SME
Orbital Mechanics · Problem type
Synodic period and launch windows
Use this for “Earth circles the Sun in 365.25 days and Mars in 687 days. How often does a Hohmann launch window to Mars open?”
Diagram
1
S
=
1
T₁
−
1
T₂
=
1
365.25
−
1
687,
so
S
=
779.9
days
The inner planet gains one lap on the outer one each synodic period: then they line up again.
This is an example. Type your own values to start a new problem.
1
S (days)
=
1
T₁ (days)
−
1
T₂ (days)
Formulas
1
S
=
1
T₁
−
1
T₂
1
779.8811
=
1
365.25
−
1
687
Assumptions
•
Circular orbits at steady speeds; the inner planet, with the shorter period T₁, is faster.
•
The same line-up, and so the same launch window, returns when the inner planet gains one lap.
Step-by-step
WE KNOW
T₁ = 365.25 days, T₂ = 687 days
FIND
synodic period (S)
Step 1 · Find synodic period (S)
Use 1 ÷ S = 1 ÷ T₁ − 1 ÷ T₂
The
inner
planet
gains
1
T₁
−
1
T₂
laps
each
day;
one
whole
lap
takes
S
.
S
=
1
1 ÷
T₁
− 1 ÷
T₂
S
=
1
1 ÷ 365.25 − 1 ÷ 687
S
=
1
0.002738 − 0.001456
S
=
1
0.001282
S
=
779.8811
days
Check
Put the numbers back in. Both sides should match.
1
779.8811
=
1
365.25
−
1
687
✓
Topic check
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