A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 × 10⁸ km a

Exercises · 7.14

1.50 \times 10^{8}\text{ km}

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Concept understanding — Kepler Third Law

Kepler's Third Law: The Harmony of the Planets

Imagine you're pushing a child on a swing. If you push harder, the swing goes higher and takes longer to come back to you. Now picture the planets: Mercury, the closest planet to the Sun, zips around in just 88 days. Neptune, way out at the edge, takes 165 years for one lap. There's a pattern here — the farther a planet is from the Sun, the slower it moves and the longer its year.

Kepler's Third Law is the mathematical rule that captures this exact relationship. It tells you: the time a planet takes to orbit the Sun (its period) is linked to its average distance from the Sun in a very specific way.

The Intuition: Why Distance Matters

Think of gravity as an invisible rope. The Sun pulls on each planet, and that pull gets weaker with distance. A planet close to the Sun feels a strong tug — it has to move fast to avoid being pulled in. A planet far away feels a weak tug — it can afford to drift slowly.

But there's a second effect: a farther planet also has a much longer path to travel (the circumference of its orbit is larger). So you have two things working together:

Weaker gravity → slower speed
Longer path → more distance to cover

Both effects push in the same direction: farther planets take dramatically longer to orbit. Kepler discovered that this isn't just a rough trend — it's a precise mathematical law.

The Precise Statement

T^2 \propto a^3

Where:

$T$ = orbital period (time for one full orbit, usually in Earth years)
$a$ = semi-major axis (average distance from the Sun, usually in Astronomical Units, where 1 AU = Earth-Sun distance)

The symbol $\propto$ means "is proportional to." So the law says: The square of the orbital period is proportional to the cube of the average distance from the Sun.

If you want an equation with a constant, it's:

T^2 = k \cdot a^3

For planets orbiting the Sun, if you measure $T$ in Earth years and $a$ in AU, the constant $k$ equals exactly 1. That makes it beautifully simple:

T^2 = a^3

What This Means in Practice

Let's test it with real planets:

Earth: $a = 1$ AU, $T = 1$ year. Check: $1^2 = 1^3$ ✓

Mars: $a \approx 1.52$ AU. Cube that: $1.52^3 \approx 3.51$ . Square root gives $T \approx \sqrt{3.51} \approx 1.87$ years. Mars's actual orbital period? 1.88 years. Almost exact.

Jupiter: $a \approx 5.2$ AU. Cube: $5.2^3 \approx 140.6$ . Square root: $T \approx 11.86$ years. Jupiter's actual period? 11.86 years. Perfect.

Note

This law works for any body orbiting a central mass — not just planets around the Sun. Moons around Jupiter, satellites around Earth, even binary stars around each other. The constant $k$ changes depending on the mass of the central object, but the $T^2 \propto a^3$ relationship always holds.

Why It's So Important

Kepler published this in 1619, decades before Newton explained why it works. Newton later showed that Kepler's Third Law is a direct consequence of his law of universal gravitation. The law lets astronomers: …

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Using Kepler’s Third Law, the orbital period ratio gives the semi-major axis ratio. Saturn’s distance from the Sun is about $1.43 \times 10^{9}\text{ km}$ .

The problem is a direct application of Kepler’s Third Law — one of the most elegant results in classical mechanics. It tells us that for planets orbiting the same star, the square of the orbital period is proportional to the cube of the semi-major axis (average distance from the star).

Why does this work? Because the gravitational force from the Sun provides the centripetal force for nearly circular orbits, and the math collapses into a simple ratio:

\frac{T^2}{a^3} = \text{constant for all planets in the system}.

So if you know the period ratio, you can find the distance ratio — no need for masses or messy constants.

Write Kepler’s Third Law in ratio form For two planets orbiting the same star:

\frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3}

Here, $T$ is the orbital period and $a$ is the average distance from the Sun.

Let Earth be planet 1 and Saturn be planet 2.

Plug in the given data

Earth’s period: $T_{\text{earth}} = 1$ year

Saturn’s period: $T_{\text{saturn}} = 29.5$ years

Earth’s distance: $a_{\text{earth}} = 1.50 \times 10^{8}\text{ km}$

We need $a_{\text{saturn}}$ .

So:

\frac{(29.5)^2}{1^2} = \frac{a_{\text{saturn}}^3}{(1.50 \times 10^{8})^3}

Solve for $a_{\text{saturn}}$ First, compute the left side:

29.5^2 = (30 - 0.5)^2 = 900 - 30 + 0.25 = 870.25

(You can also do $29.5 \times 29.5 = 870.25$ directly.)

So:

a_{\text{saturn}}^3 = 870.25 \times (1.50 \times 10^{8})^3

Take the cube root:

a_{\text{saturn}} = (870.25)^{1/3} \times 1.50 \times 10^{8}

Estimate the cube root $9^3 = 729$ , $10^3 = 1000$ . Since 870.25 is closer to 900 than to 729, try $9.5^3$ :

9.5^3 = (9.5)^2 \times 9.5 = 90.25 \times 9.5 = 857.375

That’s a bit low. Try $9.55^3$ :

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Q.A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50×108 km1.50 \times 10^{8}\text{ km}1.50×108 km away from the sun?

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Q.A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is $1.50 \times 10^{8}\text{ km}$ away from the sun?

Unlock everything free for 14 days