# What is the highest theoretical escape velocity a 2 billion metric ton vehicle can achieve using a >20° gravitational slingshot?

I want to send a flying saucer (a $$\frac{1}{10}$$ scale Independence Day City Destroyer - calculated as a sphere cap 112m high with 1.2km segment radius, $$\rho$$ about 43.3kg/m3) away from the sun as fast as possible, accelerating it from some initial velocity $$V_i$$ to escape velocity $$V_f$$ by gravity assist, and learn the $$\Delta V$$ gained. It’s mass is about $$2\times 10^9$$ metric tons. Parameters are:

• The assist will deflect the initial trajectory by no less than 20° ($$\Delta$$V is $$\delta$$ or the angle between the hyperbolic asymptotes). This allows a useable steering benefit.

• The entry vector $$V_i$$ is an indeterminate variable with less velocity than the exit vector.

• Any celestial body in our solar system can be used for the assist except the sun (the sun’s gravity can’t assist something escaping itself?)

• The mass cannot enter any atmosphere (no friction calculations or concerns) so $$\text r_0 =$$ the assist planet’s upper atmosphere.

• No thrust burn, no Oberth effect.

The entire problem is centered around the sun’s frame of reference, so the plot needs an answer with the largest rate of separation from the sun as it’s trajectory approaches that asymptote, not the assist planet. So there’s a little vector translation I need to figure out.

I suppose the most useful solution has the $$V_f = f(m)$$ and $$V_i = f(m)$$ allowing for more diverse applications. This way we know the required initial velocity along with the exit velocity achieved.

I believe the problem can be stated, find the maximum translated V$$_f$$ with $$\rho > 20°$$ as $$r_0$$ approaches the upper atmosphere from above.

# Premises:

• The saucer is an interstellar bulk carrier that wants to save fuel and travel time.

• This is for an opportunistic use not routine benefit. It’s an energy cost benefit analysis.

• I realize the diminishing returns from steering the vehicle along a vector inclined to the planet’s orbital plane.

• Jupiter seems to be the obvious choice but I don’t know if getting closer to a fast-moving planet with less atmosphere, like Mercury, gives more oomph?

• I also realize a smaller angle will allow a greater $$V_f$$, but the $$V_i$$ I feel will be prohibitive ($$\Delta$$V will get too small).

• The velocity increase of a gravitational assist comes from the vector addition of bending your trajectory closer to the trajectory of the object you're slingshotting by. You can never acquire more than twice the orbital velocity of the object being slingshot by in an unpowered slingshot, and the faster you're moving, the less the flyby can bend your trajectory. An unpowered slingshot is probably not worth it for any interstellar craft. – notovny Nov 1 '19 at 22:34
• I realize that and with 20° the approach will need to be very low, probably in most atmospheres. The angle was somewhat arbitrary but large enough to allow both acceleration and steerage. It also sounds like an assist from a faster small planet may be better than from a slower large one. – Vogon Poet Nov 1 '19 at 22:36

I don't think an unpowered gravitational slingshot against Jupiter is going to be significantly helpful for an interstellar spacecraft. Even using Jupiter as your slingshot target, your sun-relative velocity is going to need to be extremely low to get a 20° angle between your hyperbolic asymptotes

Given that Jupiter has a radius of 69.1* 106 meters, and a standard gravitational parameter of 1.26 * 1018 m3/s2

As per Wikipedia's article on Hyperbolic Trajectories, the angle between the asymptotes will be:

$$\theta = 2\nu = 2\sin^{-1}\bigg(\frac{1}{(1 + r_p * v_\infty^2/\mu)}\bigg)$$

Where $$r_p$$ is the radius of Jupiter, $$\mu$$ would be the standard gravitational parameter of Jupiter and $$v_\infty$$ = The initial relative velocity to Jupiter before the gravitational slingshot.

Solving for $$v_\infty$$ gives you

$$v_\inf = \sqrt{\frac{\mu - \mu \sin(\frac{\theta}{2})}{r_p\sin(\frac{\theta}{2})}}$$

Plugging in the values for Jupiter:

• $$r_p = 69.9 \times10^6 m$$
• $$\mu = 1.27 \times 10^{17} m^3/s^2$$
• $$\theta = \frac{\pi}{9}$$

Results in $$v_\inf = 93000 m/s$$

So to get this 20° asymptotic angle, your spacecraft cannot be moving more than 93 km/s relative to Jupiter initially. Since a gravitational assist can't change velocity by more than twice the orbital velocity of the object being slingshot by, the excess velocity picked up by a Jupiter slingshot is going to be less than 26 km/s.

At 119 km/s, a trip to Alpha Centauri, at 4.26 light-years away, is more than 10,700 years, and that neglects the effects of the sun's gravity on your interstellar trajectory It's very difficult to see such a long trip as economical for a commercial concern.

Plugging in the values for Mercury:

• $$r_p = 2.43 \times10^6 m$$
• $$\mu = 12.2 \times 10^{13} m^3/s^2$$
• $$\theta = \frac{\pi}{9}$$

Results in $$v_\inf = 15500 m/s$$

So, about 15km/s Mercury-relative. Mercury's maximum orbital velocity is 59 km sec, so even under the best circumstances, it could only change your velocity by 118 km/s.

Unpowered Planetary gravitational slingshots are great for when you're early in your space exploration and relying on chemical rockets for propulsion. When your spacecraft are capable of pulling off velocities that get you to other stars within human lifetimes, you're moving way too fast to get any significant benefit.

• Would a faster planet like Mercury do better or am I killing this with the 20°? And where is this Jupiter radius from, the atmosphere? – Vogon Poet Nov 1 '19 at 23:41
• @vogonpoet That's the polar radius. The equatorial radius is a tad larger (en.wikipedia.org/wiki/Jupiter_radius). It's a measure from its core up until where it has an atmospheric pressure of 1 bar. 1 bar mark is the pressure which you get on Earth while inside a pub on a coastal city (hence slightly lower than actual Earth sea level), which in turn is considered "sea level" for gas and ice giants. – The Square-Cube Law Nov 2 '19 at 0:32
• Ok I need to recalculate for upper atmosphere. Also it would be good to know if a smaller faster planet could offer a greater Vmax. If this is correct, then my final vector is more constrained by the atmosphere than the twice orbital velocity – Vogon Poet Nov 2 '19 at 0:35
• I added details for Mercury. Ultimately with unpowered slingshots, the faster you move by the object, the less the slingshot can help you. At speeds where interstellar travel is reasonable, you should be moving so fast that the planet might as well not be there (assuming you don't hit it, that is) – notovny Nov 2 '19 at 0:48