## Monday, August 27, 2018

### 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   VA, VP  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:

VP = h/ a(1 - e)

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

For the velocity at aphelion:

VA = h/ a (1 + e)

Then the ratio of velocities is:

(VP/VA) = (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 VP and VA, along with a and e, we can compute a, viz.

a = ½VP2 - m /a(1 - e)

at perihelion, and

a = ½VA2 - 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 2½ =

[1   +    2(4.46 x 1015 J/kg ) 2 (-4.45 x 108 m2/s2)/ (1.33 x 1020 Nm2/kg)2½ =

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:

VP = h/ a(1- e) =

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

VP = 3.03 x 104 m/s  = 30, 300 m/s

and the velocity at aphelion can be obtained using:

VA = h/ a(1 + e) =

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

VA = 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 = ½VA2 - m /a(1 +  e)

at aphelion.

a = ½VP2 - 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

4The 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