Impact Wrench Case
Impact Wrench Case
The Adventures of the Mischievous Astronaut. Can you solve the Case of the Smashed Sensor?
NASA has a sensor traveling in Earth's orbit at a considerable distance from Earth. A disgruntled, underpaid astronaut, traveling in Mars' orbit at a considerable distance from Mars, decides to destroy the sensor to avenge his low income.
He calculates that he can drop a wrench on the sensor by negating the wrench's velocity, relative to the sun, at the moment he passes through the descending node of Mars' orbit. He attaches some rockets to the wrench, programs the necessary delta-vee, and awaits its departure time with a big, evil grin.
What are the rectangular components of the sun-relative delta-vee in ecliptic coordinates?
How long does it take for the wrench to reach the sensor?
Assuming the wrench hits the sensor, what is the impact speed, relative to the sensor?
Mars' longitude of ascending node,
L = 49.57854 degrees
So the longitude of the descending node,
L' = 229.57854 degrees
Mars' argument of perihelion,
w = 286.46230 degrees
Since w is the angle, subtended at the sun, from the ascending node to the perihelion, the angle from the descending node to the perihelion would be 180 degrees less, or 106.46230 degrees. Subtracting that result from 360 degrees gives the true anomaly of the descending node,
Q = 253.53770 degrees.
Mars' semimajor axis,
a = 1.52366231 AU
Mars' orbital eccentricity,
e = 0.09341233
The heliocentric distance, r, is given as a function of true anomaly by
r = a (1 - e^2) / (1 + e cos Q)
So the heliocentric distance of Mars' descending node is
r = 1.551436 AU
Since the astronaut is on the line of nodes, he is in the ecliptic plane as well as in Mars' orbital plane. His HEC position vector is
X = r cos(L') = -1.005959 AU
Y = r sin(L') = -1.181101 AU
Z = 0
Let's define a constant,
V0 = sqrt{ GM / [ a (1 - e^2) ] } = 24235.493 m/s
There GM is the solar gravitational parameter,
GM = 1.32712440018E+20 m^3 sec^-2
And the semimajor axis was converted to meters by the unit ratio,
1 AU = 1.49597870691E+11 meters
The astronaut's velocity in Mars' orbit at its descending node is found from
Vx''' = -v0 sin Q = +23241.993 m/s
Vy''' = +v0 (e + cos Q) = -4604.067 m/s
Vz''' = 0
We rotate by Mars' argument of the perihelion, w=286.46230 degrees.
Vx'' = Vx''' cos w - Vy''' sin w = +2171.093 m/s
Vy'' = Vx''' sin w + Vy''' cos w = -23593.940 m/s
Vz'' = Vz''' = 0
We rotate by Mars' inclination to the ecliptic, i=1.85061 degrees.
Vx' = Vx'' = +2171.093 m/s
Vy' = Vy'' cos i = -23581.634 m/s
Vz' = Vy'' sin i = -761.934 m/s
We rotate by Mars' longitude of ascending node, L=49.57854 degrees.
Vx = Vx' cos L - Vy' sin L = +19360.340 m/s
Vy = Vx' sin L + Vy' cos L = -13637.608 m/s
Vz = Vz' = -761.934 m/s
And [Vx, Vy, Vz] is the sun-relative velocity (in ecliptic coordinates) of the astronaut as the wrench's rockets fire for departure. Since the objective is to neutralize the wrench's sun-relative speed, the delta-vee must be equal in magnitude and opposite in sign for all components.
dVx = -Vx = -19360.340 m/s
dVy = -Vy = +13637.608 m/s
dVz = -Vz = +761.934 m/s
Check:
sqrt(dVx^2 + dVy^2 + dVz^2) = 23693.621 m/s
sqrt[ GM (2/r - 1/a) ] = 23693.621 m/s
Those must be some very good rockets, to boost the wrench by that much delta-vee.
Since we neutralized the sun-relative velocity of the wrench, it will begin to fall radially toward the sun; i.e., along a path of constant heliocentric latitude (zero) and longitude (229.57854 degrees). The next thing to determine is the heliocentric distance of Earth's orbit where it intersects that path.
Earth's longitude (and argument) of perihelion is
we = 102.93768193 degrees
So the true anomaly of Mars' descending node in Earth's orbit is
Qe = L' - we = 126.64086 degrees
Earth's semimajor axis is
ae = 1.00000011 AU
Earth's eccentricity is
ee = 0.01671123
The heliocentric distance of that NASA sensor that the naughty astronaut is trying to ruin is
re = ae (1 - ee^2) / (1 + ee cos Qe) = 1.009792 AU
So after the rockets complete the delta-vee, the wrench will fall radially into the sun, initially from rest with respect to the sun, from r=1.551436 AU to re=1.009792 AU.
r = 2.32091522113E+11 meters
re = 1.51062690440E+11 meters
The time required for an object of negligible mass to fall radially toward the sun from r to re in a plunge orbit having a half-aphelion of a=r/2, is given by
dt = [a / sqrt(GM)] { r sqrt( 2/r - 1/a ) + 2 sqrt(a) Arctan[ sqrt(2a/r -1) ] } | (r,re)
So the wrench will require a time, dt, from the departure delta-vee to reach the NASA sensor:
dt = 7609951.1 seconds
dt = 88.078138 days
The aphelion radial velocity is, of course, zero, since it was at rest with respect to the sun immediately after the delta-vee.
We find the radial velocity at Earth orbit intercept from the Vis Viva equation:
The "semimajor axis" (or, rather the half-aphelion that plays its role) of the plunge is
a = r/2 = 0.775718 AU
Vr = sqrt[ GM (2/re - 1/a) ] = 24767.5006 m/s, toward the sun
The ecliptic rectangular components are found by noticing that the sun, with respect to the wrench, is in the same direction as Mars' ascending node, meaning that the longitude toward which the wrench is moving is
L = 49.57854 degrees
Vx = Vr cos L = +16059.3734 m/s
Vy = Vr sin L = +18855.3868 m/s
Vz = 0
Here, [Vx,Vy,Vz] is the sun-relative velocity of the wrench as it crosses Earth's orbit, in ecliptic coordinates.
The velocity of the NASA sensor is found as follows:
We define another constant,
v0e = sqrt{ GM / [ ae (1 - ee^2) ] } = 29789.0700 m/s
Vxe''' = -v0e sin Qe = -23902.5138 m/s
Vye''' = +v0e (ee + cos Qe) = -17280.2229 m/s
Vze''' = 0
We rotate by Earth's argument of perihelion, we=102.93768193 degrees.
Vxe = Vxe''' cos we - Vye''' sin we = +22193.1110 m/s
Vye = Vxe''' sin we + Vye''' cos we = -19426.8395 m/s
Vze = 0
Since for Earth's orbit the inclination and longitude of ascending node are defined as zero, we're done with the rotating, and [Vxe,Vye,Vze] is the sun-relative velocity of the sensor in ecliptic coordinates.
The impact speed of the wrench, relative to the sensor, is the vector difference:
dVx = Vx - Vxe = +16059.3734 m/s - 22193.1110 m/s
dVy = Vy - Vye = +18855.3868 m/s - (-19426.8395 m/s)
dVz = Vz - Vze = 0 - 0
dVx = -6133.7376 m/s
dVy = +38282.2263 m/s
dVz = 0
dV = sqrt(dVx^2 + dVy^2 + dVz^2) = 38770.5 m/s
There dV is the impact speed of the wrench, relative to the sensor.
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