programs throughout the worldwide community of
nations have resulted in various proposals for
avoiding impacts by Near Earth Objects
(NEOs). For success, they must addressing the overall
requirements for Detection, Prediction, and
Solvers saw in Figure 4 that simply setting off a thermonuclear blast on the surface of Égaré throws away most of its energy, and what's left irradiates the asteroid but does not show a whole lot of competence in the vectored-thrust department. Thus, we have concluded that our objective comes down to using part of the asteroid's own mass to produce the thrust.
Placement of Energy Source
Obviously, placing the thermonuclear explosion inside asteroid Égaré would be far more effective in discharging mass into space thereby generating thrust. Just imagine...
course, the expression 'effective yield' pertains
particularly to the Nuclear
Bunker Buster, which is designed to deliver
a nuclear warhead underground to destroy military
targets that have been buried in -- well,
'bunkers' -- and hardened by reinforced concrete
Solvers will find several references
with proposed features for a hardened penetrator,
from which we have made a few design choices for
the Asteroid-Buster, as
a theoretical spacecraft...
3. Depleted Uranium (D-38)
for inertial ballast inside the penetrator.
8. Propellant Tanks must be sized
for sufficient cryogenic
supplies to complete post-boost orbiting maneuvers
plus rendezvous, station-keeping and final
The length of section 4 Pneumatic Shock may actually be quite significant in determining the ultimate depth of the thermonuclear blast.
That will necessitate delivering the spacecraft from Earth to an orbit inside that of Égaré with the same eccentricity -- but slightly behind Égaré in phase. With its shorter orbital period, the satellite will gradually catch up to the asteroid. At precisely the right moment a ∆V Tangential burn must be autonomically administered to put the spacecraft into a Transfer Orbit. Finally, the spacecraft must be repositioned by Attitude Burns below the orbital plane, then turned toward the asteroid for the ∆V Normal Impact Burn.
Model for Proposed Solution
To the question posed by the Orbital Deflection puzzle poses this question...
...to which we have provided one answer in narrative form. Analyses with numerical specificity for the orbital mechanics and the detailed design of Asteroid Buster are beyond the scope of this entry in Puzzles with a Purpose. Nevertheless, in Figure 8 are a few elementary tools that some solvers may find useful for exploring these subjects further.
Let us assume that the asteroid has uniform density. Whatever the shape of the ablation boundary inside the asteroid, proportionality applies: mE / mA = (dP / dA)3, where…
dA = diameter of the asteroid; for Égaré, dA = 100 m.
dP = depth of the placement of the thermonuclear device
Using the limitation
cited above, dP Since mA
= 2.4×10 9
Using the limitation cited above,
dP= 6 m.
Since mA =
∆V = intentional change in asteroid’s velocity normal to its orbital plane
From the derivation on the puzzle page, ∆V = 11.7 m/s, thus vE = 54,000 m/s.
hY = heat energy yield from the thermonuclear explosion,
hL = heat energy loss eventually to space
...which is an estimate of net energy required for ∆V Normal and ∆i = 1/10
Base Case and Perturbations
The calculations in the model formulated above can be regarded as a Base Case. Our estimate for the required thermonuclear energy (181 KtnTNT) needs to be tested for its sensitivity to the assumptions in the model. Thus we will 'perturb' three key parameters ↕10% and ascertain the consequences by re-running the model.
Here is what is called a 'star diagram' for the results...
For Égaré, we
assumed dA to
be 100 m. As indicated in the diagram, if the
asteroid's actual diameter is 10% larger (110 m),
the required orbital deflection energy will be 321
KtnTNT, which is larger by 140 KtnTNT (77%).
But if the design of the Asteroid
Buster can increase its dP
by 10% (from 6 m to 6.6 m), the hihger
estimate for required deflection
energy might be reduced to 95 KtnTNT (53%).
Meanwhile, a reduction in energy yield (hY -
hL) by 10% (to 146 KtnTNT), will
decrease ∆V by
10% (to 10.5 m/s)
and ∆i by a