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 140000 km. Calculate the perigee altitude and the time to perigee after the second sighting.

Question

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.

Accepted Answer

We have

[latex]P_{1}: \quad r_{1}=6378+267 000=273 378 km[/latex]
[latex]P_{2}: \quad r_{2}=6378+140 000=146 378 km[/latex]
Δt = 13.5 · 3600 = 48 600s
Δθ = 5°

Since [latex]r_{1}, r_{2}[/latex] and Δθ are given, we can skip to step 3 of Algorithm 5.2 and compute

[latex]A=2.8263 imes 10^{5} km[/latex]

Then, 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)
[latex]\dot{g}=0.92241[/latex]

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 [latex]r _{1}[/latex] 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

[latex]r _{1}=r_{1} \hat{ i }=273378 \hat{ i }( km )[/latex]
[latex]r _{2}=r_{2} \cos \Delta heta \hat{ i }+r_{2} \sin \Delta heta \hat{ j }=145820 \hat{ i }+12758 \hat{ j }( km )[/latex] (b)

With (a) and (b) we obtain the velocity at [latex]P_{1}[/latex],

[latex]v _{1}=\frac{1}{g}\left( r _{2}-f r _{1} ight)[/latex]
[latex]=\frac{1}{47708}[(145820 \hat{ i }+12758 \hat{ j })-0.95846(273378 \hat{ i })][/latex]
[latex]=-2.4356 \hat{ i }-0.26741 \hat{ j }( km / s )[/latex]

Using [latex]r _{1} ext { and } v _{1}[/latex], Algorithm 4.1 yields

h = 73 105km²/s
e = 1.0506
[latex] heta_{1}[/latex] = 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

[latex]r_{p}=\frac{h^{2}}{\mu} \frac{1}{1+e \cos (0)}=6538.2 km[/latex]

which means the perigee altitude is dangerously low for a large meteoroid,

[latex]z_{p}[/latex] = 6538.2 − 6378 = 160.2km (100 miles)

To find the time of flight from [latex]P_{2}[/latex] to perigee, we note that the true anomaly of [latex]P_{2}[/latex] is

[latex] heta_{2}= heta_{1}+5^{\circ}=210.16^{\circ}[/latex]

The hyperbolic eccentric anomaly [latex]F_{2}[/latex] follows from Equation 3.42,

[latex]F_{i+1}=F_{i}-\frac{e \sinh F_{i}-F_{i}-M_{h}}{e \cosh F_{i}-1}[/latex] (3.42)

[latex]F_{2}=2 anh ^{-1}\left(\sqrt{\frac{e-1}{e+1}} an \frac{ heta_{2}}{2} ight)=-1.3347 rad[/latex]

From this we appeal to Kepler’s equation (Equation 3.37) for the mean anomaly [latex]M_{h}[/latex],

[latex]M_{h_{2}}=e \sinh \left(F_{2} ight)-F_{2}=-0.52265 rad[/latex]

Finally, Equation 3.31 yields the time

[latex]M_{h}=\frac{\mu^{2}}{h^{3}}\left(e^{2}-1 ight)^{\frac{3}{2}} t[/latex] (3.31)

[latex]t_{2}=\frac{M_{h_{2}} h^{3}}{\mu^{2}\left(e^{2}-1 ight)^{\frac{3}{2}}}=-38 396 s[/latex]

The minus sign means that 38 396 seconds (a scant 10.6 hours) remain until the meteoroid passes through perigee.

Question 5.3: A meteoroid is sighted at an altitude of 267 000km. 13.5 hou...

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