Minimum deltaV between two Orbits (re Jonathan McDowell tweet)

Question

In this Tweet Jonathan McDowell estimates the required delta V between an object and a projectile.

I have recalculated the ejection velocity of the Kosmos-2543 projectile. The delta-V between Kosmos-2543 and object 45915 is somewhere between 140 m/s and 186 m/s

For the lower bound (140 m/s) he calculates:

If instead you just calculate the minimum delta-V to change the apogee and perigee from 604 x 618 km to 505 x 784 km, ignoring all the angular variables, that's 140 m/s so it has to be at least that.

I tried to recalculate this number assuming two impulsive maneuver:

1. rising Apogee altitude from 618 km to 785 km ("positive" delta V at Perigee of a 604x618 km Orbit)
2. lowering Perigee altitude from 604 km to 505 km ("negative" delta V at Apogee of a 604x785 km Orbit)

I get 71 m/s. To much difference for rounding errors. So either me or the Tweets author is making an error. Or we made different assumptions how to change the orbit with minimum delta_V (if so, and both calculations are right, the Tweets minimum dV is not really the minimum)

QUESTION: Can someone help me, explain the difference?

MY CALCULATION

Using:

• V = sqrt(n*((2/r)-(1/a)))
• a = (r_apo + r_peri)/2
• n_earth = 398600 km³/s²
• r_earth = 6378 km

I get:

• r_604 = 6982 km
• r_618 = 6996 km
• r_784 = 7162 km
• r_505 = 6882 km
• a_604x618 = 6989 km
• a_604x784 = 7072 km
• a_784x505 = 7022.5 km

Resulting in:

• V_604x618,Peri = 7.5596 km/s
• V_604x784,Peri = 7.6037 km/s
• dV_1 = 0.0441 km/s
• V_604x784,Apo = 7.4126 km/s
• V_505x784,Apo = 7.3857 km/s
• dV_2 = 0.0269 km/s

dV = dV_1 + dv_2 = 0.071 km/s = 71 m/s

Answer 1

I could figure out, where the 140 m/s came from:

that's not the minimum delta V with two hohmann-like maneuvers. But the amount of delta V needed with one maneuver idealized in one orbital plane.

I get 146 m/s in a quick and dirty run...