6
$\begingroup$

I'm currently working on a video game that will make use of a heavily simplified patched-conic model of the Solar System.

I was wondering about some discrepancies between my own calculations as well as public data.

Take the equation derived in this post for the radius of a Hill sphere in a two-body system (approximated by L1 and L2 Lagrange points): $$ h \approx r \sqrt[3]{\frac{M_2}{3M_1}} ,$$ where $m_1$ and $m_2$ are the masses of the larger and smaller body respectively, and $r$ is the distance between the two bodies.

Now, I'm aware that the Hill sphere would/should be centered on the total barycentre of an n-body system. Therefore, to simplify, instead of basing this on an Earth-Moon Hill sphere w.r.t. the Sun, I will instead consider a Moon Hill sphere w.r.t. to the Earth:[a] $$ \begin{align} h_☾ &\approx r_{🜨-☾}\sqrt[3]{\frac{M_☾}{3M_🜨}} \\\\ &\approx 57130\mathrm{km}. \end{align} $$

Now, as far as I understand, in theory once a (relatively massless) spacecraft gets further away from the centre of mass of the Moon than this distance, the gravity of the Earth (or the common Earth-Moon barycentre in the far case) will take over and the spacecraft should leave its Moon orbit and enter an Earth orbit?

Thus I would have assumed that one could derive the escape velocity from an arbitrary circular orbit using the vis-viva equation, such that for example:[b] $$ v_1^2 = GM_☾\left(\frac{2}{r_☾ + 100\mathrm{km}} - \frac{1}{r_☾ + 100\mathrm{km}}\right) $$ for the velocity-squared of an object in a $100\mathrm{km}$ circular orbit above the surface of the Moon, and[c] $$ v_2^2 = GM_☾\left(\frac{2}{r_☾ + 100\mathrm{km}} - \frac{1}{⅔57130\mathrm{km}}\right) $$ for the velocity-squared at the same orbital height for an ellipitcal orbit with its semi-major axis equal to two-thirds of the Hill sphere radius. Please note that I've assumed for simplicity that the distance between the apses and focal points in the ellipse are very approximately equal to $½a$; it shouldn't make a significant difference in regards to the conclusion of this question. This would yield a total instantaneous velocity difference for our spacecraft of:[d] $$ \begin{align} \Delta v &= \sqrt{v_2^2 - v_1^2} \\\\ &= 1594\mathrm{m/s}. \end{align} $$

Now, if you are reading this so far, you are probably very well aware of the vis-viva-derived equation for escape velocity, where $$ v_e = \sqrt{\frac{2GM}{r}}, $$ in our case yielding an escape velocity[e] $$ \begin{align} v_e^☾ &= \sqrt{\frac{2GM_☾}{r_☾ + 100\mathrm{km}}} \\\\ &= 2310\mathrm{m/s}. \end{align} $$

Now, even though this exceeds our initial calculation by more than a factor of 3⁄2, I simply drew this down to the fact that the escape velocity equation is technically the velocity change required to achieve an orbit with eccentricity $e\ge1$, without taking gravitational influences from any other bodies into account. Therefore the initial value still seemed very correct at a glance.

That was until I checked this $\Delta v$ map on Wikipedia, which lists the $\Delta v$ between Moon escape and $100\mathrm{km}$ Low Moon Orbit as $676\mathrm{m/s}$:

So my question is how did the author of the above image derive this figure, and why does it differ so significantly from my own calculations? How can I derive the appropriate figure independently for any body in the Solar System?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
8
$\begingroup$

This is where it goes wrong:

$$\Delta v = \sqrt{v_2^2 - v_1^2}$$

You're taking the difference of the squares of the velocities at a low lunar orbit, and the perilune of the transfer orbit.

But really, this is just a simple difference of two velocity vectors, it should be:

$$\Delta v = v_2 - v_1$$

This gives me a $\Delta v$ of 641m/s, with the difference to the attached map's value of 676m/s probably being due to them doing $r_A = \infty$ calculation instead of $r_A = ⅔57130km$, not bothering with the hill sphere.

For the question asked in the title, there's no direct relationship since the mass of and the distance to the parent body are free variable as far as the escape velocity is concerned.

Improve this answer
$\endgroup$
11
  • $\begingroup$ Fascinating! I was taught elsewhere that you can't simply add together delta-v values, but you have to first take the sum of the squares of the velocities and then take the square root of that sum. (I.e. due to orbital energy invariance.) I'm guessing the exception is when you're actually calculating an instantaneous maneuver like this. In any case, thank you very much! $\endgroup$
    – Vlad
    Commented Jan 22, 2023 at 16:40
  • 1
    $\begingroup$ @Vlad You're probably referring to v^2 = v_infty^ + v_e^2. The "tricky" thing about that scenario is that the velocity vectors aren't in the same place, one is deep inside a gravity well and one is outside of it. $\endgroup$
    – SE - stop firing the good guys
    Commented Jan 22, 2023 at 17:29
  • $\begingroup$ Not exactly. I've been told that in order to find the total Δv of several discrete maneuvers (i.e. as if it was a one ideal maneuver consisting of two burns), you need to first convert each of the intervening Δv velocities into the corresponding kinetic energy, add the energies, and then convert back. As it turns out, this can be achieved regardless of the mass of the craft by simply adding the velocities-squared and finding the square root. $\endgroup$
    – Vlad
    Commented Jan 22, 2023 at 18:32
  • 3
    $\begingroup$ The root-of-squares is for cases where your velocities are orthogonal to each other. If they are in the same direction, you add them, if they are in opposite directions, you subtract them. $\endgroup$
    – Paŭlo Ebermann
    Commented Jan 23, 2023 at 0:18
  • 2
    $\begingroup$ Vector addition $\endgroup$
    – SE - stop firing the good guys
    Commented Jan 23, 2023 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.