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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

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    $\begingroup$ Unrelated: is "tweed" a term for "twitter feed" or just a typo? Signed, very old person. $\endgroup$
    – Carl Witthoft
    Commented Aug 3, 2020 at 12:13
  • $\begingroup$ @CarlWitthoft: sorry, my bad, it was just a typo $\endgroup$
    – CallMeTom
    Commented Aug 3, 2020 at 12:17
  • $\begingroup$ thanks - just wondered if I needed to learn something, $\endgroup$
    – Carl Witthoft
    Commented Aug 3, 2020 at 12:28
  • $\begingroup$ JMcD seems to have answered a question here before, might again. $\endgroup$
    – uhoh
    Commented Aug 3, 2020 at 13:10

1 Answer 1

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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...

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  • $\begingroup$ I think you are probably right. I know the rule of thumb is to double the velocity from a Hohmann transfer if done over a slow period of time. 71 times two is pretty close to 140... $\endgroup$
    – PearsonArtPhoto ♦
    Commented Aug 3, 2020 at 13:24

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