Q.The planet Mars has two moons, phobos and delmos.
(i) phobos has a period 7 hours, 39 minutes and an orbital radius of 9.4×103 km. Calculate the mass of mars.
(ii) Assume that earth and mars move in circular orbits around the sun, with the martian orbit being 1.52 times the orbital radius of the earth. What is the length of the martian year in days?
Kerala DhseTextbookSubjective· 3mImportance★★★★★est
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.
Using the force-balance (Kepler's third law) relation between orbital radius, period, and central mass, Phobos's orbit gives a mass of Mars of about 6.48×1023 kg. Comparing Mars's and Earth's orbital radii via Kepler's third law for the Sun's system gives a Martian year of about 684 days.
Part (i): Mass of Mars from Phobos's orbit
For a moon in a circular orbit, gravity supplies the centripetal force:
r2GMMarsm=mω2r=mT24π2r
Solving for the mass of Mars:
MMars=GT24π2r3
Convert the given data to SI units: T=7 h 39 min=7×3600+39×60=27,540 s, and r=9.4×103 km=9.4×106 m.
Compute r3=(9.4×106)3≈8.306×1020 m3, and T2=(27,540)2≈7.585×108 s2. Then
MMars=(6.67×10−11)×7.585×1084π2×8.306×1020=5.06×10−23.279×1022≈6.48×1023 kg
Part (ii): Length of the Martian year
Both Earth and Mars orbit the Sun, so Kepler's third law applies to compare them directly:
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KEAM 2026Set eng-2026-04184 marksMCQ
Q.Two planets revolve around the sun in elliptical orbits with their major axes in the ratio 2 : 3. Their periods of revolution around the sun are in the ratio
(A) 2 : 3
(B) 3 : 2
(C) 23:32
(D) 22:33
(E) 4 : 9
›Reveal solutionSolution
By Kepler's third law T2∝a3, so T∝a3/2 and the period ratio is (32)3/2=22:33.
Kepler's third law gives T2∝a3 (with a the semi-major axis, proportional to the major axis). Hence T∝a3/2, and …
Q.A satellite revolves around a planet in a circular orbit of radius R with the time period t. If the orbital radius is 3R, then its time period of revolution is
(A) t
(B) 2t
(C) 27t
(D) 3t
(E) 22t
›Reveal solutionSolution
Use T2∝R3 and scale the radius by 3.
By Kepler's third law T2∝R3, i.e. T∝R3/2. If the radius changes from R to 3R, …
Q.The slope of the graph plotted between square of time period of a planet T2 and the cube of its mean distance from the sun r3 is (G = Gravitational constant, M = Mass of the planet)
(A) GM4π2
(B) 4πGM
(C) M4πG
(D) G4π2M
(E) Zero
›Reveal solutionSolution
T2=GM4π2r3 is a straight line through the origin with slope GM4π2.
For a planet orbiting the Sun (mass M), equating gravitational force to centripetal force for a circular orbit gives Kepler's third law:
Q.The time period of revolution of a planet around the sun in an elliptical orbit of semi-major axis a is T. Then
(A) T2∝a2
(B) T∝a3
(C) T2∝a3
(D) T∝a31
(E) T2∝a31
›Reveal solutionSolution
Kepler's third law states the square of the orbital period is proportional to the cube of the semi-major axis: T2∝a3. …
Q.If T be the time period of a planet around the Sun and d is its mean distance from the Sun, then according to Kepler's third law
(A) T∝d
(B) T∝d2
(C) T2∝d3
(D) T2∝d
(E) T2∝d−3
›Reveal solutionSolution
Kepler's third law is T2∝d3.
Concept and Intuition
Kepler's law of periods relates a planet's orbital period to its mean distance from the Sun. It follows from Newton's gravitation applied to (nearly) circular orbits.