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Issue 146

02 Explainer · Orbital Maneuvering

Maneuvering In Space: The Basics

22 May 2026: I thought it might be helpful to do a short review on space maneuvers and their fuel requirements. First a note on why we talk about fuel usage in terms of “change in velocity” or “Delta-v” or simply “Δv” and use the unit “meters per second” (m/sec).

Author Greg Gillinger Published June 1, 2026

22 May 2026: I thought it might be helpful to do a short review on space maneuvers and their fuel requirements. First a note on why we talk about fuel usage in terms of “change in velocity” or “Delta-v” or simply “Δv” and use the unit “meters per second” (m/sec). M/sec is a universal, spacecraft-agnostic metric that describes how much orbital change you need, completely independent of the spacecraft’s mass, engine type, or propellant. Your Δv tells you what the orbit requires. In the Cosmos 2610-14 examples we described in the previous article, changing inclination by 0.8° at an average altitude of 547km costs ~106 m/sec — that’s true whether the Cosmos Satellites are 50 kg CubeSats or a 5,000 kg reconnaissance satellites. If we knew (we don’t) the mass of the Russian satellites and their propulsion (thruster) type then we could calculate the actual propellant mass required to achieve the desired Δv required for the maneuver. So Δv is the input, and the rocket equation tells you how much propellant you need for your specific vehicle.

For the curious, here is the rocket equation:

Δv=I⋅g⋅ln(m0/mf)

Where:

  • I = specific impulse (engine efficiency, in seconds)
  • g = 9.80665 m/s² (standard gravity)
  • ln stands for the natural logarithm
  • m0 = initial mass (spacecraft + fuel)
  • mf = final mass (spacecraft after burn)


Think of Δv like pricing a road trip in kilometers rather than liters of gas:

  • “It’s 705 km from DC to Boston” is universally true.
  • How many liters that takes depends on whether you’re driving a Prius or a Humvee as well as how quickly you need to arrive at your destination.


Similarly, an orbital analyst can say “that maneuver costs 50 m/sec” and every engineer on the team can independently calculate what that means for their specific spacecraft design.

The three fundamental orbital maneuver types: 1) Cross-Track; 2) Radial-Track; 3) In-Track. Of the three, Cross-Track changes are the most fuel-expensive

Graphic Comparing Δv Requirements for Cross-Track, In-Track & Radial Track Maneuvers for Circular Orbit with 550km SMA

Graphic Comparing Δv Requirements for 1° Inclination Change at Varying SMA Values

Δv Requirements for RAAN specific maneuvers varies by Orbit Inclination. Cross-Track Maneuvers include both inclination and RAAN.

Here is the math behind the energy requirement calculations specific to a 0.8° inclination change like the ones conducted by Cosmos 2610-14. I assumed Russia performed the plane change at apoapsis (apogee), where the orbital velocity is at its minimum, thus reducing the Δv cost (in this case slightly as the orbit is nearly circular).

  • ParameterValues
    • Semi-major axis (a) = 6,912 km
    • Eccentricity (e) = 0.002
    • Perigee radius = 6,898.18 km (~527.2 km altitude)
    • Apogee radius = 6,925.82 km (~554.8 km altitude)


With these parameters we can calculate the satellites velocity at apogee using the vis-viva equation:

v = orbital velocity

μ = Standard gravitational parameter (398,600.44 km³/s²)

r = radial distance from the center of the earth. Computing r at apogee = 6,912 x (1 + .002) = 6,925.82km (.002 is orbit eccentricity)

a = semi major axis (in this case 6,912km)

At apogee the orbital velocity = 7,578.76 m/sec

Now we’re ready to calculate Δv cost for changing inclination 0.8° using the pure plane change maneuver formula:

Inclination change (Δi) = 0.8°

v = satellite velocity at apogee = 7,578.76 m/sec

Therefore: Δv = 2 (7578.76) sin (0.8/2) = 105.82 m/s