Question 5.3: A meteoroid is sighted at an altitude of 267 000km. 13.5 hou...
A meteoroid is sighted at an altitude of 267 000km. 13.5 hours later, after a change in true anomaly of 5°, the altitude is observed to be 140 000 km. Calculate the perigee altitude and the time to perigee after the second sighting.
We have
P_{1}: \quad r_{1}=6378+267 000=273 378 km
P_{2}: \quad r_{2}=6378+140 000=146 378 km
Δt = 13.5 · 3600 = 48 600s
Δθ = 5°
Since r_{1}, r_{2} and Δθ are given, we can skip to step 3 of Algorithm 5.2 and compute
A=2.8263 \times 10^{5} kmThen, solving for z as in the previous example, we obtain
z = −0.17344
Since z is negative, the path of the meteoroid is a hyperbola.
With z available, we evaluate the Lagrange functions,
f = 0.95846
g = 47 708s (a)
\dot{g}=0.92241
Step 7 requires the initial and final position vectors. Therefore, for the purposes of this problem let us define a geocentric coordinate system with the x axis aligned with r _{1} and the y axis at 90° thereto in the direction of the motion (see Figure 5.6). The z axis is therefore normal to the plane of the orbit. Then
r _{1}=r_{1} \hat{ i }=273378 \hat{ i }( km )
r _{2}=r_{2} \cos \Delta \theta \hat{ i }+r_{2} \sin \Delta \theta \hat{ j }=145820 \hat{ i }+12758 \hat{ j }( km ) (b)
With (a) and (b) we obtain the velocity at P_{1},
v _{1}=\frac{1}{g}\left( r _{2}-f r _{1}\right)
=\frac{1}{47708}[(145820 \hat{ i }+12758 \hat{ j })-0.95846(273378 \hat{ i })]
=-2.4356 \hat{ i }-0.26741 \hat{ j }( km / s )
Using r _{1} \text { and } v _{1}, Algorithm 4.1 yields
h = 73 105km²/s
e = 1.0506
\theta_{1} = 205.16°
The orbit is now determined except for its orientation in space, for which no information was provided. In the plane of the orbit, the trajectory is as shown in Figure 5.6.
The perigee radius is
which means the perigee altitude is dangerously low for a large meteoroid,
z_{p} = 6538.2 − 6378 = 160.2km (100 miles)
To find the time of flight from P_{2} to perigee, we note that the true anomaly of P_{2} is
\theta_{2}=\theta_{1}+5^{\circ}=210.16^{\circ}The hyperbolic eccentric anomaly F_{2} follows from Equation 3.42,
F_{i+1}=F_{i}-\frac{e \sinh F_{i}-F_{i}-M_{h}}{e \cosh F_{i}-1} (3.42)
F_{2}=2 \tanh ^{-1}\left(\sqrt{\frac{e-1}{e+1}} \tan \frac{\theta_{2}}{2}\right)=-1.3347 radFrom this we appeal to Kepler’s equation (Equation 3.37) for the mean anomaly M_{h},
M_{h_{2}}=e \sinh \left(F_{2}\right)-F_{2}=-0.52265 radFinally, Equation 3.31 yields the time
M_{h}=\frac{\mu^{2}}{h^{3}}\left(e^{2}-1\right)^{\frac{3}{2}} t (3.31)
t_{2}=\frac{M_{h_{2}} h^{3}}{\mu^{2}\left(e^{2}-1\right)^{\frac{3}{2}}}=-38 396 sThe minus sign means that 38 396 seconds (a scant 10.6 hours) remain until the meteoroid passes through perigee.
