SME
Orbital Mechanics · Problem type
Plane-change burn
Use this for “A satellite circles at 3.075 km/s. What Δv turns its orbit plane through 28.5°? What turn does 1 km/s buy?”
Diagram
v after: 3.075 km/s at 28.5°, so v after = ⟨2.702, 1.467⟩.
−v before: 3.075 km/s at 180°, so −v before = ⟨−3.075, 0⟩.
Tip to tail: Δv = ⟨−0.3726, 1.467⟩
|Δv| = 1.514 km/s at 104.3°.
This is an example. Type your own values to start a new problem.
Δv (km/s)
=
2
×
v (km/s)
×
sin(
Δi
°
÷
2)
Formulas
Δv
=
2
×
v
×
sin(Δi°
÷
2)
1.5138 = 2 × 3.075 × sin(28.5° ÷ 2)
Assumptions
•
One impulsive burn at a node, where the old and new orbit planes cross.
•
The speed stays v, so v before, v after and Δv make an isosceles triangle.
•
Δi is the angle between the two orbit planes, in degrees.
Step-by-step
WE KNOW
v = 3.075 km/s, Δi = 28.5°
FIND
burn (Δv)
Step 1 · Find burn (Δv)
Use Δv = 2 × v × sin(Δi° ÷ 2)
Split
the
isosceles
triangle
down
its
middle:
each
half
has
Δv
2
=
v
sin(
Δi
2
).
Δv = 2 × 3.075 × sin(28.5° ÷ 2)
Δv = 2 × 3.075 × sin(14.25°)
Δv = 2 × 3.075 × 0.2462
Δv = 6.15 × 0.2462
Δv = 1.5138 km/s
Check
Put the numbers back in. Both sides should match.
1.5138 = 2 × 3.075 × sin(28.5° ÷ 2) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.