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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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$Q. A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into another circular orbit of radius 1.01 R around the earth. The period of revolution of the second satellite is larger than that of the first one by a percent of (approximately) (A) 0.5 (B) 1.0 (C) 1.5 (D) 3.0$
›Reveal solutionSolution
Kepler's third law gives $T \propto R^{3/2}$ ; a 1% increase in orbital radius produces about a 1.5% increase in period, using the binomial approximation.

Concept and Intuition

For any body in circular orbit around the Earth, gravity supplies the centripetal force, which leads directly to Kepler's third law: $T^2 \propto R^3$ , i.e. $T \propto R^{3/2}$ . For a small fractional change in radius, the fractional change in period scales by the power (3/2), via the binomial approximation $(1+x)^n \approx 1 + nx$ for small $x$ .

Step-by-Step Solution

$T \propto R^{3/2}$ , so $\dfrac{T_2}{T_1} = \left(\dfrac{R_2}{R_1}\right)^{3/2}$ .

Here $R_2 = 1.01\,R_1$ , so $\dfrac{T_2}{T_1} = (1.01)^{3/2}$ .

Using the binomial approximation: $(1+0.01)^{1.5} \approx 1 + 1.5 \times 0.01 = 1.015$ .

So $T_2$ is about 1.5% larger than $T_1$ .

Common Mistakes …
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AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$T \propto a^{3/2} g^{x} R^{y}$
›Reveal solutionSolution
Tests deriving the correct powers of $g$ and $R$ in Kepler's-third-law-type expression for orbital period by relating $GM$ to surface gravity $g$ . Answer: $x=-1/2,\ y=-1$ .

Concept and Intuition

The standard orbital period formula involves $GM$ (the planet's gravitational parameter), not $g$ and $R$ directly. But since surface gravity is $g = GM/R^2$ , we can substitute $GM = gR^2$ to re-express the period purely in terms of the given variables $a$ , $g$ , $R$ — this is a dimensional/algebraic substitution trick rather than new physics.

Step-by-Step Solution

Start from Kepler's third law (derived from gravity providing centripetal force):

$T = 2\pi\sqrt{\frac{a^3}{GM}}$

Use $g = \dfrac{GM}{R^2}$ (definition of surface gravity), so $GM = gR^2$ .

Substitute: $T = 2\pi\sqrt{\frac{a^3}{gR^2}} = 2\pi\, a^{3/2}\, (gR^2)^{-1/2} = 2\pi\, a^{3/2}\, g^{-1/2}\, R^{-1}$ …
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AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$K\sqrt{\dfrac{R^2}{GM}}$
›Reveal solutionSolution
Dimensional analysis on $T=KR^aM^bG^c$ forces $a=3/2,\,b=-1/2,\,c=1/2$ , giving $T=K\sqrt{R^3/GM}$ — the familiar Kepler third-law form.

Concept and Intuition

When a quantity depends on several others through an unknown power law, dimensional analysis pins down the powers by requiring the dimensions on both sides to match exactly, since a physical equation must be dimensionally homogeneous.

Step-by-Step Solution

Assume $T = K\,R^a M^b G^c$ (K dimensionless).

Dimensions: $[T]=\text{T}$ , $[R]=\text{L}$ , $[M]=\text{M}$ , $[G]=\text{M}^{-1}\text{L}^3\text{T}^{-2}$ .

Equate: $\text{T}^1\text{M}^0\text{L}^0 = \text{L}^{a}\,\text{M}^{b}\,\left(\text{M}^{-1}\text{L}^{3}\text{T}^{-2}\right)^{c}$ .

Collect powers: L: $a+3c=0$ ; M: $b-c=0$ ; T: $-2c=1$ . …
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AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$T^{-1}$
›Reveal solutionSolution
Combining Kepler's third law ( $R\propto T^{2/3}$ ) with the orbital kinetic energy formula ( $KE\propto 1/R$ ) shows $KE\propto T^{-2/3}$ .

Concept and Intuition

For a satellite in circular orbit, gravity supplies the centripetal force, which directly links its kinetic energy to the orbital radius. Kepler's third law separately links the orbital radius to the period. Chaining the two relations connects kinetic energy to the period.

Step-by-Step Solution

Gravity provides centripetal force: $\dfrac{mv^2}{R} = \dfrac{GMm}{R^2} \Rightarrow v^2 = \dfrac{GM}{R}$ .

Kinetic energy: $KE = \tfrac12mv^2 = \dfrac{GMm}{2R} \propto \dfrac1R$ .

Kepler's third law: $T^2\propto R^3 \Rightarrow R\propto T^{2/3}$ . …
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AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
$6.67 \times 10^6$
›Reveal solutionSolution
For a circular orbit, gravity supplies the centripetal force, giving $v=\sqrt{GM/r}$ ; plugging in the numbers gives about 7.75 km/s.

Concept and Intuition

For a satellite in a circular orbit, gravitational attraction is exactly the centripetal force needed to keep it in that orbit: $\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}$ , which gives the orbital speed $v=\sqrt{GM/r}$ — notably independent of the satellite's own mass.

Step-by-Step Solution

$GM = (6.67\times10^{-11})(6\times10^{24}) = 4.002\times10^{14}\ \mathrm{m^3 s^{-2}}$ .

$v=\sqrt{GM/r}=\sqrt{\dfrac{4.002\times10^{14}}{6.67\times10^{6}}}$ .

$\dfrac{4.002\times10^{14}}{6.67\times10^{6}}\approx6.0\times10^{7}$ . …
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AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
$= 6\times10^{24}\ \mathrm{kg}$
›Reveal solutionSolution
Kepler's third law / circular-orbit condition gives the period; converting seconds to minutes gives approximately 105 minutes.

Concept and Intuition

For a satellite in circular orbit, gravity supplies the centripetal force: $\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}=m\omega^2 r$ , leading to $T=2\pi\sqrt{r^3/GM}$ where $r$ is the orbital radius measured from Earth's centre (so altitude must be added to Earth's radius).

Step-by-Step Solution

$r = R_e + h = 6.4\times10^6 + 1\times10^6 = 7.4\times10^6$ m.

$r^3 = (7.4\times10^6)^3 = 405.224\times10^{18} = 4.05224\times10^{20}\ \mathrm{m^3}$ .

$GM = 6.67\times10^{-11}\times6\times10^{24} = 4.002\times10^{14}\ \mathrm{m^3s^{-2}}$ . …
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AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
$\dfrac{1}{R}$
›Reveal solutionSolution
If gravity instead followed $F \propto 1/R^3$ , balancing this against the centripetal force requirement for a circular orbit shows $T \propto R^2$ — a much steeper dependence than Kepler's usual $T \propto R^{3/2}$ .

Concept and Intuition

For any central force law $F = k/R^n$ providing the centripetal force for a circular orbit, we can derive how the orbital period scales with radius by equating $F$ to $m\omega^2 R$ and solving for $\omega$ , then converting to $T = 2\pi/\omega$ .

Step-by-Step Solution

Required centripetal force for circular motion: $F = m\omega^2 R$ .

Given force law: $F = \dfrac{k}{R^3}$ (inverse cube).

Equate: $m\omega^2 R = \dfrac{k}{R^3} \Rightarrow \omega^2 = \dfrac{k}{mR^4} \Rightarrow \omega \propto R^{-2}$ . …
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AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
$Q. Two planets, A and B orbit around a star such that time period of A is 8 times the time period of B. The ratio of orbital velocities of the planets A and B is (A) 4:1 (B) 1:4 (C) 2:1 (D) 1:2$
›Reveal solutionSolution
Combine Kepler's third law with the definition of orbital speed to relate speed ratio to period ratio.

Concept and Intuition

For circular orbits around a common star, Kepler's third law gives $T^2\propto r^3$ . Orbital speed is $v=\dfrac{2\pi r}{T}$ . Eliminating $r$ in favor of $T$ lets us directly compare speeds using only the period ratio.

Step-by-Step Solution

Kepler's third law: $T^2 \propto r^3 \Rightarrow r \propto T^{2/3}$ .

Orbital speed: $v = \dfrac{2\pi r}{T} \propto \dfrac{T^{2/3}}{T} = T^{-1/3}$ .

Given $T_A = 8T_B$ : $\dfrac{v_A}{v_B} = \left(\dfrac{T_A}{T_B}\right)^{-1/3} = 8^{-1/3} = \dfrac12$ . …
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AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
$Q. The orbital period of a geostationary satellite is (A) 2 h (B) 5 h (C) 24 h (D) 12 h$
›Reveal solutionSolution
"Geostationary" means the satellite appears stationary relative to a fixed

point on Earth's surface, which requires its orbital period to exactly match

Earth's rotation period — 24 hours.

Concept and Intuition

A satellite in a geostationary orbit orbits in the equatorial plane, in the

same direction as Earth's rotation, at exactly the altitude (about 35,786 km)

where Kepler's third law gives an orbital period equal to Earth's sidereal

rotation period. Because the satellite's angular position tracks Earth's

rotation exactly, it appears to hang motionless over one longitude — which is

precisely why it must have a period of 24 hours (approximating the sidereal

day as 24 h for this level of question).

Step-by-Step Solution

"Geostationary" is defined operationally: the satellite must remain above the same point on Earth's surface at all times.

This is only possible if the satellite's angular velocity equals Earth's angular velocity of rotation.

Earth completes one rotation (one day) in 24 hours. …
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AP EAPCET 2021Set ap-2021-10-05-FN1 markMCQ
$7R$
›Reveal solutionSolution
Applying Kepler's third law ( $T^2\propto r^3$ ) between the geostationary orbit (radius $8R$ , $T=24$ h) and the new orbit (radius $2R$ ) gives a new period of 3 hours.

Concept and Intuition

A geostationary satellite orbits Earth with a period equal to Earth's rotation period, $T=24$ hours, at an orbital radius corresponding to a height of about $7R$ above the surface — i.e., orbital radius $r_1 = R + 7R = 8R$ from Earth's centre.

Kepler's third law states that for objects orbiting the same central body (Earth here), $T^2 \propto r^3$ , i.e., $\dfrac{T_2^2}{T_1^2} = \left(\dfrac{r_2}{r_1}\right)^3$ . This lets us find the new period once the new orbital radius is known, without needing to know $G$ or $M$ explicitly.

Step-by-Step Solution

Original orbital radius (geostationary): $r_1 = R + 7R = 8R$ , with $T_1 = 24$ h.

New orbital radius (height $R$ above surface): $r_2 = R + R = 2R$ . …
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