Solutions To Celestial Mechanics Problems

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

Solution:  We're given:

a = ½V_P² - m /a(1 - e)

at perihelion, and 

a = ½V_A² - m /a(1 +  e)


at aphelion.

Calculating:

½ V_P ² - m /[a(1-e)]   =   

½{3.03 x 10⁴ m/s}²   - (1.33 x 10²⁰ Nm²/kg) / [1.496 x 10¹¹ m(1 - 0.016)]

=
-4.45 x 10⁸ m²/s²

Similarly:

½V_A² - m/[a(1+ e)]   =   

  ½{2.93 x 10⁴ m/s}²  -   (1.33 x 10²⁰ Nm²/kg) / [1.496 x 10¹¹ m(1 + 0.016)]

=    -4.45 x 10⁸ m²/s²

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.

Solution:

The vis viva equation is:   V² = m (2/r - 1/a)  or:  V =   [ m (2/r - 1/a)] ^½

We let:  r1 =   1.01671 AU  (Aphelion distance)

And r2= 0.98329 AU         (Perihelion distance)

Then the difference in velocity between the two points will be:

D V =   V2 - V1    where:

 m= 1.33 x 10²⁰ Nm²/kg 

For aphelion:  V1  =  [ m (2/r1 - 1/a)] ^½

For perihelion: V2 =  [ m (2/r2 - 1/a)] ^½

 Substituting in the requisite radius vector values (r2, r1, but converted to meters, e.g.

1.01671 AU =(1.01671) (1.496 x 10¹¹ m) =   1.52 x  10¹¹ m )


 We find:

 V2 =  3.03 x 10⁴ m/s  = 30, 300 m/s

V1  =  2.93 x 10⁴ m/s    = 29, 300 m/s

  Then:  V2 - V1 =  30, 300 m/s   -    29, 300 m/s  =    1,000 m/s 

 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.

Solution:

We note that the constants c1, c2  and c3 can be obtained from:

c1/ h = sin W sin (i)

c2/ h = - cos W sin (i)

c3/h = cos(i)

Since  the inclination of Earth's orbit to the ecliptic  (i) is known (23.5 deg) and   W =   11.26 deg. But these angles must first be converted to radians, i.e. by dividing each by 57.3 deg,  given 57.3 degrees per radian measure. We compute c3 first since there is only one angle unknown (i):

c3 = h    cos(i)

Also, h was computed independently in the post, i.e.

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

c3 = h    cos(i) =  (4.46 x 10¹⁵ N-m/kg ) (cos (0.41) = 

4.09 x 10¹⁵ J/kg 

 Also:  c2/ h = - cos W sin (i)  so that:

c2  = - h cos W sin (i)

=  - (4.46 x 10¹⁵ N-m/kg ) (cos (0.197) sin(0.41)) 

c2 =    - 1.74  x 10¹⁵ J/kg 

And:  c1/ h = sin W sin (i)

So:  c1 = h  sin W sin (i) = 4.46 x 10¹⁵ N-m/kg ) (sin (0.197) sin(0.41))

c1 =  3.48 x 10¹⁴ J/kg 

Now, assuming the constants are already known, it would be the two angles defining the orientation of the orbital plane one needs to compute. So we  would first obtain the angle of inclination i from:

cos i = c3/ h 

From which i = arc cos [c3/h]

=  arc cos [(4.09 x 10¹⁵ J/kg )/ (4.46 x 10¹⁵  J/kg)]

i = arc cos (0.917) =  0. 41 radians 

So in degrees:  i  =   (0.41 rad) (57.3 deg/ rad) = 23.5 deg (approx.)

Similarly,   sin W   =  c1/ h sin i  =   c1 / h sin (0.41)

So that: W   =  arc sin [c1 / h sin (0.41)]

with analogous substitutions  which are left to the reader to confirm.

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

Solution:

For an ellipse, the eccentricity is defined in celestial mechanics:

e =   [1   +   (2 h ²  a)/ m ² ] ^½

Then, to get h one only needs to solve for it algebraically:

So first square both sides, viz.

 e²   =   1   +   (2 h ²  a)/  m ²

Subtract '1' from each side,

e²   -   1   =     (2 h ²  a)/ m ²

Then:

(m ² )  [  e²   -   1 ]  =     (2 h ²  a)

Substitute:   a =   - m / 2a   (energy constant)

(m ² ) [ e²   -   1 ]  =     (2 h ² ) ( - m / 2a)  

Or, simplifying, dividing both sides by  m :

m [  e²   -   1 ]  =   -  h ² / a 

Or:   h ² =   - a m [ e²   -   1 ]

So h = + [ m  a(1 - e²)] ^½

Now, compute:   h =  [c1 ² +  c2 ²   +  c3 ²]  ^½

h =

  [( 3.48 x 10¹⁴ J/kg ) 2   + (- 1.74  x 10¹⁵ J/kg) 2  +   (4.09 x 10¹⁵ J/kg )2 ] 1/2

=    4.46 x 10¹⁵ J/kg


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. (M = 1.99 x 10³⁰ kg)


Solution:

 In this problem we are dealing with Jupiter (planet) and its 5th satellite, with all the respective parameters given. Then we  have to pair two different ratios in a, T (semi-major axis, period)  i.e. of (1) the satellite (moving around Jupiter) to (2) Jupiter (moving around the Sun). We simply apply the same type of approach as shown with P1, P2 to a1, a2, thus: 

For the planet and Sun,  using Kepler's 3rd law:

T = 2π (a³/m)^½ 

For the planet and satellite (Jupiter and its 5th satellite):

T' = 2π [(a')³/m')^½

Then, dividing the bottom form by the top:

T'/T = {[(a')³/ a³ ] [m/m']}^½

Or:

m/m’ = (T'/T)² (a/a')³

where m = (M + m) and m' = (m + m') 

Given G (Newtonian gravitational constant) cancels out, M= Sun's mass, m = Jupiter's mass, and m' = satellite's mass.

 This is simplified by reducing a, T to AU and yrs, so that: a = 5.203 AU, a' = 0.001207 AU, T = 11.86 yrs. and T' = (0. 4982/365.25) yr = 0.00136 yr. Then: 

(M + m)/(m + m') =

 (0.00136/11.86)² (5.203/0.001207)³

Now the left side can be posed (by appropriate algebraic manipulation):

(M + m)/(m + m') = (M/m) [(1 + m/M)/(1 + m'/m)]

and in the limit of m/M >>> 1 we can write:

M/m = (0.00136/11.86)² (5.203/0.001207)³

= 1.059 x 10³ 

or: M = (1.059 x 10³) m

So, the mass of the Sun is (1.059 x 10³ Jupiter's) or inverting:

m/M = 1/(1.059 x 10³) = 9.438 x 10^-4

I.e. The ratio of Jupiter's mass to that of the Sun. 
-----
 We can check this accuracy by noting the mass of the Sun is given as 1.99 x 10³⁰ kg, so the mass of Jupiter  based on the computed ratio would be:

m = (9.438 x 10^-4) (1.99 x 10³⁰ kg) = 1.878 x 10²⁷ kg

which compares to the (tabulated) mass of 1.89 x 10²⁷ kg 

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

Solution:

a) The mass is found using the mass ratio. Since m(c)/m(P) = 0.12 and m(P) =1.27 x 10²² kg  then:

m(C) = 0.12 (1.27 x 10²² kg ) = 1.52 x 10²¹ kg

b) This ratio is obtained from:

(V_P/V_A)  = (1 + e)/ (1 - e)

and we know, e = 0.0020, so:

(V_P/V_A)  = (1 + 0.0020)/ (1 - 0.0020) =

 (1.0020)/ (0.998) = 1.004

Therefore:  (V_P/V_A)  = 1.004

c) The period is obtained from: T = 2π (a³/m)^½

where a = 19, 450 km = 1.94 x 10⁷ m

For preserving unit consistency this is one time we decline to use units of AU and years, and instead use the distance in meters to conform with the required units for G. Then: T =

2π [(1.94 x 10⁷ m)³/[6.7 x 10^-11 Nm²/kg²) (1.422 x 10²² kg)]^½

T = 5.5 x 10⁵ s = 6.37 days

The velocity is:

V = 2πa/T = 2π( 1.94 x 10⁷ m )/(5.5 x 10⁵ s) =

 221.6 ms^-1