Problems on Orbits
Problem :
In Solving the Orbits we derived the
equation:
From this, derive the expression we stated for
1/r. Hint: Define
y = 1/r
and use the fact that
=  .
Making the
y substitution, we have:
We can then complete the square on the righthand side and we have:
Then we let
p = y  and we have:
=  p^{2} + 1 + =  p^{2} + C 

where we have just defined
C in terms of the constant
E, G, M, m, L. We can
now separate variables:
= dθ'âá’cos^{1}(p'/)^{p}_{p1} = θ  θ_{1}âá’p = cos(θ  θ_{1})  cos^{1}(p_{1}/) = cos(θ  θ_{0}) 

Finally, recalling
p = (1/r)  , and we can pick an axis such
that
θ_{0} = 0:
The quantity under the radical is defined a
ε, the eccentricity.
Problem :
Using the expression we derived for (1/r), show that this reduces to x^{2} = y^{2} = k^{2} 2kεx + ε^{2}x^{2}, where k = ,
ε = , and cosθ = x/r.
We have:
= (1 + εcosθ)âá’1 = (1 + ε)âá’k = r + εx 

We can solve for
r and then use
r^{2} = x^{2} + y^{2}:
x^{2} + y^{2} = k^{2}–2kxε + x^{2}ε^{2} 

which is the result we wanted.
Problem :
For 0 < ε < 1, use the above equation to derive the equation for an
elliptical orbit. What are the semimajor and semiminor axis lengths? Where
are the foci?
We can rearrange the equation to
(1  ε^{2})x^{2} +2kεx + y^{2} = k^{2}. We can divide through by
(1  ε^{2}) and complete the square in x:
Rearranging this equation into the standard form for an ellipse we have:
+ = 1 

This is an ellipse with one foci at the origin, the other at
(, 0), semimajor axis length
a = and semiminor axis length
b = .
Problem :
What is the energy difference between a circular earth orbit of radius 7.0×10^{3} kilometers and an elliptical earth orbit with apogee 5.8×10^{3} kilometers and perigee 4.8×10^{3} kilometers. The mass
of the satellite in question is 3500 kilograms and the mass of the earth is
5.98×10^{24} kilograms.
The energy of the circular orbit is given by
E =  = 9.97×10^{10} Joules. The equation used here can also be applied to elliptical
orbits with
r replaced by the semimajor axis length
a. The semimajor axis
length is found from
a = = 5.3×10^{6} meters. Then
E =  = 1.32×10^{11} Joules.
The energy of the elliptical orbit is higher.
Problem :
If a comet of mass 6.0×10^{22} kilograms has a hyperbolic orbit around
the sun of eccentricity
ε = 1.5, what is its closest distance
of approach to the sun in terms of its angular momentum (the mass of the sun is
1.99×10^{30}
kilograms)?
Its closest approach is just
r_{min}, which is given by:
r_{min} = = (6.44×10^{67})L^{2} 
