# Why did BepiColombo use an Earth gravity assist?

Since BepiColombo didn't receive another gravity assist from a different orbiting body before reencountering Earth, wouldn't it reenter Earth's SOI with the same hyperbolic excess velocity as it initially left with?

• Earth gravity assists are common for interplanetary spacecraft. For example, Lucy is using 3. Its first one is happening very close to 1 year after launch and it will gain ~5km/s of heliocentric velocity from it. With respect to the flyby planet, they are just hyperbolic flybys, so the spacecraft velocity before and after the flyby have equal magnitude but rotated. But with respect to the Sun, velocity magnitude can also increase or decrease Nov 3 at 16:19
• @AlfonsoGonzalez Which should immediately raise the question: Why not wait to launch the probe until the flyby time? The same hyperbolic speed should mean the same escape delta-v cost, just without all this silly waiting time in space. Nov 3 at 16:55
• @SE-stopfiringthegoodguys Exactly. Thank you for elaborating. My guess is the course wasn't ballistic, but I haven't found anything to back this up. Nov 3 at 17:30
• Would this site contain the justification you are looking for? It is less condensed than @Alfonso Gonzalez explanation. In a nutshell, the Earth flyby changes the velocity wrt to the Sun (not Earth). It says that 450Kg of fuel (Xenon) is saved this way. Nov 3 at 17:44

## 1 Answer

You make a good guess in the comments:

My guess is the course wasn't ballistic, but I haven't found anything to back this up

The BepiColombo Fact Sheet states that the spacecraft uses solar electric propulsion:

The BepiColombo trajectory employs a solar electric propulsion system so that a combination of low-thrust arcs and flybys at Earth, Venus and Mercury are used to reach Mercury with low relative velocity.

A quick look the trajectory on JPL SSD Horizons shows that the spacecraft left Earth with $$C3=12 \frac{km^2}{s^2}$$ and at the Earth flyby had $$C3=16 \frac{km^2}{s^2}$$ (greater hyperbolic speed) with no plane change.

The launch mass of 4100 kg (from fact sheet) and the Ariane 5 (ECA version according to Wikipedia) performance given by this answer suggests that it probably couldn't launch (on Ariane 5) at $$C3=16 \frac{km^2}{s^2}$$.