Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital

Exercises · 7.3

Q. Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

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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

Watch out

Many students think the law says $T \propto a^{3/2}$ — which is true — but then assume that doubling the distance doubles the period. It doesn't. Doubling $a$ multiplies $T$ by $2^{3/2} \approx 2.83$ . The period grows faster than the distance.

The Big Picture

Kepler's Third Law is the key that unlocks the solar system's scale. Before Kepler, astronomers knew the relative distances of planets (e.g., Mars is about 1.5 times farther than Earth), but not the absolute distances. Once you measure one planet's period and distance in real units (say, Earth's 1 year and 1 AU), the law gives you every other planet's distance in kilometers — just by timing their orbits.

It also works in reverse: observe a star's wobble caused by an orbiting planet, measure the planet's period, and you can calculate how far the planet is from the star. This is how most exoplanets are discovered.

Final takeaway: Kepler's Third Law is a simple, beautiful relationship — $T^2 \propto a^3$ — that connects how long an orbit takes to how far out it is. It works because gravity follows an inverse-square law, and it lets us measure the masses of astronomical objects.

Looking up "Kepler's Third Law: definition, formula & real-world examples" is a good habit before an exam, and it is worth knowing that Kepler's Third Law is drawn directly from the Gravitation coverage of the NCERT/CBSE Class 11 Physics syllabus and recurs often in JEE Main and NEET papers. Cross-checking this explanation against the relevant NCERT Physics chapter and solving a few past-year questions will round out your preparation.

For a planet orbiting the Sun, Kepler’s third law ties orbital period and size. If the planet’s period is half that of Earth, its orbital radius is about 0.63 times Earth’s orbital radius — roughly two-thirds the size.

The key here is Kepler’s third law — the square of the orbital period is proportional to the cube of the semi-major axis (orbital size). This law is a direct consequence of the gravitational force being central and inverse-square, and it holds for all planets orbiting the same central body (the Sun).

The problem says the planet goes around the Sun “twice as fast” as Earth. That means its orbital speed is double? Not quite — “goes around twice as fast” in everyday language usually means it completes one orbit in half the time. So the orbital period $T$ of this planet is half of Earth’s period $T_E$ .

Let’s work it out.

State Kepler’s third law For any planet orbiting the Sun,

T^2 \propto a^3

where $T$ is the orbital period and $a$ is the semi-major axis (orbital radius, assuming near-circular orbits).

If we take Earth as reference:

T_E^2 \propto a_E^3

Relate the planet’s period to Earth’s The planet’s period is half of Earth’s:

T = \frac{T_E}{2}

Apply the proportionality For the planet:

T^2 \propto a^3 \quad \Rightarrow \quad \left(\frac{T_E}{2}\right)^2 \propto a^3

\frac{T_E^2}{4} \propto a^3

For Earth:

T_E^2 \propto a_E^3

Dividing the planet’s relation by Earth’s:

\frac{T_E^2/4}{T_E^2} = \frac{a^3}{a_E^3}

\frac{1}{4} = \left(\frac{a}{a_E}\right)^3

Solve for the ratio Take the cube root:

\frac{a}{a_E} = \sqrt[3]{\frac{1}{4}} = \frac{1}{\sqrt[3]{4}}

Numerically, $\sqrt[3]{4} \approx 1.5874$ , so

\frac{a}{a_E} \approx 0.63

Watch out

A common mistake is to think “twice as fast” means orbital speed is double. That would give a different answer (using $v \propto 1/\sqrt{a}$ from circular orbit dynamics). But the phrase “goes around twice as fast” refers to completing the orbit in half the time — period, not speed. Always check what “fast” means in context.

Tip

You can also think: if period halves, $T^2$ becomes one-fourth, so $a^3$ must be one-fourth, meaning $a$ is the cube root of one-fourth. No need to remember numbers — just the cube root of 1/4.

✓Final answer

0.63

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

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 mark
$Q. State Kepler's third law of planetary motion.$
›Reveal solutionSolution

Kepler's third law (the law of periods) states that the square of a planet's orbital period is proportional to the cube of the semi-major axis of its orbit.

Statement: The square of the time period (T) of revolution of a planet around the Sun is directly proportional to the cube of the semi-major axis (a) of its elliptical orbit.

Mathematically: T² ∝ a³, or T²/a³ = constant (the same constant for all planets orbiting the Sun).

This law can be derived from Newton's law of gravitation combined with the centripetal force requirement for (approximately) circular orbits, and it lets us compare the orbital sizes of different planets once their periods are known.

✓Final answer
$Kepler's third law: T² \propto a³ — the square of a planet's orbital period is proportional to the cube of its orbit's semi-major axis.$
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 mark
$Q. Question number 33 (continued) is a case-study based question. Read the case carefully and answer the question that follows: Recently, on 7th September 2025, a lunar eclipse was observed on the day of full moon. Some people in a village avoided eating or cooking during the eclipse, believing it could bring ill effects. A young boy, who had studied about eclipses in science, knew that solar and lunar eclipses are natural celestial phenomena caused by the motion of the Earth, Moon, and Sun and have no harmful effects on living beings. When he explained this to his mother, she ignored his reasoning and followed the traditional belief. On his way to a friend's house, he noticed people performing prayers and rituals to ward off the so-called "ill effects" of the eclipse. The boy felt disheartened at the superstitions and decided to study more about the scientific reason behind such phenomena.$
(iii) The time period of Jupiter around the Sun is 10 years. Calculate its distance from the Sun, given that the Earth's distance from the Sun is R and its time period is 1 year. OR Alternative case study (Or), continued: Ajay was always fascinated by the flights of airplanes and rockets. He often wondered how rockets could travel into space and not return, while everything he threw upward would always fall back due to Earth's gravitational pull. Curious about this, he discussed the matter with his elder brother, who explained that rockets are launched based on a scientific principle. When the rocket is projected with increasing speed, a point comes when its speed becomes high enough to overcome Earth's gravitational pull — this speed is called the escape velocity. At this speed, the rocket can move out of Earth's gravitational field and never return on its own. Understanding this concept inspired Ajay to dream of becoming an astronaut in the future.
(iii) Find the ratio of escape velocity to orbital velocity for a satellite revolving close to the earth's surface.
›Reveal solutionSolution

Jupiter's distance from the Sun is about 4.64R, where R is the Earth's distance from the Sun.

(This is the primary part of the case-study question; the alternative — the ratio of escape velocity to orbital velocity — is not required since the primary is fully answerable.)

By Kepler's third law, T² ∝ r³ for planets orbiting the Sun, where T is the orbital period and r is the (mean) orbital radius.

For Earth: T_E = 1 year, r_E = R.

For Jupiter: T_J = 10 years, r_J = ?

Taking the ratio:

(T_J / T_E)² = (r_J / R)³

(10/1)² = (r_J/R)³

100 = (r_J/R)³

r_J/R = 100^(1/3) ≈ 4.64

So r_J ≈ 4.64 R.

✓Final answer
$Jupiter's distance from the Sun is about 4.64R (obtained from Kepler's third law, (T_J/T_E)² = (r_J/R)³, using T_J = 10 years and T_E = 1 year).$

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Q.Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

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