# Hyperbolic orbit: correct?

I'm writing a sci-fi story. The protagonist wants to strap rockets to an asteroid and send it rocketing to earth, into orbit, to harvest the metals, etc. An acquaintance (i lost contact with him but he had possession of the equations) told me that if the protag chooses an asteroid, say 2.5 astronomical units from the sun in the asteroid belt, 90 degrees ahead of the earth, it could be launched at a velocity of 23 km. per second, directly at the sun, it will arrive the required (by my plot) 3 months later. Is this correct? Any other details (velocity at arrival to earth, velocity to kill to get it into LEO, etc.) would be much appreciated.

• It would be better if he used a Hohmann transfer to bring it to earth. If he was using a hyperbolic trajectory in the asteroid belt (assuming sun’s gravity is dominant), it would conceivably send the asteroid out of the solar system. Also, he would have to decrease the velocity of the asteroid to send it on a trajectory to earth, not increase. – Paul Oct 1 '18 at 17:39
• I started poking around with the online pork chop plottter, and it does seem like you might need something 23 km/s of delta-V is you insist on getting there in 90 days. I didn't have time to develop a proper answer. If you don't mind it taking a couple of years you can do with massively less. I – Steve Linton Oct 1 '18 at 17:43
• Drat the luck, getting the body into earth vicinity within 90 days is crucial to my plot, otherwise the Hohmann orbit does save fuel. Kill=decrease; to about 7km/s in the earth's frame for the LEO. – chasrob Oct 2 '18 at 1:39
• Remember relativity. Hyperbolic in terms of what :)? Earth approach yes- in terns of sun, no. – Magic Octopus Urn Oct 3 '18 at 6:13
• @chasrob Changing the question after an answer is posted is strongly discouraged in Stack Exchange. I understand that you're doing it for the best of reasons, but it's not the way this site works. You can post the substantial addition information and orbit instead in a new question, and ask about it there, and include a link to this question. To get it, don't worry, just click "edit" and scroll down and you'll see your edit in the previous versions. I used the "rollback", edits are not lost, the previous versions are all still there. – uhoh Oct 4 '18 at 19:03

90 days to Earth? Yes!

Hyperbolic? No.

I did a quick check and yes if you both cancelled the 19 km/s orbital velocity as well as added a 23 km/s radial velocity toward the Sun, you'd intercept 1 AU in 90 days. Without the Sun's gravitational acceleration, the trip would have taken 23 days longer.

You'd then be going 40 km/s and have a heck of a deceleration in store for you if you wanted to be captured in Earth orbit.

However, at this set of speeds and distances, the net energy is still negative. You would still be gravitationally bound to the Sun, so the orbit would not be called hyperbolic. Kinetic and potential energies per unit mass are as follows:

$$\frac{E}{m} = \frac{1}{2} v^2 - \frac{GM_S}{r^2}$$

or about 265 - 354 = -89 million Joules/kg, where the standard gravitational parameter for the Sun $$GM_S$$ is 1.327E+20 m³/s².

Here's a short Python script for the calculation.

def deriv(X, t):
x, v = X
acc = -GMs/x**2
return np.array([v, acc])

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint

GMs = 1.327E+20

AU  = 150E+06*1000. # meters
km  = 1000.         # meters
day = 24.*3600.     # sec

v0  = -23000        # m/s
x0  = 2.5 * AU      # m/s

vorb = np.sqrt(GMs/x0)
print "vorb: ", vorb

X0 = np.array([x0, v0])

ke = 0.5 * v0**2    # Joules/kg
pe = -GMs/x0
e  = ke + pe
print "ke: ", ke
print "pe: ", pe
print "e:  ", e

time = day * np.arange(0, 90, 0.1)

answer, info = ODEint(deriv, X0, time, full_output=True)

x_free = x0 + v0 * time
v_free = v0 * np.ones_like(time)

plt.figure()
plt.subplot(2, 1, 1)
plt.plot(time/day, x/AU, linewidth=1.5)
# plt.plot(time/day, x_free/AU, '--k')
plt.title('Distance from Sun (AU) vs time (days)', fontsize=16)
plt.subplot(2, 1, 2)
plt.plot(time/day, v/km, linewidth=1.5)
# plt.plot(time/day, v_free/km, '--k')
plt.title('speed (km/s) vs time (days)', fontsize=16)
plt.show()

• It would be a hyperbolic trajectory of the Earth, @chasrob, an additional change would be required at Earth to turn it into an orbit, and it's likely more than can be acheived with a combination of aerobraking and lunar gravity assist. – user20636 Oct 2 '18 at 14:45
• Trying to visualize this... as the asteroid approaches, the earth is moving at ~30km/s, head-on, the asteroid 40 km/s, a roughly 7km/s orbit of earth is needed, so a retro burn of 30+40-7... about 60km/s is required to put it into a earth orbit? – chasrob Oct 2 '18 at 15:12
• The mass(fuel) > mass(asteroid). :( A 90 day travel time is not going to work. Rewrite time... – chasrob Oct 2 '18 at 15:43
• @chasrob. It's not head on. The Asteroid is moving at right angles to the Earth on arrival,, so if neither asteroid nor Earth had gravity, relative velocity on arrival is about 50 km/s, using those numbers. Earth's gravity well will nudge things a little higher, and wanting an Earth orbit rather than a dead stop will nudge things lower, but it's not going to reach 70 km/s – notovny Oct 2 '18 at 15:56
• @chasrob consider a coincidence then? Have a look at all of the references and listen to the conference call described in the question Have there been any documented mini-moons since 2006 RH120? Many believe that from time to time near-earth asteroids will get temporarily trapped in the Earth-Moon system in what I call "squiggle orbits". If there's some way to take advantage of an unusual opportunity, that would be a nice way to do it. However they are usually small because in general the number of asteroids go up quickly with decreasing size. – uhoh Oct 2 '18 at 18:38

Big problem here: There's no way this can actually be done.

Lesser issue: That's a mighty hot rocket, at a minimum it's fusion based. You need a total of about 100 km/sec to put it Earth orbit in the time given. (Look at uhoh's answer--and note that he's only getting a flyby, not a capture.)

Main issue: The basic problem here is that asteroids are generally quite weak and without a lot of study you can't be sure there are no weaknesses in your target asteroid. You're planning some big burns that must be done fast if you want to meet your 90 day time limit--almost certainly the result is the asteroid comes apart.

Side issue: Earth is going to take a very dim view of what you're up to. Expect a military response. Make a little mistake with your navigation and you've got anywhere from a city killer to an extinction event depending on the size of your rock.