How will you 'weigh the sun', that is estimate its mass? The mean orbital radius of the earth around the sun i

Exercises · 7.13

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

Yanam CbseNCERTSubjective· 3mImportance★★★★★est

31% · 21/67 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Kepler's Third Law

Kepler's Third Law: The Harmony of the Planets

Imagine you're watching two planets orbiting the Sun. One is close in — Mercury, zipping around in just 88 days. Another is far out — Saturn, taking nearly 30 years to complete one lap. You'd expect the farther planet to take longer, but here's the surprising part: the relationship isn't just "farther = slower." It's much more precise, and it reveals a deep truth about gravity itself.

The Intuition

Think of a planet as a runner on a circular track. The farther out the track, the longer the lap — that's obvious. But Kepler noticed something subtler: if you double the distance from the Sun, the orbital period doesn't just double. It increases by a factor of about $2.8$ (which is $\sqrt{8}$ ). Triple the distance, and the period grows by about $5.2$ (which is $\sqrt{27}$ ).

There's a pattern here. The period seems to grow as the 3/2 power of the distance. Why? Because gravity weakens with distance, so a farther planet feels a weaker pull and moves more slowly — not just because the track is longer, but because it's moving slower along that track.

The Precise Statement

T^2 \propto a^3

For planets orbiting the Sun, if you measure $T$ in Earth years and $a$ in astronomical units (AU, where 1 AU = Earth's average distance from the Sun), the constant of proportionality is exactly 1:

T^2 = a^3

So for Earth: $T = 1$ year, $a = 1$ AU, and $1^2 = 1^3$ — it checks out.

For Mars: $a \approx 1.52$ AU, so $T^2 = (1.52)^3 \approx 3.51$ , giving $T \approx 1.87$ years. That's about 687 days — exactly right.

Note

This law applies to any body orbiting a much more massive central body: moons around planets, satellites around Earth, binary stars around each other. The constant of proportionality changes depending on the mass of the central body.

Why It Works (The Physics)

Newton later showed that Kepler's Third Law is a direct consequence of his law of gravitation. For a circular orbit (a good approximation for most planets), the centripetal force needed to keep the planet in orbit is provided by gravity:

\frac{GMm}{r^2} = \frac{mv^2}{r}

Here $M$ is the Sun's mass, $m$ the planet's mass, $r$ the orbital radius, and $v$ the orbital speed. The speed is related to the period by $v = 2\pi r / T$ . Substituting and simplifying:

\frac{GM}{r^2} = \frac{4\pi^2 r}{T^2}

Rearranging:

T^2 = \frac{4\pi^2}{GM} r^3

The quantity $4\pi^2/(GM)$ is a constant for all planets orbiting the Sun. So $T^2 \propto r^3$ — exactly Kepler's law.

Important

The constant $4\pi^2/(GM)$ depends only on the mass of the central body. This means: if you know the period and distance of any moon or planet, you can calculate the mass of the body it orbits. This is how astronomers "weigh" stars, black holes, and galaxies.

A Common Mistake …

Start your 14-day free trial to unlock the full solution →

We use Kepler’s Third Law in Newton’s form: the orbital period and radius of Earth give the Sun’s mass directly. The result is $M_\odot \approx 2.0 \times 10^{30}\ \text{kg}$ .

The idea is beautifully simple. Earth orbits the Sun because of gravity. If we know how far away Earth is and how long it takes to go around once, Newton’s law of gravitation and circular motion let us solve for the Sun’s mass. No need to visit the Sun — just use the orbit.

Kepler’s Third Law originally said “the square of the period is proportional to the cube of the semi-major axis.” Newton later gave it physical meaning: the constant of proportionality involves the mass of the central body. For a planet orbiting a much heavier star, the law becomes:

T^2 = \frac{4\pi^2}{GM_\odot} a^3

where $T$ is the orbital period, $a$ is the mean orbital radius (semi-major axis), $G$ is the gravitational constant, and $M_\odot$ is the Sun’s mass. Rearranging gives $M_\odot = \frac{4\pi^2 a^3}{G T^2}$ .

Let’s plug in the numbers step by step.

Get the data in consistent units. The mean orbital radius is given as $a = 1.5 \times 10^8\ \text{km}$ . Convert to metres:

a = 1.5 \times 10^8 \times 10^3 = 1.5 \times 10^{11}\ \text{m}.

Earth’s orbital period is one year. In seconds:

T = 1\ \text{year} = 365.25 \times 24 \times 3600 \approx 3.156 \times 10^7\ \text{s}.

(We use 365.25 days to account for the leap-year cycle — it’s precise enough.)

Recall the gravitational constant.

G = 6.67 \times 10^{-11}\ \text{N·m}^2/\text{kg}^2.

Compute $a^3$ .

a^3 = (1.5 \times 10^{11})^3 = 3.375 \times 10^{33}\ \text{m}^3.

Compute $T^2$ .

T^2 = (3.156 \times 10^7)^2 \approx 9.96 \times 10^{14}\ \text{s}^2.

Plug into the formula.

M_\odot = \frac{4\pi^2 \times 3.375 \times 10^{33}}{6.67 \times 10^{-11} \times 9.96 \times 10^{14}}.

First, the numerator: $4\pi^2 \approx 39.48$ , so

$39.48 \times 3.375 \times 10^{33} \approx 1.332 \times 10^{35}.$ …

Start your 14-day free trial to unlock the full solution →

←Previous question7.12 — A rocket is fired from the earth towards the sun. At what distance fro…→Next question7.14 — A saturn year is 29.5 times the earth year. How far is the saturn from…

Q.How will you 'weigh the sun', that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5×108 km1.5 \times 10^{8}\text{ km}1.5×108 km.

🎓Unlock everything free for 14 days

Q.How will you 'weigh the sun', that is estimate its mass? The mean orbital radius of the earth around the sun is $1.5 \times 10^{8}\text{ km}$ .

Unlock everything free for 14 days