# Minimum delta V required to return a piece of metallic asteroid 16 Psyche to Earth?

Suppose somebody sent a rover to Psyche 16 and found massive concentrated deposits of platinum-group metals near the surface, and turned it into a 10 ton sphere of platinum/iridium/osmium/gold alloy before giving it the exact minimum push to make it crash into an ocean on earth at high speed. (BTW how much of the 10 tons would be recoverable?)

I can find lots of info on the delta V required to reach various objects from earth, but not so much on the return trip. The delta V for the return trip should be a lot smaller when you don't care about the impact velocity. The orbital velocity of 16 Psyche is 8700m/s, so that's an upper bound, because if you just stopped it at the right time it could hit the earth while falling directly towards the sun. But we don't need to stop it completely, we just need to get it into any elliptical orbit that intersects earth's orbit. So, what would be the minimum delta V to make a piece of 16 Psyche intersect earth's orbit?

• A ten ton sphere of 20 g/cm^2 goodies will be 1 meter in diameter. I think that rapid heating could cause it to boil or explode if it comes in from the wrong direction, so recovery from the bottom of the ocean would be complicated, so this will need some thought.
– uhoh
Jan 22, 2021 at 1:05
• Hey future answerers (hopefully present by now!), if you could include the general methodology for calculating the minimum delta-v needed for changing from an orbit with one set of parameters to an orbit with another set around a different body, that would be greatly appreciated! Jan 22, 2021 at 2:33
• Given the average density of Psyche 16 is 4.2 g/cm3 & the density of your rich sphere would around 21.5 g/cm3, you would be very lucky to find such a rich density deposit of that size. But if you did it could worth around \$516 billion if it didn't flood the market.
– Fred
Jan 22, 2021 at 7:17
• There is no reasonable expectation of finding any high concentration deposits of Platinum Group Metals - the nature of asteroid formation means they ended up well mixed with Nickel-Iron. Going by meteorite samples, at 10's of ppm, up to 100ppm. See - space.stackexchange.com/questions/27431/… Jan 25 at 23:58

Given the relatively low inclination of 16 Psyche (3.095°), and the fact that the line of apside is relatively close to the node line anyway, this can be approximated as a planar transfer.

Thus, this can be solved using the powerful vis-viva equation

$$v = \sqrt{\mu \left(\frac{2}{r} - \frac{1}{a}\right)}$$ # Option 1: Direct transfer

(red ellipse)

The most favourable way of doing this is to do a burn at aphelion, lowering the perihelion down to Earth's orbital radius radius.

The required velocity after Psyche escape is then the difference between the aphelion velocity of psyche (15140 m/s) and the aphelion velocity of the transfer ellipse:

$$v_{\infty} = \sqrt{\mu \left(\frac{2}{r_{A_{Psyche}}} - \frac{2}{r_{A_{Psyche}} + r_{P_{Psyche}}}\right)} - \sqrt{\mu \left(\frac{2}{r_{A_{Psyche}}} - \frac{2}{r_{A_{Psyche}} + r_{Earth}}\right)} = 4046 m/s$$

Combined with the escape (180 m/s), we have

$$\Delta v = \sqrt{v_e^2 +v_{\infty}^2} = 4050 m/s$$

# Option 2: Jupiter flyby

(green and violet ellipse)

We can also do a burn at perihelion, raising the aphelion slightly beyond Jupiter's orbit (1.195 times Jupiter's orbital radius). The extra distance is needed so our relative velocity to Jupiter at the encounter is equal to the relative velocity of a Earth transfer ellipse.

Another vis-viva difference:

$$v_{\infty} = \sqrt{\mu \left(\frac{2}{r_{P_{Psyche}}} - \frac{2}{r_{P_{Psyche}} + r_{Jupiter}}\right)} - \sqrt{\mu \left(\frac{2}{r_{P_{Psyche}}} - \frac{2}{r_{A_{Psyche}} + r_{P_{Psyche}}}\right)}$$

$$v_{\infty} = 2367 m/s$$

$$\Delta v = \sqrt{v_e^2 +v_{\infty}^2} = 2374 m/s$$

• Huh? I see the number 4046 in both approaches but it's a different ellipse. And I'm frankly amazed, I didn't realize it took anything like that to bring something back from the asteroid belt. Jan 25 at 3:30
• @LorenPechtel Was a leftover from typing it up, but the correct number was used on the next line. Jan 25 at 5:18