## Which Planets on Which Orbits?

The short-form of the axis-period equation:

... does not refer to the mass of either the star or the planet, because generally the mass of the planet is so small compared to that of the star that it becomes negligible. This need not always be the case, however (for instance, a brown dwarf/gas giant pair; see below), and the full version of the equation [1]:

... takes this into account, where

Thus, by rearranging, it is possible to arrive at a “required” mass figure for a planet occupying a given orbit:

*MP*is the mass of the planet and*MS*is the mass of the star, both given in solar units. In both equations,*a*is expressed in Astronomical Units, and*P*is in Earth years.Thus, by rearranging, it is possible to arrive at a “required” mass figure for a planet occupying a given orbit:

… but, there is something well worth noting, here:

This equation requires that an orbital period (in Earth years) be specified for the planet in order to calculate a mass for the planet; thus, the mass of the planet isnotsolelydependent on the orbital distance, but uponbothorbital parameters: how far from the star it orbits,and alsothe time it takes to do so. Assuming the orbital distance remains the same, any change in the chosen orbital period will cause a corresponding, usually drastic inverse change in the calculated necessary mass of the planet (see below).

As an extreme example, the ratio of the mass of the

Let’s look at the case of Jupiter: it orbits the Sun at 5.202 AU and its mass is 0.0009546 that of the Sun. We know empirically that its orbital period is 11.8618 Earth years, but let’s calculate it using the long form of the equation:

*least massive stars*(at 80 Jupiter masses) to that of the*most massive planets*(weighing in at ~13 times the mass of Jupiter, or ~4131.4 Earth-masses) is only a factor of about 6.15:1; in these cases, the mass of the planet is a*significant*fraction of the mass of the star, and thus it might be useful (and relevant) to work backward from orbital distances to calculate the masses of planets orbiting such a small star.

Let’s look at the case of Jupiter: it orbits the Sun at 5.202 AU and its mass is 0.0009546 that of the Sun. We know empirically that its orbital period is 11.8618 Earth years, but let’s calculate it using the long form of the equation:

… which is close (99.9968%) to its measured period of 11.8618 years; we’ll use our calculated value in the subsequent examples.

Working backward:

Working backward:

… we can see that the equation does return a usable value for the mass of the planet (expressed in solar masses), given the semi-major axis and the orbital period.

So, what happens if we change Jupiter’s orbital period by a minuscule amount to, say, 11.0 Earth years exactly?

So, what happens if we change Jupiter’s orbital period by a minuscule amount to, say, 11.0 Earth years exactly?

The mass of the planet has increased by a factor of

What if we halve its orbital period to 5.926 Earth years (leaving it in its current orbit)?

*171.247 times*, based on a*very small change*in the orbital period—less than one year!What if we halve its orbital period to 5.926 Earth years (leaving it in its current orbit)?

Wow. Jupiter has now become the primary gravitational body in the system!

As can be seen, changes in the orbital period have effects on the resultant calculated mass for the planet, and the larger the change in the orbital period, the more drastic the change in the planet’s mass,

Note, also, that setting the orbital period to

As can be seen, changes in the orbital period have effects on the resultant calculated mass for the planet, and the larger the change in the orbital period, the more drastic the change in the planet’s mass,

*if the orbital distance remains the same*. In this case, the planet has actually grown to three times the size of the star it's orbiting!!!Note, also, that setting the orbital period to

*any value*greater than Jupiter’s known orbital period results in an impossible*negative mass*for the planet; for instance, let’s raise the orbital period minutely, to exactly 12.0 Earth years:We arrive at a negative mass for the planet!

*: a planet’s “default” orbital period is the upper limit for its semi-major axis. In order for the planet to orbit more slowly, the semi-major axis of the orbit***This is true for all known planets***must*increase. We can calculate the size the orbit would have to have by rearranging our equation to:… and plugging in the appropriate numbers:

… so, the orbit doesn’t have to increase by much (1.00792 times) to increase the orbital period from 11.852 years up to a mere 12.0 years, but it

This shows that the relationship between orbital distance, orbital period, and the mass of the orbiting body is a delicate one:

Does this mean that Jupiter cannot orbit at, say, Venus’s distance from the Sun? No, of course it can: in fact, many exoplanets orbit far closer to their stars. As mentioned in the previous blog, Kepler 42c, for instance, has a semi-major axis of ~0.00585 AU from its star, or less than one-sixtieth the distance of Mercury from the Sun.

What it does mean, though, is that to assume that Jupiter would have

Using the simple axis-period relation:

*must*increase if the planet’s mass remains the same and its orbital period becomes longer.This shows that the relationship between orbital distance, orbital period, and the mass of the orbiting body is a delicate one:

- Reducing Jupiter’s orbital period by a mere 0.852 years increased its mass by over 170 times,
- Reducing its orbital period by half caused Jupiter to become three times more massive than the Sun; and,
- Increasing Jupiter’s orbital period without increasing its orbital axis caused it to have an impossible
*negative mass*.

Does this mean that Jupiter cannot orbit at, say, Venus’s distance from the Sun? No, of course it can: in fact, many exoplanets orbit far closer to their stars. As mentioned in the previous blog, Kepler 42c, for instance, has a semi-major axis of ~0.00585 AU from its star, or less than one-sixtieth the distance of Mercury from the Sun.

What it does mean, though, is that to assume that Jupiter would have

*the exact same orbital period*as Venus on the same orbit as Venus may not be a valid assumption.

Using the simple axis-period relation:

… and applying Venus’ semi-major axis, a = 0.723, we obtain:

… which is very close (99.92%) to its measured period of 0.615198 years.

Because mass isn’t involved in the short form of the equation, a Jupiter-mass planet (318 times more massive than Venus), would be assigned the same orbital period if assigned to the same orbit.

Instead, let’s calculate Jupiter’s orbital period on Venus’ orbit using the long form of the equation:

Because mass isn’t involved in the short form of the equation, a Jupiter-mass planet (318 times more massive than Venus), would be assigned the same orbital period if assigned to the same orbit.

Instead, let’s calculate Jupiter’s orbital period on Venus’ orbit using the long form of the equation:

… and we see that while Jupiter’s orbital period on Venus’ orbit isn’t significantly different (99.87%), it is, nevertheless, not

*exactly*the same.As I have noted elsewhere, while it is technically possible for any planet to occupy any orbit, exoplanet systems so far discovered seem to show a tendency for inner-system planets to be terrestrials and outer-system planets to be giants. (See the illustrations at Cosmic Diary and Centauri Dreams.)

## Small star and big planet

Let's imagine an instance of the lowest mass star being orbited by the highest mass planet, as mentioned above. The mass of the star is 80 Jupiter masses, and the mass of the planet is 13 Jupiter masses. One Jupiter-mass is 0.0009546 solar masses, so the masses of these two bodies in solar masses are 0.076 for the star and 0.012 for the planet. If the planet orbits at Mercury's distance of 0.387 AU from its star, what is its orbital period?

Using the simple version of the equation first:

Using the simple version of the equation first:

... we calculate its orbital period at 87.964 days (which, of course, is Mercury's orbital period).

Using the long version:

Using the long version:

... the orbital period increases to 296.526 days, a value 3.371 times longer than that produced by the simple form of the equation.

What orbital distance would be required for the planet to have an orbital period of 0.241 years?

What orbital distance would be required for the planet to have an orbital period of 0.241 years?

... or 44.51% of Mercury's orbit.

## A Note About The 1.0 Earth-Year Orbital Period

In the "Designing System Orbits" blogs 1 through 5, we learned various ways to calculate planetary orbits, always keeping in mind that we might want a planet to orbit at precisely 1.0 AU, and that we would (for some methods) have to shoe-horn it into whatever set of calculated orbits we generated.

**PLEASE be aware that a**

*semi-major axis*of 1.0 AU equals an*orbital period*of 1.0 Earth-years*IF**AND**ONLY*__the mass of the central star(s) is__*IF**exactly*1.0 times that of the Sun__the mass of the planet orbiting at 1.0 AU is__*and**exactly*1.0 times that of the Earth.

**Under**

__other conditions the simple formula will not return a correct orbital period for a planet orbiting its star(s) at 1.0 AU.__*ANY*In other words, the formula:

satisfies this equality:

**...**

__IF____AND____ONLY____IF__the mass of the central star(s) is*equal to that of the Sun*and the mass of the planet orbiting the central star(s) at 1.0 AU is*equal to that of the Earth*.This is because it is a simplification of the full equation in which

*MS = 1.0*and*MP*is so small comparatively to be, to all intents and purposes, zero, and the equation becomes: ... which, of course, is just the square root of 1—which is 1.

Thus, any time the value of

Thus, any time the value of

*MS*not 1.0, the full equation must be used, or at least the equation in the form:When

*MS*is any value other than 1.0 and*MP*is so small as to be negligible.Note also that the greater the magnitude of the value of MS from 1.0—either smaller or larger—the greater will be the error in the calculated orbital periods; in other words, ignoring an MS of 1.1 will produce a smaller error than ignoring an MS of 2.5 would.

Let's explore this in more detail.

Assume a central star—Orban—with a mass of 1.5 times that of the Sun.

Orbiting Orban at 1.0 AU is a planet—Saýas—with the same mass as the Earth.

If we use the simple equation to calculate Saýas' orbital period, we will arrive at a value of 1.0 Earth-years, because:

Assume a central star—Orban—with a mass of 1.5 times that of the Sun.

Orbiting Orban at 1.0 AU is a planet—Saýas—with the same mass as the Earth.

If we use the simple equation to calculate Saýas' orbital period, we will arrive at a value of 1.0 Earth-years, because:

However, what is the calculated orbital period for Saýas if we use the full version:

This requires a little extra effort on our part.

The mass of the star (

Fortunately, we've said that Saýas' mass is the same as Earth's so we need only divide the mass of the Earth in grams (5.972E+27) by the mass of the Sun in grams (1.989E+33) to get Saýas' (and Earth's) mass in solar units:

The mass of the star (

*MS*) and the mass of the planet (*MP*) need to be expressed in terms of the Sun. The value of*MS*in the equation above is straightforward: it is 1.5, because we've already said that Orban is 1.5 solar units in mass. However, Saýas' mass also needs to be expressed in terms of that of the Sun.Fortunately, we've said that Saýas' mass is the same as Earth's so we need only divide the mass of the Earth in grams (5.972E+27) by the mass of the Sun in grams (1.989E+33) to get Saýas' (and Earth's) mass in solar units:

Yes, it's an exceedingly small number (0.0000030025 expanded to ten decimal places), but it

Thus, we'll use 3.0025E-06 as

*does*make a difference.Thus, we'll use 3.0025E-06 as

*MP*in the above equation, which will look like this:... and work out to:

... which demonstrably is

If Saýas' mass were, perhaps, 0.62 Earth-masses, then we'd need to calculate its mass in terms of Earth's mass expressed in solar units:

**1.0 Earth-years.***not*If Saýas' mass were, perhaps, 0.62 Earth-masses, then we'd need to calculate its mass in terms of Earth's mass expressed in solar units:

... and our equation becomes:

... which is quite close to our previous result, but still a far cry from being 1.0 Earth-years.

What to do?

Well, if we don't care that Saýas' orbital period is shorter than Earth's, we can stop here.

However, if we want to be sure that Saýas' orbital period is 1.0 Earth-years, then we need to

Our derived equation, from above, for calculating a semi-major axis based on the orbital period and the masses of the bodies, is:

What to do?

Well, if we don't care that Saýas' orbital period is shorter than Earth's, we can stop here.

However, if we want to be sure that Saýas' orbital period is 1.0 Earth-years, then we need to

**its semi-major axis based on the that fact (and the sum of the masses of Orban and Saýas).***calculate*Our derived equation, from above, for calculating a semi-major axis based on the orbital period and the masses of the bodies, is:

... and works out for Orban and Saýas as:

... which is demonstrably

*1.0 AU, and this is the value we need to shoe-horn into whatever other orbits we calculated using one of the formulae.***not**

**Summation**So, if the following are true:

- The central star(s) of our system have a mass (or combined mass) which is
*anything* - We know we want a planet (regardless of its mass) to have an orbital period of 1.0 Earth-years,

*must*:- Use the full version of the equation to calculate what are the orbital periods for the planets orbiting at the calculated semi-major axes; or,
- Use the derivation of the full version to calculate at what semi-major axis a planet with an orbital period of 1.0 Earth-years
*must*orbit. This result, then, is what we would shoe-horn into whatever other orbits we obtained from one of the formulas.

## A Further Note about 1.0 solar luminosity

The discussion above makes clear that for any stellar mass other than 1.0 solar masses, the long version of the axis-period relation must be used to calculate correct orbital periods for a give orbit. It also shows how to calculate the correct semi-major axis for a planet having a 1.0 Earth-year orbital period around a star (or stars) with any single or combined mass other than 1.0 solar-masses.

This is because a star of any mass other than 1.0 solar-masses will also have a luminosity other than 1.0 solar-luminosity.

Recall that both gravity and electromagnetic radiation obey the inverse-square law:

**HOWEVER . . .**

It should be noted that an orbital period of 1.0 Earth-years around a star (or stars) with any single or combined mass other than 1.0 solar-massesIt should be noted that an orbital period of 1.0 Earth-years around a star (or stars) with any single or combined mass other than 1.0 solar-masses

__does not guarantee that the stellar irradiance on that planet will be the same as that which the Earth receives from the Sun__at 1.0 AU.This is because a star of any mass other than 1.0 solar-masses will also have a luminosity other than 1.0 solar-luminosity.

Recall that both gravity and electromagnetic radiation obey the inverse-square law:

... such that a given force diminishes according the the square of the distance from the source.

For example, if we use mass in place of force in the above equation:

For example, if we use mass in place of force in the above equation:

... we can readily see that, specifying a mass of one solar mass (

*M = 1.0*) and a distance of 1.0 AU (*d = 1.0*) results in an a gravitational intensity of 1.0:But, if we keep the mass at 1.0 solar-masses (

*M = 1.0*) and double the distance to 2.0 AU (*d = 2.0*), then the resultant force of gravity is reduced to one-quarter:Similarly, if the distance remains at 1.0 AU (

*d = 1.0*), but the mass of the star is halved (*M = 0.5*), the resultant pull of gravity is reduced by one-half:We can rearrange the equation to tell us at what distance a planet would have to orbit a 0.5 solar-mass primary in order to be subject to 1.0 solar-gravity:

... such that, with half a solar mass (

*M = 0.5*) and a desired pull of gravity of 1.0 solar gravities (*I = 1.0*), the orbital distance needs to be:... just less than 71% of the former distance.

W can also use the same form of the calculation to determine intensities of stars at give distances.

W can also use the same form of the calculation to determine intensities of stars at give distances.

Here, intensity is a measure of the strength of the light received by a planet from its star(s), termed the

A star of 1.0 solar-luminosity at a distance of 1.0 AU produces the same insolation as the Sun does on the Earth:

*insolation*.A star of 1.0 solar-luminosity at a distance of 1.0 AU produces the same insolation as the Sun does on the Earth:

A star of 0.5 solar-luminosity (

*L = 0.5*) at a distance of 1.0 AU (*d = 1.0*), produces one half the insolation:BUT . . .

A star of 1.0 solar-luminosity (

A star of 1.0 solar-luminosity (

*L = 1.0*) at 0.5 AU (*d = 0.5*), produces*four times*the insolation:Here's the rub:

Here's what I mean: Let's say you want to have a star (Dahel) of 0.85 solar-masses, but you want your planet (Aldat) to have a mass of 1.0 Earth-mass and an orbital period of 1.0 Earth-years. At what distance must Aldat orbit Dahel?

**Because changes in the mass or luminosity of your central star(s) affect their luminosity or mass, respectively, the only way to have your planet possess an orbital period of 1.0 Earth-years**__and__receive the same insolation from its star(s) as the Earth does from the Sun is for your planet to have a mass of 1.0 Earth-masses and your star(s) to have a single or combined mass of 1.0 solar-masses, and a single or combined luminosity of 1.0 solar-luminosities.Here's what I mean: Let's say you want to have a star (Dahel) of 0.85 solar-masses, but you want your planet (Aldat) to have a mass of 1.0 Earth-mass and an orbital period of 1.0 Earth-years. At what distance must Aldat orbit Dahel?

So, if Aldat orbits Dahel at 0.94727 AU, it will have an orbital period of 1.0 Earth-years.

However,

However,

*it will not have a stellar irradiance (insolation) equal to that received by the Earth from the Sun*, because a star of 0.85 solar-masses (mass regime 2) has a luminosity of:... which, at 0.94727 AU produces a stellar insolation of:

... just less than 60% of the irradiance receive by the Earth from the Sun.

If we change the intensity of Dahel to ensure 1.0 solar-insolation at 0.94727 AU distance, then the mass of the star changes, and the orbital period of Aldat is no longer 1.0 Earth-years.

Here are the calculations:

The necessary luminosity for 1.0 solar-insolation at 0.94727 AU is:

If we change the intensity of Dahel to ensure 1.0 solar-insolation at 0.94727 AU distance, then the mass of the star changes, and the orbital period of Aldat is no longer 1.0 Earth-years.

Here are the calculations:

The necessary luminosity for 1.0 solar-insolation at 0.94727 AU is:

This puts Dahel in luminosity regime 2, so its mass is:

... which, for Aldat's mass (3.003E-06, the same as Earth's mass in solar units) at 0.94727 AU results in an orbital period of:

... about 350.707 days--

*1.0 Earth-years.***not**The

*solution which allows Dahel to have 0.85 solar-masses and 0.89732 solar-luminosities,***only***Aldat to have an orbital period of 1.0 Earth-years at an orbital distance of 0.94727 AU is for the mass of Aldat to be something other than 1.0 Earth-mass:***and**... or 58761.239 Earth-masses, which is—frankly—ridiculous.

If the desire is to have a planet with an orbital period of 1.0 Earth-years and a mass of 1.0 Earth-masses, then it will be necessary for it to orbit a star or stars with a single or combined luminosity of 1.0 solar-luminosities and a single or combined mass of 1.0 solar-masses.

**Summation**If the desire is to have a planet with an orbital period of 1.0 Earth-years and a mass of 1.0 Earth-masses, then it will be necessary for it to orbit a star or stars with a single or combined luminosity of 1.0 solar-luminosities and a single or combined mass of 1.0 solar-masses.

## Conclusion

The short form of the axis-period relation is useful only when the mass of the star(s) is 1.0, and it is of no concern whether or not any of the orbital periods comes out to exactly 1.0 Earth-years.

However, if

However, if

*any*of the following are true:- You have a planet with a mass totaling a significant percentage of the mass of your star(s);
- You have a central star (or stars) with a total mass of
*anything*other than 1.0 solar units; - You have a planet you know
*must*have an orbital period of precisely 1.0 Earth-years;

1. Isaac Newton's precise version of the equation is: