A star like the sun has several bodies moving around it at different distances. Consider that all of them are moving in circular orbits. Let $r$ be the distance of the body from the centre of the star and let its linear velocity be $v$, angular velocity $\omega$, kinetic energy $K$, gravitational potential energy $U$, total energy $E$ and angular momentum $l$. As the radius $r$ of the orbit increases, determine which of the above quantities increase and which ones decrease.

A star like the sun has several bodies moving around it at different distances. Consider that all of them are

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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 …

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For a body in a circular orbit under a central gravitational force, as $r$ increases: $v$ , $\omega$ , $K$ , and $E$ decrease; $U$ increases (becomes less negative); $l$ increases. The key is that angular momentum grows with $r$ even though speed drops.

The entire analysis hinges on one idea: the gravitational force provides the centripetal force. That single equality ties $v$ , $\omega$ , $r$ , and the masses together, and from it every other quantity follows.

Let the star have mass $M$ and the orbiting body have mass $m$ . For a circular orbit of radius $r$ , Newton’s law of gravitation and centripetal force give:

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

Cancel $m$ and one factor of $r$ :

\frac{GM}{r} = v^2

That’s the master relation. Everything else is derived from it.

Linear speed $v$

From $v^2 = \frac{GM}{r}$ , we see $v \propto 1/\sqrt{r}$ . As $r$ increases, $v$ decreases.
Angular speed $\omega$

Since $v = \omega r$ , substitute $v = \sqrt{GM/r}$ to get $\omega = v/r = \sqrt{GM/r^3}$ . So $\omega \propto 1/r^{3/2}$ — it decreases even faster than $v$ .
Kinetic energy $K$

$K = \frac12 m v^2 = \frac12 m \cdot \frac{GM}{r} = \frac{GMm}{2r}$ . Clearly $K \propto 1/r$ , so $K$ decreases as $r$ grows.
Gravitational potential energy $U$

$U = -\frac{GMm}{r}$ . This is negative, and its magnitude shrinks as $r$ increases. Since $-1/r$ becomes less negative, $U$ increases (it goes from a larger negative number toward zero).
Total energy $E$ …

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