SME
Oceanography · Problem type
Spring and neap tides from the Sun and Moon
Use this for “The Sun is 1.989 × 10³⁰ kg at 1.496 × 10¹¹ m and the Moon 7.35 × 10²² kg at 3.844 × 10⁸ m. How big is the Sun’s tide next to the Moon’s?”
Diagram
Spring
tide:
the
Sun’s
tide
is
r
=
0.4591
of
the
Moon’s,
so
F
=
√
1 +
r
2
+ 2r cos 2θ
=
1.459
times
the
Moon’s
range
alone.
This is an example. Type your own values to start a new problem.
r
=
(
MS (kg)
÷
MM (kg)
)
×
(
dM (m)
÷
dS (m)
)
3
k
Spring range ÷ neap range
calculated
θ
Moon’s angle from the Sun
example · °
F
Range ÷ the Moon’s alone
worked out (not typed)
Formulas
r
=
(
MS
MM
)
×
(
dM
dS
)
3
0.4591
=
(
1.989 ×
10
30
7.35 ×
10
22
)
×
(
3.844 ×
10
8
1.496 ×
10
11
)³
k
=
1 +
r
1 −
r
2.6975
=
1 + 0.4591
1 − 0.4591
F
=
√(1
+
r
2
+
2
×
r
×
cos(2
×
θ
°))
1.4591
=
√(1
+
0.4591
2
+
2
×
0.4591
×
cos(2
×
180°))
Assumptions
•
The tide-raising force grows with the mass and falls as the distance cubed, so the nearer Moon wins.
•
In line, at new and full Moon, the two tides add (spring tides); at the quarter Moons they partly cancel (neap tides).
•
Coastlines and basins make real ranges differ from these ratios.
Step-by-step
WE KNOW
MS = 1.989 × 10³⁰ kg, dS = 1.496 × 10¹¹ m, MM = 7.35 × 10²² kg, dM = 3.844 × 10⁸ m, θ = 180°
FIND
Sun’s tide ÷ Moon’s tide (r), spring range ÷ neap range (k), range ÷ the Moon’s alone (F)
Step 1 · Find Sun’s tide ÷ Moon’s tide (r)
Use r = (MS ÷ MM) × (dM ÷ dS)³
Each body’s tide goes as its mass over its distance cubed; take the Sun’s over the Moon’s.
r
=
(
1.989 ×
10
30
7.35 ×
10
22
)
×
(
3.844 ×
10
8
1.496 ×
10
11
)³
r
=
1.989 ×
10
30
7.35 ×
10
22
×
0.0025695
3
r
=
1.989 ×
10
30
7.35 ×
10
22
×
(1.6965
×
10
-8
)
r
=
(2.7061
×
10
7
)
×
(1.6965
×
10
-8
)
r
=
0.4591
Step 2 · Find spring range ÷ neap range (k)
Use k = (1 + r) ÷ (1 − r)
At spring tides the two add; at neap tides the Sun’s takes away from the Moon’s.
k
=
1 + 0.4591
1 − 0.4591
k
=
1.4591
0.5409
k
=
2.6975
Step 3 · Find range ÷ the Moon’s alone (F)
Use F = √(1 + r² + 2 × r × cos(2 × θ°))
The
two
bulges
add
like
arrows
at
twice
the
Moon’s
angle:
1
+
r
in
line,
1
−
r
at
right
angles.
F
=
√(1
+
0.4591
2
+
2
×
0.4591
×
cos(2
×
180°))
F
=
√(1
+
0.4591
2
+
2
×
0.4591
×
cos(360°))
F
=
√
1 + 0.2108 + 2 × 0.4591
F
=
√
1 + 0.2108 + 0.9182
F
=
√
2.129
F
=
1.4591
Check
Put the numbers back in. Both sides should match.
0.4591
=
(
1.989 ×
10
30
7.35 ×
10
22
)
×
(
3.844 ×
10
8
1.496 ×
10
11
)³
✓
2.6975
=
1 + 0.4591
1 − 0.4591
✓
1.4591
=
√(1
+
0.4591
2
+
2
×
0.4591
×
cos(2
×
180°))
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Waves and tides
Main lesson
Tsunami speed and travel time
Use this for “A tsunami crosses 3,000 km of ocean 4,000 m deep. How fast does it travel, and how many hours does it take?”
Wave energy and power
Use this for “Waves 2 m high arrive every 10 s. How much energy does each square meter of sea hold, and how much power reaches each meter of coast?”