SME
Grade 12 Science · Earth & Space Science
The solar system: formation, planets and small bodies
Use this for “Mars orbits at 1.52 AU with an eccentricity of 0.093. What are its period and its closest and farthest distances?”
Diagram
Perihelion a(1 − e) = 1.52 × (1 − 0.093) = 1.38 AU; aphelion a(1 + e) = 1.66 AU.
Each shaded sector is swept in 1/8 of the period: equal areas in equal times, so the planet moves fastest near the sun.
T
2
=
a
3
:
T
=
√
1.52
3
=
1.87
years
This is an example. Type your own values to start a new problem.
(
T (years)
)
2
=
(
a (AU)
)
3
e
Eccentricity
example
q
Perihelion distance
calculated · AU
Q
Aphelion distance
calculated · AU
Formulas
q
=
a
×
(1
−
e
)
1.379 = 1.52 × (1 − 0.093)
Q
=
a
×
(1
+
e
)
1.661 = 1.52 × (1 + 0.093)
T
2
=
a
3
1.87
2
=
1.52
3
Assumptions
•
First law: each planet moves on an ellipse with the Sun at one focus.
•
Second law: the line to the Sun sweeps equal areas in equal times, so the planet is fastest at perihelion.
•
Third law: T² = a³, with T in years and a in AU.
•
T² = a³ holds only for bodies orbiting the Sun.
•
1 AU = 150 million km, Earth’s distance from the Sun: divide million km by 150 (Mars’s 228 million km is 1.52 AU).
Step-by-step
WE KNOW
a = 1.52 AU, e = 0.093
FIND
perihelion distance (q), aphelion distance (Q), period (T)
Step 1 · Find perihelion distance (q)
Use q = a × (1 − e)
Closest: a less the Sun’s offset from the center, ae.
q
=
1.52
×
(1
−
0.093)
q
=
1.52
×
0.907
q
=
1.379
AU
Step 2 · Find aphelion distance (Q)
Use Q = a × (1 + e)
Farthest:
a
plus
the
offset
ae.
Q
=
1.52
×
(1
+
0.093)
Q
=
1.52
×
1.093
Q
=
1.661
AU
Step 3 · Find period (T)
Use T² = a³
Kepler’s
third
law:
T
is
the
square
root
of
a
3
.
T
=
√
a
3
T
=
√
1.52
3
T
=
√
3.512
T
=
1.87
years
Check
Put the numbers back in. Both sides should match.
1.379 ≈ 1.52 × (1 − 0.093) ✓
1.661 ≈ 1.52 × (1 + 0.093) ✓
1.874
2
=
1.52
3
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.