SME
Grade 12 Science · Problem type
The habitable zone and a planet’s temperature
Use this for “A star gives off 0.25 times the Sun’s light. Where is its habitable zone, and is a planet at 0.5 AU in it?”
Diagram
The
zone
runs
from
0.95 ×
√
0.25
= 0.475 AU
to
1.37 ×
√
0.25
= 0.685 AU.
At
0.5
AU
the
planet
is
inside
the
zone,
T = 278 × ⁴
√
0.25
÷
√
0.5
= 278 K.
This is an example. Type your own values to start a new problem.
L
Star’s luminosity
example · L☉
d₁
Inner edge of the zone
worked out (not typed) · AU
d₂
Outer edge of the zone
worked out (not typed) · AU
a
Planet’s orbit
example · AU
T
Planet’s temperature
calculated · K
Formulas
d₁
=
0.95
×
√
L
0.475
=
0.95
×
√
0.25
d₂
=
1.37
×
√
L
0.685
=
1.37
×
√
0.25
T
=
278
×
L
1
4
÷
√
a
278
=
278
×
0.25
1
4
÷
√
0.5
Assumptions
•
Between d₁ and d₂ a planet like Earth could keep liquid water.
•
Light spreads out as the square of distance, so the zone moves out as √L.
•
T leaves out clouds and greenhouse gases: it gives Earth 278 K, but Earth averages about 288 K.
Step-by-step
WE KNOW
L = 0.25 L☉, a = 0.5 AU
FIND
inner edge of the zone (d₁), outer edge of the zone (d₂), planet’s temperature (T)
Step 1 · Find inner edge of the zone (d₁)
Use d₁ = 0.95 × √L
Nearer than this, a planet like Earth grows too hot and loses its water.
d₁
=
0.95
×
√
0.25
d₁
=
0.95
×
0.5
d₁
=
0.475
AU
Step 2 · Find outer edge of the zone (d₂)
Use d₂ = 1.37 × √L
Farther than this, its water freezes.
d₂
=
1.37
×
√
0.25
d₂
=
1.37
×
0.5
d₂
=
0.685
AU
Step 3 · Find planet’s temperature (T)
Use T = 278 × L1/4 ÷ √a
A planet at 1 AU from the Sun comes to 278 K; more light warms it, more distance cools it.
T
=
278
×
∜
L
÷
√
a
T
=
278
×
∜0.25
÷
√
0.5
T
=
278
×
0.7071
0.7071
T
=
278
K
(0.475
<
0.5
<
0.685
AU:
in
the
zone)
Check
Put the numbers back in. Both sides should match.
0.475
=
0.95
×
√
0.25
✓
0.685
=
1.37
×
√
0.25
✓
278
=
278
×
0.25
1
4
÷
√
0.5
✓
Topic check
See what you know: 5 questions on this topic
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