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 8). Triple the distance, and the period grows by about 5.2 (which is 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
T2∝a3
The square of the orbital period T is proportional to the cube of the semi-major axis a of the orbit.
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:
T2=a3
So for Earth: T=1 year, a=1 AU, and 12=13 — it checks out.
For Mars: a≈1.52 AU, so T2=(1.52)3≈3.51, giving T≈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:
r2GMm=rmv2
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=2πr/T. Substituting and simplifying:
r2GM=T24π2r
Rearranging:
T2=GM4π2r3
The quantity 4π2/(GM) is a constant for all planets orbiting the Sun. So T2∝r3 — exactly Kepler's law.
Important
The constant 4π2/(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∝a3/2 — which is true — but then assume that doubling the distance doubles the period. It doesn't. Doubling a multiplies T by 23/2≈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 — T2∝a3 — 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.
Concept: Kepler’s Third Law — the square of the orbital period is proportional to the cube of the semi-major axis (orbital size).
Let TE and RE be Earth’s period and orbital radius. For the planet, TP=21TE (twice as fast means half the period).
Kepler’s third law: TE2TP2=RE3RP3.
Substitute: (21)2=RE3RP3⟹41=RE3RP3.
Take cube root: RERP=(41)1/3=341.
✓Final answer
The planet’s orbital size is 341 times that of Earth’s orbit.
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 TE.
Let’s work it out.
State Kepler’s third law
For any planet orbiting the Sun,
T2∝a3
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:
TE2∝aE3
Relate the planet’s period to Earth’s
The planet’s period is half of Earth’s:
T=2TE
Apply the proportionality
For the planet:
T2∝a3⇒(2TE)2∝a3
So
4TE2∝a3
For Earth:
TE2∝aE3
Dividing the planet’s relation by Earth’s:
TE2TE2/4=aE3a3
41=(aEa)3
Solve for the ratio
Take the cube root:
aEa=341=341
Numerically, 34≈1.5874, so
aEa≈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∝1/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, T2 becomes one-fourth, so a3 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
The orbital size of the planet is about 0.63 times that of Earth, i.e., a≈0.63aE.
Step 1: 'Twice as fast' means the orbital period is halved: TP=21TE.
Step 2: Apply Kepler's third law, T2∝R3, for both planets around the same Sun: (TETP)2=(RERP)3.
Step 3: Substitute: (21)2=41=(RERP)3.
Step 4: Take the cube root: RERP=(41)1/3=341≈0.63 — the faster planet's orbit is smaller.