Use Of Basic Celestial Mechanics To Obtain Some Earth Orbital Parameters

In this post, basic celestial mechanics will be used to obtain some basic information on Earth's orbital dynamics, including: velocity at aphelion and perihelion points, magnitude of energy given by energy constant a,  eccentricity e, specific relative angular momentum h, and derivation of the "vis viva" equation which enables one to obtain the velocity at any point in the orbit. (The latter is also known as the "orbital energy invariance law" because the total energy is the same despite how the velocity v varies over the orbit. )

To fix ideas we focus on the diagram shown above. Let the points A and P denote the aphelion (farthest point) and perihelion (closest point) to the Sun (S), respectively. We let   V_A, V_P  be the respective velocities at those orbital extrema. As may be deduced here, points A and P are the only ones in the whole orbit for which the velocities are truly tangential or at right angles to the radius vectors for those positions. Consequently, we can write:

V = (2π/T) r

where r is the radius vector at the point, and T is the period.

 If Kepler's 2nd (equal areas)  law holds at every point (i.e. equal areas swept out in equal intervals of time) we also have:

r² (2π/T) = h 

where 'h' is a constant ('specific relative angular momentum') which is twice the rate of area description (i.e. by the radius vector). Thus, if the radius vector is r1, then h = 2A1, when A1 = π(r1)². Hence, at aphelion and perihelion only we have: 

V = h/r    for which r = a(1   +  e)  OR:   r =  a (1  -  e)

For the perihelion velocity we have:

V_P = h/ a(1 - e)

where a is the semi-major axis, and e is the eccentricity.

For the velocity at aphelion:

V_A = h/ a (1 + e)

Then the ratio of velocities is:

(V_P/V_A) = (1 + e)/ (1 - e)

The correct energy ("vis viva")  equation can be written:

½V² - m /r = a

where a is an energy integration constant.   More conventionally it is written
 without  the energy constant (see derivation at end of post):

V² = m   (2/r - 1/a )


Energy constants in celestial mechanics are very useful for quickly coming to terms with specific properties of an orbit such as shown in the  more detailed accompanying sketch- designating a generic orbit in x-y-z space, e.g.. 


In the diagram, w   is the argument of the perihelion, W is the longitude of the ascending node , f is the true anomaly and i is the inclination of the orbit. The critical or key parameter here is h, the angular momentum vector for the orbiting system.

Getting specific, assuming r and r' are r (radius vector) and d r/dt, respectively, the magnitude h, of the angular momentum vector is:

h = r x r’ =

(y z’ - z y’)

(z x’ - x z’) = (c1 c2 c3)

(x y’ - y z’)

so:   (r x r’) = (c1/ h, c2/ h, c3/h)

Where c1, c2 and c3  are integration constants that determine the orientation of the orbital plane.

Inserting angular orbital elements (i, W) one finds:

c1/ h = sin W sin (i)

c2/ h = - cos W sin (i)

c3/h = cos(i)

Now since the inclination of Earth's orbit to the ecliptic  (i) is known (23.5 deg) and therefore cos(i) can be determined, then sin(i) can be as well.  Also, h can be determined, since: h = c3 / cos(i) .  (Also h =  [c1 ² +  c2 ²   +  c3 ²]  ^½)   We also know  W =  11.26 deg.

Since for any bound system of masses m1 and m2, m = G (m1 + m2), where G is the Newtonian gravitational constant (G = 6.7 x 10^-11 Nm²/kg²) then if we know V_P and V_A, along with a and e, we can compute a, viz.

a = ½V_P² - m /a(1 - e)

at perihelion, and

a = ½V_A² - m/a(1 - e)

at aphelion. 

For the Earth-Sun system :

m= 1.33 x 10²⁰ Nm²/kg

(Note: for m, we already know G and m1= 1.99 x 10³⁰ kg (Sun's mass) and m2 = 6.4 x 10²⁴ kg, (Earth's mass)

Also: a (semi-major axis)  = 1.496 x 10¹¹ m

Then h = + [m a(1 - e²)]^½ =    4.46 x 10¹⁵ N-m/kg = 4.46 x 10¹⁵ J/kg 

The energy constant a =   - m/ 2a   for an elliptical orbit

So:   

a =    -(1.33 x 10²⁰ Nm²/kg) / 2 (1.496 x 10¹¹ m)  =  = -4.45 x 10⁸ m²/s² 

The eccentricity of the orbit e, can now be obtained from:

e =   [1   +   (2 h ²  a )/ m ² ] ^½ =


[1   +    2(4.46 x 10¹⁵ J/kg ) ² (-4.45 x 10⁸ m²/s²)/ (1.33 x 10²⁰ Nm²/kg)² ] ^½ =

0.016

What about the velocities at perihelion and aphelion?

Since we have obtained h and e,    the velocity at perihelion is easy to calculate

from:

V_P = h/ a(1- e) =

4.46 x 10¹⁵ J/kg / [1.496 x 10¹¹ m(1 - 0.016)]

V_P = 3.03 x 10⁴ m/s  = 30, 300 m/s

and the velocity at aphelion can be obtained using:

V_A = h/ a(1 + e) =

4.46 x 10¹⁵ J/kg / [1.496 x 10¹¹ m(1 + 0.016)]

V_A = 2.93 x 10⁴ m/s    = 29, 300 m/s

 Now, how would the vis viva equation (given earlier) be derived? 

From the earlier energy constant equations (at aphelion, perihelion):

a = ½V_A² - m /a(1 +  e)

at aphelion. 

a = ½V_P² - m /a(1 - e)

at perihelion.

Then,   we may write without  loss of generality:

 ½V² - m/r   = a  =   - m/ 2a

Or:

½V² =   m/r    - m/ 2a

And:   

V² =   2 [  m/r    - m/ 2a ]    

Whence:

V² = m   (2/r - 1/a)

Comprehension Problems:

1) Show that the energy constant  a   is the same at aphelion and perihelion.

2)  The Earth's aphelion distance is 1.01671 AU and its perihelion distance is 0.98329 AU. Use the vis viva equation to obtain the difference in velocity between the two points.

3) Calculate the three integration constants applicable to the orientation of the Earth's orbital plane.: c1, c2 and c3,    In standard practice these are already computed, then used to obtain the longitude of the ascending node, W   and the inclination, i.  Show how this could be done.

4) Derive the independent expression for h of the form:

h = + [m a(1 - e²)]^½

And show it is equal to:  h =  [c1 ² +  c2 ²   +  c3 ²]  ^½

4) The orbital period of Jupiter's 5th satellite is 0.4982 days about the planet. Its orbital semi-major axis is 0.001207 AU. The orbital period and semi-major axis of Jupiter are 11.86 yrs. and 5.203 AU. Estimate the ratio of the mass of Jupiter to that of the Sun.

5)  For the Pluto-Charon system,  the orbit of Pluto's moon Charon has an eccentricity e = 0.0020. The semi-major axis of the orbit is 19, 450 km. The mass of Pluto = 1.27 x 10²² kg and the mass ratio (Charon to Pluto) is found to be m(c)/m(P) = 0.12. From this information, find:

a) The mass of Charon

b) The ratio of the velocity of Charon at perihelion to aphelion

c) The period of Charon, and its velocity

Another Crankster Exposed With Kepler's 2nd Law

In a post from a year and a half ago, I wrote:

"One thing that makes volunteering at 'All Experts' interesting is a particular class of questions received: usually when a person believes he or s he has disproven a fundamental physical or astronomical law or principle."

Thus, as an expert in astronomy and astrophysics the past 13 years (on 'All Experts') I have beheld not only cranks who claim to have overturned Einstein's theories of relativity (special and general) , but also basic laws -principles such as Kepler's Second law or area law. If Kepler's 2nd law holds at every point (equal areas swept out in equal intervals of time) we have for an elliptical orbit described over some interval T::

(2πab /T) = h =   2πa² Ö (1 - e ² )  / T

Where a = the semi-major axis and b is the semi minor axis, and h is a constant known as the 'specific relative angular momentum'.  Recall my takedown of one of these characters, which I discussed here:

http://brane-space.blogspot.com/2015/08/resolving-basic-astronomical-error-in.html


Now another crank has  recently surfaced with equally ridiculous claims, writing.

"Kepler's area law says r*Vp=Ct.  Newton's universal attraction force says this force is radial F=Fr and a perpendicular force (Fp)  to the radial, also a side force component does not exist.  So: m*dVp/dt=Fp=0.  Then dVp/dt=0 and with integration we get Vp=Ct.     f Vp=Ct is correct, elliptical orbits theory has to be modified to a new theory, new math. And the motion equation should be r=-4*t^2+4*t*T-4*T^2/6 .This equation does not indicate an ellipse but a parabolic vortex spiral."

Note several points:

1) He has not expressed the 2nd law in proper form as I showed in the preceding link and also below..   Indeed it makes no sense at all.   The proper  basis form (to obtain h)  is:

  d/dt (r^  x  a^) =  0  or:  r^  x  a^ =  constant  = h

And note the related case: r^  x  v^ =  constant   applies to the conservation of angular momentum.

But even that still leaves out factors.  Also, he has not defined 'C', and if it is the same as h he doesn't say so one can't just assume it.

2) His equation:  m*dVp/dt=Fp=0  is incorrect. As I note in the development below the correct form for the relevant forces is:

(m1)  d²   r1/ dt ² =  G m1m2 (r^)/ r ³   = (m2) d²   r2/ dt ²

Hence, he's failed to distinguish between a non-existent force, i.e.  perpendicular to the orbit (since the gravitational force of attraction acts through the mass centers)  and one which  exists but is zero in a given situation.   For example, a mass m in an orbiting spacecraft for which g = 0  (in reality the mass falling at the same rate as the space craft in its orbit) so mg = 0, i.e. weightlessness. But if the space craft is imparted a rotation then artificial gravity  can be created so we have an acceleration g'  and mg' is non-zero. But one cannot magically induce a perpendicular force "Fp" to an orbit where one could never exist before.

In effect, his integration result, e.g. Vp=Ct  is spurious and his claim for a "new theory, new math"  and "parabolic vortex spiral" falls of its own vacuous weight.

Then his actual question:

"Now which expression is correct.r*Vp=Ct ? or Vp=Ct?  Kepler's law or Newton's law.If Vp=Ct is correct ,elliptical orbits theory has to be modified to a new theory, new math."

Is specious because of the reasons given.  Crank hood is evident in that the question is also posed as a false option. What he thinks is Newton's "law" (Newton's 2nd law) is actually nothing of the sort because it's based on a non-existent entity. Newton's 2nd law only applies to real forces even if they might be zero in certain situations.  (E.g. weightlessness) 

I had at first attempted to use similar arguments (and diagrams) to those given in the earlier link. He then responded that I "didn't  even use Newton's laws"  to form the physical basis so I presented an alternative argument based on those.

I referred him to the diagram labelled Fig. 1 a. above

Here the two masses shown are  planet P (m1 ) and the Sun S (m2). For m2 in the field of m1 we can write:   F21 = G m1m2 (r^)/ r ³

Similarly for m1 in the field of m2:     F12 =  G m2m1 (r^)/  r ³

By Newton’s 2^nd law used in the given context:

(m1)  (d²   r1/ dt ²) =  G m1m2 (r^)/ r ³

And:

(m2) (d²   r2/ dt ²) = -  G m1m2 (r^)/ r ³

But by Newton’s 3^rd law, e.g. F12 = - F21. Or:

(d²   r1/ dt ²) =  G m1m2 (r^)/ r ³   = (m2) (d²   r2/ dt ² )

Integration of the above leads to:

(m1) d(r1)/dt  +  (m2) d(r1)/dt  =  p

Where p  is momentum and v1 = d(r1)/dt     and v2 = d(r1)/dt 

Or:   m1r1 + m2r2 = pt  + q^

Where q^ is a constant displacement vector

With further working for a center of mass system consisting  of (m1 + m2) at R we can write:

m1r1 + m2r2 = (m1 + m2)R

Combining the expressions with the 2^nd derivatives seen earlier:

(d²   r / dt ²) =   G m1m2 (r^)/ r ³ 

Note the acceleration (LHS) is always anti-parallel to the vector r

It can then be shown, letting   m = G (m1 +  m2):

r^  x  a^ =  (m / r ³ ) (r^) x r^ = 0

Since the vector product of any vector with itself is zero:

d/dt (r^  x  a^) =  0  or:  r^  x  a^ =  constant

This  quantity is exactly h used in the Kepler 2^nd law, and a constant of the motion.  In effect, with a bit of further working:

dA/ dt =   h z^/ 2

This is   the proper form for the 2nd law that the latest challenger  seems not to have heard about.  As I showed,  what he describes as Kepler's  2nd law is nothing of the sort. The most hysterical part of his claim - after being informed that if his "theory" was correct no space craft would ever reach another planet (since trajectory computation is partly based on the 2nd law) - is the following:

"Anyhow,  sending celestial probes to the Moon,or Mars or to any body will still be successful even if the orbits are triangular. It does not depend on  the form of the planet's trajectory . It is controlled from the earth."

In other words, he's saying you just need to fire the rocket into space, and the human controllers on Earth (like at Johnson Space Center) will do the rest,  to "steer" it to the destination.   No computations required, such as depicted in the recent film 'Hidden Figures'. Totally out of crank left field!

Another way the 2nd law can be derived is shown below, based on Figure 1a.


Fig. 1 (b) showing how to derive the 2nd law

This  can also be used  to show the geometrical significance of h. The diagram shows the rate at which the radius vector joining an origin O to a moving point P sweeps out a surface. This rate is called the "areal velocity relative to the origin O.

Let   D A  =    ½  [r   x (r +  D  r )] =   r^  x   Dr  

Now, the areal velocity at P  is by definition:

lim Dt _®0  (D A /D t)  

Then :  dA/ dt =  ½ r   x  r'  =   ½ h  

The moral of this story is that if you're going to challenge the validity of the Areal law at any serious level, you will need to have all your 'ducks' in a row. And especially -  if your math and physics are proven wrong -  you need to admit the putative space craft will not get to its destination - as opposed to saying:   "Sending celestial probes to the Moon  or Mars or to any body will be successful even if the orbits are triangular It does not depend on  the form of the planet's trajectory .It is controlled from the earth." 

Such a remark disqualifies you instantly from being a serious planetary theoretician, and also merits being put onto my growing astronomical crank list. Never mind, by next month another crank will offer his pet idea of why the universe and astronomical, physical laws aren't operating the way they should. And he (or she) is the only one to have the "correct" form!

On the upside, at least I don't have to toss away some twenty or so "scientific papers"  sent by cranks per week as MIT physicist Philip Morrison once wryly observed he had to do.