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

Question

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?

Answer 1

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:

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

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

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