__Question__-

Hello

I am designing a planetary system and want a moon to orbit a planet in

such a way that, when it is viewed from the surface of the planet, the

moon is seen as stationary in the sky.(Moons orbital period is same as

planetary rotational period) I believe this is called an acentric orbit?

How do I go about determining the mass/distance relationship? It is my

intention to have two moons, one behind the other, both stationary as seen

from the planets surface. Can this be done?

__Answer__:

The type of orbit that you described, with the moon (or satellite)

stationary in the sky with respect to position on the planet, is called

**'geostationary'**.

The first thing you need to do, is to obtain or set the mass (M) for your

planet. This is so you can get its acceleration of gravity, g.

You also want to know the radius (r) of your planet.

If you know M, and r, then:

g = (G M)/ r

^{2}

where G is the universal constant of gravitation, or

G = 6.7 x 10

^{-11 }(note that this is the value compatible with using metric

units of meters for r, and kilograms or kg for M).

Thus, in the case of the Earth, for which M = 6 x 10

^{24}kg, and r = 6.4 x

10

^{6}m, we have:

g = [(6.7 x 10

^{-11 }) (6 x 10

^{24 })]/ (6.4 x 10

^{6})

^{ 2}

which you can check on any good multi-function calculator, and which will

be found to be g = 9.8 m/s

^{2 }approx.

For a geo-stationary condition, the distance R of a moon or satellite from

the CENTER of the planet is:

R = {[g* (r

^{2})*T

^{2}]/ 4p

^{2}}

^{1/3}

where all the quantities are as defined above, and T is the period of your

planet - in seconds- to make one rotation. (Note that the whole quantity

is raised to the 1/3 power, e.g. the cube root)

We can use p = 3.14

Let's apply it to the case of the Earth, for reference:

R = {[(9.8)* (6.4 x 10

^{6})

^{ 2 }(86,400)

^{2}]/ 4 * (3.14)

^{ 2 }}

^{1/3}

which comes out to:

R = 4.23 x 10

^{7}m

or R = 42,300 km (since 1 km = 1000 m or 10

^{3}m)

In other words, this is the distance that a moon would have to be from

Earth to be stationary to an observer on the surface.

Note here that: R = d + r where r is the radius of the planet (Earth) and

d is the distance from surface to Moon center.

Thus, if r = 6400 km, then

d = R - r = 42,300 km - 6400 km = 35, 900 km

or R = (35, 900 km ) 0.625 = 22, 400 miles

(Since 1 km = 5/8 or 0.625 of a mile)

All of this can be applied to your own planet and moon, simply by changing

the quantities as needed when making the substitutions.

One final point. You want to make sure that your moon is beyond the Roche

limit, i.e. the limit wherein the tidal forces of the planet can tear it

asunder.

This is typically:

2.423 x the planet's radius, assuming both planet and moon have the SAME

density.

In that instance, you need to use the formula for the Roche Limit:

RL = [(2.423)* r* (density of planet)]/ (density of Moon)

We know, from a geodetic table, that:

density of Earth = 5.51 g/ cc (grams per cubic centimeter)

density of Moon = 3.34 g/cc

Then: RL = [2.423* (6.4 x 10

^{6 }m)*5.51]/ 3.34

Note that we needn't worry about density unit compatibility since the same

factor will apply whether in kg/ m

^{3}or g/cm

^{3 })

Thus:

RL = 2.423 * (6.4 x 10

^{6 })* 1.65 = 2.55 x 10

^{7}m

or 25, 500 km (or 40, 939 miles)

Since a geo-stationary 'moon' for Earth would be at R = 42,300 km

and

RL < R (25, 500 km < 42, 300 km)

then the condition is met, at least for the case of the Earth .

This is what you will have to do with your own 'designed' system to make

sure it conforms to realistic physical conditions.

Now, you also mentioned having "an additional moon", so that one is behind

the other and "both stationary as seen from the planets surface".

Alas, you are seeking to have 'your cake and eat it'.

The point here is that

__only one moon__can occupy a given geo-stationary

orbit (i.e. for one planet) and conform to those orbital parameters.

Thus, you would virtually have to have one moon 'sitting' adjacent to the

other in orbit, but this is clearly impossible since the Roche Limit

prohibits it!

Thus, the realistic position for a 2nd moon would have to be such that

it's OUTSIDE the Roche Limit of the inner moon.

Say, your first moon's radius is 1000 km, then the 2nd can be no closer

than:

2.423 x (1000 km) = 2, 423 km

assuming both have the same density. But, now - 2,423 km further out, one

has (say for the case of Earth and a geo-stationary moon):

R' = 42,300 km + 2, 423 km = 44, 723 km

and since R <> R' (not equal R') the moon can no longer be stationary for

the same observer on the planet (Earth).

Hopefully, you will find this information helpful in designing your

system!

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