# Patched conics Earth to Mars transfer, find Vinf at arrival at Mars

I've tried to solve a patched conic problem found on the 8th chapter of Fundamentals of Astrodynamics, Bate, Mueller and White, Dover 1971. Attached is the problem, which concerns the sub part (e), requiring the calculation of hyperbolic excess speed at the arrival at Mars. I've managed to get the matching answers to the sub parts of the question up to (e), where my answer seems to diverge from what is given in the text. So I thought of giving it a try here.

8.13 (This problem is fairly long, so work carefully and follow any suggestions.) We wish to travel from the Earth to Mars. The mission will begin in a 1.5 DU$$_{\bigoplus}$$ circular orbit. The burnout speed after thrust application in the circular orbit is to be 1.5 DU$$_{\bigoplus}$$ /TU$$_{\bigoplus}$$ The thrust is applied at the perigee of the escape orbit. The transfer orbit has an energy (with respect to the Sun) of -0.28 AU$$^{2}$$ /TU$$_{\bigodot}^{2}$$.

(a) Find the hyperbolic excess speed, V (Ans. 0.956 DU$$_{\bigoplus}$$ /TU$$_{\bigoplus}$$)
(b) Find the hyperbolic excess speed in heliocentric units. (Ans. 0.254 AU /TU$$_{\bigodot}$$)
(c) Find the velocity of the satellite with respect to the Sun at its departure from the Earth. (Ans. 1.2 AU /TU$$_{\bigodot}$$)
(d) Find V at arrival at the Mars orbit (Ans. 0.867 AU /TU$$_{\bigodot}$$)
(e) Find the hyperbolic excess speed upon arrival at Mars. (Hint: Find $$\phi_{1}$$ from the Law of Cosines, then find h, $$\phi_{2}$$ and $$V_{mv}$$ in that order.) (Ans. $$V_{mv}$$= 0.373 AU /TU$$_{\bigodot}$$)
(f) What will be the re-entry speed at the surface of Mars? (Ans. 3.39 DU$$_{mars}$$ /TU$$_{mars}$$)

Following was my line of thought.

1. Since,

$$V_{\infty1}$$= 0.956 DU$$_{\bigoplus}$$/TU$$_{\bigoplus}$$; $$from$$ $$(a)$$

$$V_{cs1}$$= $$\sqrt{\frac{\mu_{\bigoplus}} {1.5DU_{\bigoplus}} }$$ = 0.816 DU$$_{\bigoplus}$$/TU$$_{\bigoplus}$$
$$V_{bo}$$= 1.5 DU$$_{\bigoplus}$$/TU$$_{\bigoplus}$$

Therefore, using the law of cosines,

$$V_{\infty1}^{2}$$ = $$V_{cs1}^{2}$$ + $$V_{bo}^{2}$$ - 2$$V_{cs1} V_{bo}$$ cos $$\phi_{1}$$

cos $$\phi_{1}$$ = 0.81668

1. Now, for the heliocentric transfer orbit, we find the angular momentum, $$h$$. Since from (c), we have found $$V_{sv1}$$ = 1.2 AU/TU$$_{\bigodot}$$.

$$h$$ = $$r_{1}V_{sv1}$$ cos $$\phi_{1}$$ = (1 AU) 1.2 AU/TU$$_{\bigodot}$$ (0.81668) = 0.98 AU$$^{2}$$/TU$$_{\bigodot}$$

1. Assuming the angular momentum of the heliocentric transfer orbit is conserved, we can write, for the arrival at the Mars,

$$h$$ = $$r_{2}V_{sv2}$$ cos $$\phi_{2}$$, where $$r_{2}$$ = 1.524 AU, $$V_{sv2}$$ = 0.867 AU/TU$$_{\bigodot}$$ already calculated in (d). This leads to find cos $$\phi_{2}$$ = 0.7414

1. Finally, the law of cosines can again be applied to find the hyperbolic excess speed upon the arrival at Mars.

$$V_{\infty}^{2}$$ = $$V_{cs2}^{2}$$ + $$V_{sv2}^{2}$$ - 2$$V_{cs2} V_{sv2}$$ cos $$\phi_{2}$$

$$V_{cs2}$$= $$\sqrt{\frac{\mu_{\bigodot}} {1.524 AU} }$$ = 0.81 AU/TU$$_{\bigodot}$$

$$V_{\infty}^{2}$$ = $$0.81^{2}$$ + $$0.867^{2}$$ - ($$2$$)($$0.81$$)($$0.867$$) ($$0.7414$$) AU$$^{2}$$/TU$$_{\bigodot}^{2}$$

$$V_{\infty}$$ = $$V_{mv}$$= $$0.605$$ AU/TU$$_{\bigodot}$$

But this does not seem to be the correct answer as per the answer given in the text. Any ideas ? Thanks a lot in advance !

Update : Added the text version above.

• Thanks for the comment. Text added now. Dec 6 '20 at 2:07
• That's quite an effort; thank you very much!
– uhoh
Dec 6 '20 at 2:26

In the subsection (e) you are asked to find the hyperbolic excess speed upon arrival at Mars. (Hint: Find $$\phi_1$$ from the Law of Cosines, then find $$h$$, $$\phi_2$$ and $$V_{mv}$$ in that order.)

The first step, following the provided hint, is to find $$\phi_1$$. This angle is the flight-path angle (the angle between the velocity and the local horizontal plane, see the Figure below, extracted from the same book).

This angle is useful because it will allow us to obtain $$h$$ knowing the magnitude of $$\vec{r}$$ and $$\vec{v}$$, since $$\gamma$$ y $$\phi$$ are complementary angles:

$$\vec{h} = \vec{r}\times\vec{v}$$

$$h = rv\sin(\gamma)=rv\cos(\phi)$$

Thus, we calculate $$\phi_1$$ with the data we already have (when departing from Earth), which allows us to compute $$h$$ (from the transfer orbit) and as this magnitude is constant in the two-body problem, we can use it again to calculate $$\phi_2$$ (when arriving at Mars). This means that we have to make these calculations using the tranfer orbit around the Sun, i.e., using its speed and position at departure and at arrival.

The trick is that the Earth's speed and the departure speed with respect the Sun form the same $$\phi_1$$ angle, because as the Earth's orbit is considered to be circular $$r_{\oplus}$$ and $$v_{\oplus}$$ are perpendicular and then $$v_{\oplus}$$ is parallel to the local horizon of the satellite. The same happens when arriving at Mars. Therefore, as we know that $$\vec{v_{dep}}=\vec{v_{\oplus}} + \vec{v_{\infty}}$$, we have the triangle of velocities that allows us to clear $$\phi_1$$ at the departure. The equivalent triangle can be used to obtain the hyperbolic excess speed at the arrival.

You were trying to use te cosine law not in the heliocentric transfer orbit but in the hyperbolic orbit around Earth. But in such case, the triangle of velocities you were trying to build does not exist, as the hiperbolic excess speed is reached when $$r\rightarrow\infty$$, and is related to its perigee velocity (the burnout speed) through the energy equation. The parking speed is related to the hyperbolic orbit speet at its perigee through the $$\Delta v$$ that it should be applied to change from the parking orbit to the hyperbolic orbit. But this last magnitude is not asked.

• Looks great, thanks!
– uhoh
Jan 12 at 1:15

Well, I guess after some thinking and trial-and-error I've gotten closer to the answer. I thought the fact that I applied the law of cosines, while being in the frame of reference of Earth for the departure from Earth, was perhaps erroneous and hence applied the law of cosines in the frame of reference of Sun. After all, in (b) it is asked to compute V$$_{\infty1}$$ with respect to the Sun.

This led me to find $$cos$$ $$\phi_{1}$$ $$=$$ $$0.9897850$$, $$h$$ $$=$$ $$1.187742$$ AU$$^{2}$$/TU$$_{\bigodot}$$ and consequently to find $$cos$$ $$\phi_{2}$$ $$=$$ $$0.8989138$$.

Applying again the law of cosines for the arrival at Mars in the frame of reference of the Sun, $$V_{\infty2}$$ = $$V_{mv}$$= $$0.381$$ AU/TU$$_{\bigodot}$$, which is closer to the given answer.

Subsequently, I answered (f), with $$3.3338$$ DU$$_{mars}$$ /TU$$_{mars}$$.

Now, I understand that one must apply the law of cosines in the frame of reference of the Sun for the departure, taking into consideration, the circular velocity of the Earth and the velocity of the satellite with respect to the Sun on the transfer orbit, at the burn-out point, as the two vectors that define the angle $$\phi_{1}$$. And NOT the velocity of the parked orbit around the Earth with respect to the frame of reference of the Earth at the burn-out point to define the angle $$\phi_{1}$$ .

Nevertheless, I'd be much grateful if someone could explain WHY and HOW this choice of opting for the reference frame of Sun instead of that of the Earth matters. Thanks in advance !