Hyperbolic orbit: correct?

Question

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.

Answer 1

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, v = answer.T
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()

Answer 2

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.