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