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.

TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
$11.2~\mathrm{km~s^{-1}}$
›Reveal solutionSolution
Escape speed depends on mass and radius. Given constant density, radius scales as the cube root of mass, so escape speed scales as $M^{1/3}$ . With mass 8 times Earth’s, escape speed doubles to $22.4\ \mathrm{km\,s^{-1}}$ .

The escape speed from a planet’s surface is the minimum speed needed for an object to leave its gravitational pull forever. The formula is $v_e = \sqrt{\frac{2GM}{R}}$ , where $M$ is the planet’s mass and $R$ its radius. The key insight: if the mean density $\rho$ is the same as Earth’s, then mass and radius are linked through $\rho = \frac{M}{\frac{4}{3}\pi R^3}$ , so $R \propto M^{1/3}$ . That lets us find how $v_e$ changes when only $M$ changes, without needing numerical values.

Write the escape speed formula.

For Earth: $v_e = \sqrt{\frac{2GM}{R}}$ . For the other planet: $v_e' = \sqrt{\frac{2G M'}{R'}}$ , where $M' = 8M$ and $\rho' = \rho$ .

Relate radius to mass using constant density.

Since $\rho = \frac{M}{\frac{4}{3}\pi R^3}$ is the same for both,

$\frac{M}{R^3} = \frac{M'}{R'^3} \quad\Rightarrow\quad \frac{R'}{R} = \left(\frac{M'}{M}\right)^{1/3} = 8^{1/3} = 2.$
So the planet’s radius is twice Earth’s.

Substitute into the escape speed ratio.

$\frac{v_e'}{v_e} = \sqrt{\frac{M'}{M} \cdot \frac{R}{R'}} = \sqrt{8 \cdot \frac{1}{2}} = \sqrt{4} = 2.$
Therefore $v_e' = 2 \times 11.2\ \mathrm{km\,s^{-1}} = 22.4\ \mathrm{km\,s^{-1}}$ .

Watch out

A common mistake is to forget that radius changes with mass when density is fixed. If you blindly use $v_e \propto \sqrt{M}$ (assuming same radius), you’d get $v_e' = 11.2 \times \sqrt{8} \approx 31.7\ \mathrm{km\,s^{-1}}$ , which is not among the options — a clue that radius must adjust.

Tip

For constant density, escape speed scales as $M^{1/3}$ (since $R \propto M^{1/3}$ and $v_e \propto \sqrt{M/R} \propto M^{1/3}$ ). Doubling mass multiplies escape speed by $2^{1/3}$ , but here mass is $8 = 2^3$ times, so $v_e$ doubles — a clean shortcut.

✓Final answer
$22.4\ \mathrm{km\,s^{-1}}$
TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
$\dfrac{1}{10}$
›Reveal solutionSolution
The escape velocity depends on both mass and radius. Equating the given ratios leads to the planet’s radius being $\frac{R}{5}$ .

The escape velocity from a celestial body is the minimum speed needed for an object to break free from its gravitational pull without further propulsion. The formula is derived from energy conservation: kinetic energy at launch equals the work done against gravity to infinity. For a planet of mass $M$ and radius $R$ , the escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$ , where $G$ is the universal gravitational constant.

The key insight: escape velocity scales as the square root of mass over radius. So if we know how mass and escape velocity compare between two planets, we can solve for the unknown radius ratio.

Let’s denote Earth’s mass as $M_e$ and radius as $R_e = R$ . The planet’s mass is $M_p = \frac{1}{10} M_e$ . Its escape velocity $v_{e,p}$ is given as $\frac{1}{\sqrt{2}}$ times Earth’s escape velocity $v_{e,e}$ .

Write the escape velocity for Earth:

$v_{e,e} = \sqrt{\frac{2GM_e}{R}}$

Write the escape velocity for the planet:

$v_{e,p} = \sqrt{\frac{2G M_p}{R_p}} = \sqrt{\frac{2G \cdot \frac{1}{10} M_e}{R_p}}$

The problem states:

$v_{e,p} = \frac{1}{\sqrt{2}} v_{e,e}$
Substitute the expressions:
$\sqrt{\frac{2G \cdot \frac{1}{10} M_e}{R_p}} = \frac{1}{\sqrt{2}} \sqrt{\frac{2GM_e}{R}}$

Square both sides to remove square roots:

$\frac{2G \cdot \frac{1}{10} M_e}{R_p} = \frac{1}{2} \cdot \frac{2GM_e}{R}$

Simplify. Cancel $2G M_e$ (non-zero) from both sides:

$\frac{1/10}{R_p} = \frac{1}{2} \cdot \frac{1}{R}$ $\frac{1}{10 R_p} = \frac{1}{2R}$

Cross-multiply:

$2R = 10 R_p$ $R_p = \frac{2R}{10} = \frac{R}{5}$
Watch out

A common mistake is to forget that the escape velocity formula has $M$ in the numerator and $R$ in the denominator. If you accidentally invert the ratio, you might get $5R$ instead of $R/5$ . Always check: a smaller mass reduces escape velocity, but a smaller radius increases it — the given factor $\frac{1}{\sqrt{2}}$ tells you the radius must be smaller than Earth’s.

✓Final answer
$\boxed{\frac{R}{5}}$
TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
$x$
›Reveal solutionSolution
The key idea is conservation of angular momentum: if Earth’s radius becomes $x$ times larger, its moment of inertia increases by $x^2$ , so its rotation slows by $x^2$ , making the new period $24x^2$ hours.

We start with the concept: Earth’s rotation is determined by its angular momentum, which is conserved if no external torque acts. When the radius changes, the mass distribution changes, altering the moment of inertia. Since angular momentum $L = I \omega$ stays constant, the angular speed $\omega$ must adjust inversely to $I$ . The period $T = 2\pi / \omega$ then changes proportionally to $I$ .

Moment of inertia of a sphere For a solid sphere of mass $M$ and radius $R$ , the moment of inertia about its axis is $I = \frac{2}{5} M R^2$ . If the radius becomes $x$ times the present value, the new radius is $R' = xR$ , so the new moment of inertia is

$I' = \frac{2}{5} M (xR)^2 = \frac{2}{5} M R^2 \cdot x^2 = I \cdot x^2.$

Conservation of angular momentum No external torque acts (we assume the change happens internally or gradually), so

$L = I \omega = I' \omega'.$
Substituting $I' = I x^2$ gives
$I \omega = (I x^2) \omega' \quad \Rightarrow \quad \omega' = \frac{\omega}{x^2}.$

Relation between period and angular speed The period $T = 2\pi / \omega$ . The new period is

$T' = \frac{2\pi}{\omega'} = \frac{2\pi}{\omega / x^2} = x^2 \cdot \frac{2\pi}{\omega} = x^2 T.$
The present period of Earth’s rotation is $T = 24$ hours. Hence
$T' = 24 x^2 \text{ hours}.$
Watch out

A common mistake is to think the period changes linearly with radius (like $T' = 24x$ ). That would be true only if the moment of inertia scaled as $R$ , but it scales as $R^2$ for a sphere.

Tip

Notice that the factor $\frac{2}{5}$ cancels out — the exact shape factor doesn’t matter as long as the mass distribution is similar. Any object whose moment of inertia scales as $R^2$ will have its period scale as $R^2$ .

✓Final answer
$The correct option is (C) . ANSWER: C$
TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
$Q. Which of the following statement is false? (A) All planets move in elliptical orbits with sun at one of the foci. (B) The square root of time period of revolution of planet is proportional to the cube of semi major axis of the ellipse. (C) Line that joins any planet to the sun sweeps equal areas in equal intervals of time. (D) The measurement of G has refined by Cavendish’s experiment.$
›Reveal solutionSolution
Kepler’s three laws describe planetary motion; statement (B) misstates the third law (it says “square root” instead of “square”), making it the false statement.

The question tests your knowledge of Kepler’s laws of planetary motion and a famous experimental result. Kepler’s laws are:

Law of Ellipses – planets move in ellipses with the Sun at one focus.

Law of Equal Areas – a line joining a planet to the Sun sweeps out equal areas in equal times.

Law of Harmonies – the square of the orbital period is proportional to the cube of the semi-major axis.

Statement (D) refers to Cavendish’s experiment, which measured the gravitational constant $G$ . That is a true historical fact. So the false statement must be among (A)–(C). Let’s check each carefully.

Statement (A): “All planets move in elliptical orbits with sun at one of the foci.”

This is exactly Kepler’s first law. It is true.

Pitfall: Some might think orbits are perfectly circular, but Kepler showed they are ellipses (circles are a special case of ellipses, but the Sun is at a focus, not the center).

Statement (B): “The square root of time period of revolution of planet is proportional to the cube of semi major axis of the ellipse.”

Kepler’s third law says: $T^2 \propto a^3$ , where $T$ is the period and $a$ is the semi-major axis.

Taking square roots: $\sqrt{T^2} = T \propto a^{3/2}$ . That is not “square root of time period proportional to cube of semi-major axis.” The statement says $\sqrt{T} \propto a^3$ , which is false.

Watch out

A common mistake is to misremember the exponent: it’s $T^2 \propto a^3$ , not $T^{1/2} \propto a^3$ . The phrase “square root of time period” means $\sqrt{T}$ , which is wrong.

Statement (C): “Line that joins any planet to the sun sweeps equal areas in equal intervals of time.”

This is Kepler’s second law, true for all planets.

Statement (D): “The measurement of G has refined by Cavendish’s experiment.”

Cavendish’s torsion balance experiment in 1798 gave the first accurate measurement of the gravitational constant $G$ . This is historically correct.

Thus, the only false statement is (B).

✓Final answer
$The correct option is (B). ANSWER: B$

←Previous question7.2 — Choose the correct alternative: (a) Acceleration due to gravity increa…→Next question7.4 — Io, one of the satellites of Jupiter, has an orbital period of 1.769 d…

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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