_{}

^{}

*Solution*: We're given:

a = ½V

_{P}

^{2}- m /a(1 - e)

at perihelion, and

a = ½V

_{A}

^{2}- m /a(1 + e)

at aphelion.

Calculating:

½ V

_{P}

^{2}- m /[a(1-e)] =

½{3.03 x 10

^{4}m/s}^{2}- (1.33 x 10^{20}Nm^{2}/kg) / [1.496 x 10^{11}m(1 - 0.016)]-4.45 x 10

^{8}m

^{2}/s

^{2}

Similarly:

½V

_{A}^{2}- m/[a(1+ e)] =
½{2.93
x 10

^{4 }m/s}^{2}- (1.33 x 10^{20}Nm^{2}/kg) / [1.496 x 10^{11}m(1 + 0.016)]_{}

^{}

^{8}m

^{2}/s

^{2}

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: V

^{2}= 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 10

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

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

Substituting in the requisite radius vector values (r2, r1, but

We find:

V2 = 3.03 x 10

V1 = 2.93 x 10

^{20}Nm^{2}/kgFor 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 10*

^{11 }m) = 1.52 x 10^{11 }m )We find:

V2 = 3.03 x 10

^{4}m/s = 30, 300 m/sV1 = 2.93 x 10

^{4 }m/s = 29, 300 m/sThen: 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)

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 - e

^{2})]^{½}= 4.46 x 10^{15}J/kg
c3 = h cos(i) = (4.46 x 10

^{15}N-m/kg ) (cos (0.41) =
4.09 x 10

^{15}J/kg_{}^{}
Also: c2/ h = - cos W sin (i) so that:

c2 = - h cos W sin (i)

= - (4.46 x 10

^{15}N-m/kg ) (cos (0.197) sin(0.41))
c2 = - 1.74 x 10

^{15}J/kg
And: c1/ h = sin W sin (i)

So: c1 = h sin W sin (i) = 4.46 x 10

^{15}N-m/kg ) (sin (0.197) sin(0.41))
c1 = 3.48 x 10

^{14}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 10

^{15}J/kg )/ (4.46 x 10^{15}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 =

*h*of the form:h =

__+__[ m a(1 - e^{2})]^{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.

e

^{2 }= 1 + (2 h^{2}a)/ m^{2}
Subtract '1' from each side,

e

^{2 }- 1 = (2 h^{2}a)/ m^{2}
Then:

(m

^{2 })^{ }[^{ }e^{2 }- 1 ] = (2 h^{2}a)
Substitute: a = - m / 2a (energy constant)

(m

^{2 })^{ }[^{ }e^{2 }- 1 ] = (2 h^{2}) ( - m / 2a)
Or, simplifying, dividing both sides by m :

m [

^{ }e^{2 }- 1 ] = - h^{2}/ a
Or: h

^{2}= - a m [^{ }e^{2 }- 1 ]So h =

__+__[ m a(1 - e

^{2})]

^{½}

Now, compute: h =

**[c1**^{2}+ c2^{2 }+ c3^{2}]^{ ½}
h =

^{14}J/kg ) 2 + (- 1.74 x 10

^{15}J/kg) 2 + (4.09 x 10

^{15}J/kg )2 ] 1/2

_{}

^{}

_{}

^{}

= 4.46 x 10

^{15}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 10

^{30}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 (a

^{3}/m)

^{½}

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

T' = 2π [(a')

Then, dividing the bottom form by the top:

T'/T = {[(a')

Or:

m/m’ = (T'/T)

T' = 2π [(a')

^{3}/m')^{½}Then, dividing the bottom form by the top:

T'/T = {[(a')

^{3}/ a^{3}] [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:

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)

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 }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 10

^{3 }or: M = (1.059 x 10

^{3}) m

So, the mass of the Sun is (1.059 x 10

^{3}Jupiter's) or inverting:

m/M = 1/(1.059 x 10

^{3}) = 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 10

^{30}kg, so the mass of Jupiter based on the computed ratio would be:

m = (9.438 x 10

^{-4}) (1.99 x 10

^{30}kg) = 1.878 x 10

^{27}kg

which
compares to the (tabulated) mass of 1.89 x 10

^{27}kg^{22}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 10

^{22}kg then:

m(C) = 0.12 (1.27 x 10

^{22}kg ) = 1.52 x 10

^{21}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

c) The period is obtained from: T = 2π (a

where a = 19, 450 km = 1.94 x 10

Therefore: (V

_{P}/V_{A}) = 1.004c) The period is obtained from: T = 2π (a

^{3}/m)^{½}where a = 19, 450 km = 1.94 x 10

^{7}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 10

T = 5.5 x 10

^{7}m)^{3}/[6.7 x 10^{-11}Nm^{2}/kg^{2}) (1.422 x 10^{22}kg)]^{½}T = 5.5 x 10

^{5}s =__6.37 days__
The velocity is:

V = 2πa/T = 2π( 1.94 x 10

V = 2πa/T = 2π( 1.94 x 10

^{7}m )/(5.5 x 10^{5}s) =__221.6 ms__

^{-1}
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