Selected Questions- Answers From All Experts Astronomy Forum (Relative Distances of Superior, Inferior Planets)

Question: Can you explain how one can find the relative distance to an inferior or superior planet, i.e. in the solar system?

Answer:  Let's first define geometrically the inferior and superior planet based on the diagram shown below:
















As seen from the diagrams, an inferior planet defines one which is interior to Earth's orbit, relative to the Sun. A superior planet defines one which is exterior to the Earth's orbit.

While  it's easy to apply Kepler's 3rd law to obtain the basic dimensions of an orbit, namely the semi-major axis (a)  of the orbit - or the mean distance from the Sun, things become somewhat more difficult when we seek to find the distance say, of the Earth to the planet.

So, we consider two cases:

(A) Inferior planet (see the diagram A)

(B) Superior planet (Diagram B)

Case (A):

The maximum elongation occurs when the planet's geocentric radius vector (pE in diagram A) is perpendicular to the planet's heliocentric radius vector, pS. Then,  by a careful measurement of the angle SEp, say over a series of nights around maximum elongation, one can obtain a value for the angle of maximum elongation. At such time the angle SEp is right-angled hence:

Sp/ SE = sin(SEp)

The quantity Sp/SE is therefore the distance of the planet from the Sun in terms of Earth units (or AU, astronomical units, where 1 AU = 1.5 x 10¹¹  m). Let Sp = R and SE = a(E) the semi-major axis for Earth's orbit, then:

Sp = R = sin(SEp) [a(E)]

Example:

If the angle SEp = 60 deg, find the planet's distance from the Sun.

Then: sin(SEp)= sin (60) = Ö3 /2 = 0.866

So: R = 0.866[a(E)]= 0.866 AU  

Case (B):

Here, we let the planet p be in opposition at some given time with the Earth and Sun (e.g. showing the alignment S-E-p in diagram (B). As we know, with opposition, the elongation is a straight angle or 180 degrees. Then after t days have elapsed the Earth's radius vector SE has moved ahead of the planet's as shown in comparing SE1 to Sp1. As can be seen, this reduces the angle of elongation from 180 degrees at opposition to angle SE1p1. This is then measured.

Now, over t days, the angle ESp will have increased from 0 (at opposition) to a value Θ given by:

Θ = [n - n(p)] t

where n, n(p) are the mean daily motions of the Earth and the planet, respectively. Using relations for the periods (from the previous answer) we may write:

Θ = 360 (1/P - 1/P') t

where P and P' are the sidereal periods for the Earth and the planet, respectively.  Then, it follows by the synodic/ sidereal relations seen in previous answer. 

Θ = 360 (1/S)

So, since t and S are both known, Θ can be obtained - that is, angle E1Sp1 is calculated. Hence, angle E1p1S can be found from:

angle E1p1S = 180 - angle SE1p1 - angle E1Sp1

From plane trigonometry we then obtain:

sin(p1E1S)/ Sp1 = sin(E1p1S)/SE1

or:

Sp1/ SE1 = sin(p1E1S)/ sin(E1p1S)

Again, giving the distance from the Sun in terms of Earth's distance unit. 


To fix ideas here is a further  example problem:

Estimate the distance of Venus from the Sun at its most recent maximum elongation, if the angle of maximum  elongation was 46 degrees.

Solution:

The diagram is shown below with the angle of maximum elongation.


We know, from the geometry:

SV/ SE = sin (SEV)

Then Venus' distance is:

SV = (sin(SEV)) SE

where: angle SEV = 46 deg and SE = a(E) = 1 AU

then:

SV = sin (46) 1 AU = (0.719) 1 AU = 0.719 AU

Selected Questions-Answers From All Experts Astronomy Forum (Sidereal and Synodic Periods of a Planet)

Question: Can you please explain the difference between the sidereal and synodic periods for a planet and how they can be computed? Thanks!

Answer:

The sideral period is the time interval for a planet's revolution about the Sun reckoned against the fixed stars. We photograph a planet at point 'x' in its orbit, with the stars determining a specific background format, then again when it reaches the same point x, relative to that background.

Of particular interest also in these computations are what we call the 'mean motions'.  For reference I use the diagram below:


Let P1 and P2 be the sidereal periods of revolution of two planets  (p1 and p2) around the Sun. Then the mean motions are defined:

n1 = 360 deg/ P1

n2 = 360 deg/P2

where we assume for example, that n1 > n2

in which case the radius vector for the (interior) planet p1 gains on the radius vector for planet p2.

As an example, let us consider the earth (P2 = 365¼ days) vs. Venus (P1 = 224.7 days).

Then:

n2 = 360 deg/ (365¼ days) = 0.985647 deg/day

n1 = 360 deg/(224.7 days) = 1.602130 deg/day

which confirms that a planet nearer the Sun has a smaller period of revolution than one further away so that P1 < P2, and also: n1 > n2.(By virtue of the  SMALLER PERIOD it has a more rapid mean motion.)

And, in effect, in the case above, Venus' radius vector gains on the Earth's by (n2 - n1) degrees per day, or [(1.602130 deg/day) - (0.985647 deg/day)] = 0.61646 deg/day.

By inspection of the diagram let the alignment Sp1p2 denote the positions of the Sun, Venus and Earth at a particular time or "epoch" in the parlance of celestial mechanics. Then, it is clear on the same inspection that one synodic period S will have elapsed by the time the two are in the new alignment Sp1'p2'. So, during the time interval,  Venus'  radius vector (Sp1) will have gained 360 degrees on the Earth's.

Now, since Sp1 gains on Sp2 by 0.61646 deg/day then it gains 360 degrees in a time S =

S x (0.61646 deg/day) = 360 deg

Or: S = 360 deg/ (0.61646 deg/day) = 583.9 days

Which is exactly the synodic period for Venus!

Hence, in general:

S x (n1 - n2) = 360 deg or

S = 360 deg/ (n1 - n2)

Or, by reference to the earlier formulations for the n1, n2 mean motions:

S (360/P1 - 360/ P2) = 360 deg

And:

S = 360 deg/ [360/P1 - 360/ P2]

Or:

1/ S = 1/P1 - 1/P2

In most problem -solving scenarios a planet's period is sought when Earth's is known. The key to the solution lies in determining whether the unknown planet is an inferior one (e.g. interior to Earth's orbit) or a superior one, e.g. exterior to the Earth's orbit. Thus there emerge two cases:

(a) The planet is inferior (like Venus), then P1 refers to the planet's sidereal period and P2 refers to Earth's.

(b) The planet is superior (e.g. Jupiter) then P1 refers to the Earth's sidereal period and P2 to the planet's.


Example Problem:

The synodic period of Venus is found to be 583.9 days. If the length of the Earth's year is 365¼ days, find the sidereal period of Venus:

1/S = 1/P1 - 1/P2

Venus is an inferior planet, so that Earth's period is inserted for P2.

Then:

1/(583.8 d) = 1/P1 - 1/(365¼ d)

Or:

1/P1 = [365¼ + 583.9 d]/ [583.9 x 365¼ ]


yielding P1 = 224.7 days 


Another problem with the roles of P1, P2 reversed:

If the synodic period of Saturn is 1.03513 years, find its sidereal period. 


Solution:

Saturn is a superior planet (orbit exterior to Earth's) so we apply:


1/ S = 1/P1 - 1/P2

but now with P1 = Earth's sidereal period, and P2 = Saturn's. Thus we have:

1/P2 = 1/P1 - 1/S

where S = 1.03513 years

1/P2 = 1/1 - 1/1.03513 = 1 - 0.966 = 0.0339

P2 = 1/ 0.0339 = 29. 4 yrs. 


yielding P1 = 224.7 days

Tackling Intermediate Astronomy Problems (5)

One of the key tasks for planetary astronomers - amateur or professional - is to ascertain the point at which a superior planet's motion changes from being retrograde to direct. We are referring here to the planet's geocentric (Earth-centered or referenced) angular velocity, and identifying when it becomes direct - between opposition and quadrature (see previous blog instalments for these configuration definitions).

Clearly, if such a change occurs, there will be some point the motion is neither retrograde or direct and this is called a "null" point or better, "stationary point". The problem then, is to obtain an expression for the planet's elongation E at a stationary point in terms of the distances of the planet and Earth from the Sun. Using such an expression and the measured elongation, the planet's heliocentric distance can be computed. (In addition, we can also obtaibn an analytical expression giving the angle Θ between the helicocentric radius vectors of planet and Earth at the stationary point) Further, such an analytic expression together with the synodic period of the planet, allows us to predict its next stationary point.

To fix ideas here, we use the accompanying diagram and let the positions of Earth and planet at the stationary point be denoted by E and P. The velocity of P relative to E must lie along the geocentric radius vector (EP) if the planet appears stationary. As we learned in the "Basic Physics" blogs, we can form a "parallelogram" of velocities - just as we can for forces. In this case, we can form the velocity parallelogram PABC, with the resultant given by PB.(E.g. the resultant of the vectors (-V_E) and the planet's tangential velocity V_p. Note that: PA = V_p and PC = -V_E.

Since we assume the orbits are circular (a close enough approximation) the angle Θ between the heliocentric radius vectors must be angle PCB between the velocity vectors. In addition, note that angle APB = 90 + φ. Now, if we extend CP to meet the line SE (produced at D) then angle DPA = Θ and angle EPD = 90 - (Θ + φ). Focusing now on the triangle PCB, we may use the sine formula of trigonometry to write:

sin[90 - (Θ + φ)]/ V_p = sin (90 + φ)/ V_E

or (using trigonometric identities):

cos (Θ + φ) = (V_p/V_E) cos φ

Making use of another trig identity and triangle SEP, we may write:

SP = PE cos φ + SE cos Θ

or:

(I) b = PE cos φ + a cos Θ

Applying it once more to triangle SEP we get:

SE = SP cos Θ + PE cos (SEP)

or:

(II) a = b cos Θ + PE cos (Θ + φ)

From the geometry of the configuration, since angle SEP = 180 - (Θ + φ), we can write equations (I) and (II) as:

(Ia) b - a cos Θ = PE cos φ

and:

(IIa) a - b cos Θ = PE cos (Θ + φ)

Dividing through eqn. (Ia) by eqn. (IIa):

[b - a cos Θ]/ [a - b cos Θ] = cos φ/ [cos (Θ + φ)]

Now, cos (Θ + φ) = (V_p/V_E) cos φ

and from the previous instalment, we saw:

(V2/V1) = (a1/a2)^½

Combining these in (Ia) and (IIa) we have:


[b - a cos Θ]/ [a - b cos Θ] = (V_E/ V_p) = (b)^½/ (a)^½

Algebraically re-arranging we find:

cos Θ = [a^½ b^½ ( a^½ + b^½)]/ (a^3/2 + b^3/2)

The preceding can be vastly simplified if all units are normalized to those in terms of Earth's to the Sun. Thus, a = 1 and b is in terms of a. So let:

Q ^½=a^½ b^½

(1 + Q^½) = a^3/2 + b^3/2

Then re-write as:

cos Θ = [Q ^½ (1 + Q^½) ]/ (1 + Q^3/2)

As may be discerned from the diagram, when Earth is at point E' where angle ESP = 90 degrees, we have another stationary point. It is of interest here to compute the total time during which a planet will be seen to move retrograde and this is just the time it takes Earth's radius vector to advance through an angle equal to 2 Θ, with respect to the planet's radius vector. This will be given by t( Θ), where:

t( Θ) = 2 Θ/ 360 x S X (ΘS)/ 180

where S is the synodic period.

The time that elapses between opposition and the next stationary point is just t( Θ)/2.

Meanwhile, the time interval during a synodic period that a planet's motion is direct is t(D) where:

t(D) = (360 - 2 Θ)/ 360 X S X (1 - Θ/ 180)S


Example Problem:

Mars has a synodic period S = 779.9 days. At a particular stationary point Mars' radius vector is determined to be 1.52 AU. Find the angle Θ that Earth would have advanced to reach this point and time elapsed between opposition and the next stationary point .

Solution:

We have:

cos Θ = [Q ^½ (1 + Q^½) ]/ (1 + Q^3/2)

where:

Q ^½=a^½ b^½ = (1) (1.52)^½ = 1.23


(1 + Q^3/2) = a^½ + b^3/2 = 1 + (1.52)^3/2 = 1 + 1.87 = 2.87

Then:

cos Θ = [(1.23) ^½ (2.23) ]/ (2.87)

cos Θ = [(1.10)(2.23)]/ (2.87) = 1.64/2.87 = 0.571

Θ = arc cos [0.571] = 55.1 deg

The time that elapses between opposition and the next stationary point is just:

t( Θ)/2= ½ [2 Θ/ 360 x S X (ΘS)/ 180 ]

Where: S = 779.9d/ 365.25 d = 2.13 yrs

So:


t( Θ)/2= ½ [(110.2)/ 360 x (2.13) X (55.1 x 2.13)/ 180 ]

t( Θ)/2= ½ [(0.306) (2.13) (0.652)] = 0.21 yrs.

Other problems:

1) Mars reaches a stationary point 36 ½ days after opposition. Its elongation is then measured to be 136.2 deg. Given that the sidereal period is 687 days, find the distance of Mars from Earth in AU. Also, find the time to its next stationary point.

2) Find the length of time Jupiter has retrograde motion in each synodic period given its heliocentric distance is 5.2 AU and its sidereal period is 11.86 years.

3) In the Epsilon Eridani star system a planet designated Epsilon Eridani III is determined to have the exact same orbital parameters as Earth (e.g. a, e, i etc.).

In the same system, another exoplanet designated Epsilon Eridani IV is found to have V_P = 4.5 km/s.

a) Using your knowledge of the known parameters, plus the diagram shown, construct an appropriate parallelogram of velocities and hence obtain the angles: Θ and φ.

b) Hence, or otherwise, estimate the time planet Epsilon Eridani IV will be moving retrograde relative to Epsilon Eridani III, and also the time between its opposition and the next stationary point.

c) Obtain the time during Epsilon Eridani IV's synodic period that it is moving direct.