animations do not reproduce the finer details of the Ptolemaic and Copernican systems. For example, the center of the orbit of the epicycle of Venus is not offset from the Earth. These details are not important in illustrating of the phases of Venus, retrograde motion, and bounded elongation as predicted by the two models.
  • The sizes of Earth, Venus, and the Sun are all greatly exaggerated so that they can be easily viewed on the scale of the orbits.
  • The daily rotation of the Earth on its axis is left out of the animations entirely. Hence, your virtual observations are not limited to the hours between sunset and sunrise.
  • The fixed stars in the background of the animations are randomly-generated patterns included to provide a backdrop against which to "measure" the motions of Venus and the Sun. Any resemblance to actual stars and constellations is purely accidental.
  • The fixed stars appear to be moving when viewed from Earth. This apparent motion is a consequence of maintaining a fixed line of sight between the Earth and Sun. It is not due to the rotation of the Earth on its axis, which has been left out entirely, and would lead to a much faster apparent motion.
  • Variations in planetary sizes shown by the animations are not correct. The "camera" in the view from Earth is not actually located on Earth. Instead it looks through the Earth (made invisible) at the Sun from a distance greater than the Earth-Sun distance. This is necessary in order for the entire orbit of Venus to fit into the field of view, but as a result, Venus does not vary in apparent size as much as it does if viewed from Earth.
  • The radii of the orbits in the animation of the Ptolemaic system are based on mean distances given in Ptolemy's Planetary Hypotheses [Go67, VH85], while distances in the animation of the Copernican system are based on currently accepted values [NSSDC].

Suggested Activities

Compare the two views of each animation to get a sense of the two models. Carefully describe any differences you see. You may want to add to your response as you make further observations.

Use the view from Earth to observe the phases of Venus over a few orbits. Make a sketch of the progression of phases predicted by each model. Galileo reported observations of a complete cycle of Moon-like phases of Venus. Does Galileo's observation of the phases of Venus favor one model over the other? Explain.

Planets are observed to move relative to the fixed stars. Sometimes their motion is in the same direction as the apparent motion of the stars, and at other times they move in the opposite direction. The opposing motion is called retrograde motion . Use the view from Earth in each animation to determine whether or not each model predicts retrograde motion. Does the empirical observation of retrograde motion favor one model over the other? Explain.

The angular distance of a planet from the Sun in the sky is called its elongation . The inner planets, Mercury and Venus, are observed to have bounded elongation . That is, they each seem to have a maximum elongation, whereas the outer planets are observed to move through all possible angular separations from the Sun. Use the view from Earth in each animation to determine whether or not each model predicts bounded elongation. Does the empirical observation of bounded elongation favor one model over the other? Explain.

The Ptolemaic model places the Earth at the center of the solar system, while the Copernican model places the Sun at the center. What kind of observation(s) would settle this specific dispute? Explain.

Aside : Absolute Distances

The following discussion of absolute distances has nothing to do with the phases of Venus. However, if you are curious about the absolute scales of the models, read on.

If you observe a distant object (a planet, the Moon, a star, e.g.) in the sky with a telescope, it is not very difficult to measure its angular position, which can be expressed as two angles, similar to the latitude and longitude angles used to identify geographic locations on the Earth. However, the distance between the object and the Earth is not at all easy to measure directly.

The distance of an object from the Earth can be determined indirectly by measuring its parallax . The parallax of an object is the difference in its angular position in the sky when viewed from two different locations. If the both the parallax and the distance between the locations of the observations are known, trigonometry can be used to calculate distance from the Earth to the object. The parallax of a planet viewed from opposite sides of the Earth is illustrated in Figure 3.

Figure 3 : The parallax of a planet viewed from two different locations is the difference between its apparent angular positions at these locations relative to the "backdrop" of very distant stars. In this illustration, the two locations are on opposite sides of the Earth. (The diameter of the Earth and, and hence the resulting parallax, is greatly exaggerated in the diagram.)

Despite many attempts, a reliable measurement of the Earth-Sun distance was not made until much later than one might suspect. The solar parallax was first measured with reasonable precision during subsequent transits of Venus across the Sun in 1761 and 1769. Measurements of the apparent angular positions of Venus at its time of alignment with the Sun, viewed from many different locations on Earth, yielded high-precision values for the solar parallax [VH85].

The measurements of the solar parallax in 1761 and 1769 led to an astonishing result. The Earth-Sun distance is about 20 times those proposed by Ptolemy and Copernicus. The fact that an error of this magnitude survived for so many centuries illustrates the difficulty of determining absolute distances between objects in our solar system, and how irrelevant those distances are in predicting the angular positions which we can measure directly.

Further Reading


[Dr57] Galilei, Galileo (1613), Letters on Sunspots , In Discoveries and Opinions of Galileo , Stillman Drake, Trans. and Ed., New York: Doubleday, 1957.

[Dr67] Galilei, Galileo (1632), Dialogue Concerning the Two Chief World Systems, Ptolemaic and Copernican , Stillman Drake, Trans., Berkeley : University of California Press, 1967.

[NSSDC] Williams, David R., "Planetary Fact Sheets," National Space Science Data Center, 13 April 2004 <>.

[VH85] Van Helden, Albert, Measuring the Universe , Chicago : The University of Chicago Press, 1985. Sidereus Nuncius or The Sidereal Messenger , Albert van Helden, Trans., Chicago : The University of Chicago Press, 1989. -->

Here's How Astronomers Found Seven Earth-Sized Planets Around A Dwarf Star

There are seven planets orbiting a small dwarf star known as TRAPPIST-1. In 2016, three planets were discovered around the star, and today four more were announced. All of these worlds are roughly the size of Earth, and three are potentially habitable (as astrophysicist Ethan Siegel discusses here). On its own, TRAPPIST-1 would be easy to overlook. It's a dim, 18th magnitude star, 40 light years away in the constellation of Aquarius. So what led astronomers to look for planets around this unassuming star, and how did they find them?

TRAPPIST-1 is known as an ultra-cool dwarf star. It has only 8% the mass of our Sun, or about 84 times the mass of Jupiter. If it were much smaller it wouldn't have enough mass to fuse hydrogen in its core, and would instead be a brown dwarf. Although it's about 80 times more massive than Jupiter, it isn't much larger than Jupiter. That's because the star is much more dense than a planet due to its gravitational weight. Although it is a star, it's size makes it somewhat Jupiter-like.

Artist's depiction of an ultra-cool dwarf star like TRAPPIST-1 (left) with brown dwarfs of 65 and 30 . [+] Jupiter masses (center) and Jupiter (left). (Credit: NASA/IPAC/R. Hurt (SSC))

In some ways Jupiter can be seen as a miniature planetary system. Eight of its moons have nearly circular orbits, and four of them are the size of our Moon or larger. Other gas planets in our solar system have large moons in nearly circular orbits, so some astronomers speculated that a small star like TRAPPIST-1 might have Earth-sized planets orbiting close to the star. Although ultra-cool dwarf stars are dim and cool, a close planet might be warm enough to support liquid water on its surface. And unlike larger dwarf stars, ultra-cool dwarfs might lack large solar flares that would threaten the habitability of close planets.

So a team of astronomers started looking for evidence of planets around small dwarf stars. Using the TRAPPIST–South (TRAnsiting Planets and PlanetesImals Small Telescope–South) telescope at ESO’s La Silla Observatory in Chile, they observed the changing brightness of stars like TRAPPIST-1, hoping to catch a planet as it passed in front of the star. This is known as the transit method of planetary detection, since the planet transits in front of the star from our vantage point.

Animation showing the transit of a planet. (Credit: NASA/Hubble.)

TRAPPIST-1 is so small and distant that we can't see a planet transit directly, such as when Venus has transited the Sun. Even viewed with a telescope like TRAPPIST–South, the star looks like a point of light. But when a planet passes in front of the star, it blocks some of its light. We see this as a slight dimming of the star. It's a simple method in principle, but in practice can be quite complicated. Transiting planets aren't the only thing than can cause a star to dim. So can things such as starspots and other solar activity. There have been cases where what initially looked like a planet turned out to be a false positive.

This plot shows the varying brightness of TRAPPIST-1 during an unusual triple transit event. . [+] (Credit: ESO/M. Gillon et al.)

To make sure a dip in brightness is actually a planet, astronomers need to make multiple observations to ensure they follow a regular pattern. The more transits they observe, the more confident they can be that it really is a planet. That's part of the reason only the three closest planets were announced last year, and the outer four planets were announced today.

Although the only evidence for these planets is the dimming of TRAPPIST-1, there is still quite a bit we know about them. To begin with, the size of a planet determines the amount of dimming during a transit. Larger planets block more of the star, and therefore the dip in brightness is larger. Knowing the size of TRAPPIST-1, the team could measure the transit dimming due to each planet to determine its size. This is how we know all seven planets are roughly the size of Earth. Some a bit larger, and some a bit smaller.

Artist's illustrations showing the size of the TRAPPIST-1 planets compared to those of our solar . [+] system. (Credit: NASA)

We also have a good idea of their orbits. The time between transits tells us how long each planet takes to orbit the star, known as its orbital period. The orbital period of a planet depends upon the mass of its star and the planet's distance from the star. Since we know the mass of TRAPPIST-1 reasonably well, we can calculate the distance of each planet. By measuring the length of each transit we can also get an idea of each planet's speed. Since these speeds agree with the speed of a circular orbit, we know the orbits of these planets are fairly circular, just like planets in our solar system.

It's an amazing system, but could it be a habitable system? That's a completely different question, and one that must be explored more cautiously. Three of these planets are at a potentially habitable distance, meaning that theoretically liquid water could exist on their surface. While we know the size of these planets is similar to Earth, we don't know their mass. We also have no data on whether they have atmospheres, or if they are wet or dry. All of this affects their potential habitability. But even without this information we can make some educated guesses, given what we know about other planets and planetary systems. It's a fascinating topic, and one you can read all about in a post by my colleague and astrophysicist Ethan Siegel.

Paper: M. Gillon et al. Seven temperate terrestrial planets around the nearby ultracool dwarf star TRAPPIST-1. Nature (2017)