I have a big chunk of water ice that needs to go from (say) one of Mars's poles to its equator.
Neglecting the atmosphere, how would you calculate the delta-v necessary for this trajectory, and what would the resulting value be?
Question inspired by @SteveLinton's comment.
Given separation angle $θ$, body mass $M$ and body radius $r$, before finding delta-V we must find the ellipse major axis - or more precisely, half its length (distance from ellipse center to most extreme point):
$a = {(1+\sin{θ \over 2}) }{r\over2 }$
Given that, delta-V is given by the following equation:
$\Delta v = \sqrt {G M ({2 \over r} - { 1\over a})}$
The launch should be performed at angle $\alpha$:
$\alpha ={{\pi-θ}\over 4 }$
For powered landing, double the delta-V requirements.
Credit goes to Hop's Blog - Travel on airless worlds. The article contains derivation of the above equations and a link to a helpful spreadsheet where you can calculate the hop parameters for any celestial body and any chosen separation angle:
For the trip from the pole to equator (90 degrees separation angle), on Mars, the launch delta-V would be 3228 m/s. The launch would be optimally performed at 22.5 degrees.
The closest solution i can find comes from Hale’s (1994) Introduction to Spaceflight, wherein Chapter 9 discusses range equations for such ballistic bodies. He derives an equation
$$cot(\frac{\Psi}{2})=\frac{2}{Q_{bo}}csc(2\phi_{bo}) - cot(\phi_{bo})$$
where
$$Q_{bo}=\frac{V_{bo}^2r_{bo}}{\mu}$$
is a dimensionless quantity that is roughly measures twice the ratio of kinetic to potential energy at the burnout point (subscript “bo”). $\mu$ is the standard gravitational parameter and $\Psi$ is the range angle and $\phi_{bo}$ is the launch angle.
What you want is to have $\Phi=90^0$ and $r_{bo}=$ the radius of mars, assuming an impulse thrust at the planet’s surface. Then you can play with the burnout velocity and launch angle until you get a feasible solution. Note that even though many launch angle will give a burnout velocity, not all will. Even so, some of the solutions are infeasible because they may, for example, rely on the orbit traversing through the interior of the planet.
Keep in mind that this equation is based on a lot of simplifying assumptions: non rotating earth, no atmosphere, a spherical planet, symmetrical trajectory, and a an insignificant freefall range.
The answer poasted by @SF. based on Hop's blog checks out!
Here's a numerical verification. It's not pretty, but continuing the orbits (shown for both Mars and Earth) for 55% of their period nicely intersects 90 degrees away from the staring point, yay!
body a(km) dv(m/s) alpha(deg)
----- ----- ------- ----------
Earth 5438 7199 22.5
Mars 2893 3235 22.5
def deriv(X, t):
x, v = X.reshape(2, -1)
acc = -GM * x * ((x**2).sum())**-1.5
return np.hstack((v, acc))
def Hops_hop(theta, R, GM):
a = (1. + np.sin(0.5*theta)) * 0.5 * R
dv = np.sqrt(GM * (2./R - 1./a))
alpha = 0.25 * (pi - theta)
return a, dv, alpha
import numpy as np
from scipy.integrate import odeint as ODEint
import matplotlib.pyplot as plt
halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
# standard gravitational parameter
GMe = 3.986E+14 # m^3/s^2
GMm = 4.283E+13 # m^3/s^2
Re = 6371000. # meters
Rm = 3389500. # meters
pairs = (Re, GMe), (Rm, GMm)
theta = halfpi # 90 degrees
answers = []
for R, GM in pairs:
a, dv, alpha = Hops_hop(theta, R, GM)
T = twopi * np.sqrt(a**3/GM)
time = np.linspace(0, 0.55*T, 500)
x0 = R * np.array([ np.sin(0.5*theta), np.cos(0.5*theta)])
v0 = dv * np.array([ np.cos(alpha), -np.sin(alpha) ])
X0 = np.hstack((x0, v0))
answer, info = ODEint(deriv, X0, time, full_output=True)
answers.append(answer)
theta = np.linspace(0, twopi, 361)
unit_circle = [f(theta) for f in (np.cos, np.sin)]
sqrt2 = np.sqrt(2.)
if True:
plt.figure()
for answer, (R, GM) in zip(answers, pairs):
x, y = answer.T[:2]
plt.plot(x, y)
plt.plot(x[:1], y[:1], 'ok')
for answer, (R, GM) in zip(answers, pairs):
x, y = [R*thing for thing in unit_circle]
plt.plot(x, y, '-k')
plt.plot([0, Re/sqrt2], [0, Re/sqrt2], '-k')
plt.plot([0, Re/sqrt2], [0, -Re/sqrt2], '-k')
plt.show()