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?

Orbital radius 4× geostationary radius ⇒ period = 8× geostationary period = 8 days.

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

Given $r' = 4r_{geo}$: $\dfrac{T'}{T_{geo}}=\left(\dfrac{r'}{r_{geo}}\right)^{3/2}=4^{3/2}=8$.

Since $T_{geo}=24\,\text{hours}=1$ day, $T'=8\times1=8$ days.

Kepler's third law: the square of the orbital period is proportional to the cube of the semi-major axis, so T ∝ R³ᐟ².

Kepler's law of periods (third law) states that for planets orbiting the sun (or, more generally, any satellite in a near-circular orbit):

$T^2 \propto R^3$

Taking the square root of both sides:

$T \propto R^{3/2}$

This can also be derived from Newton's law of gravitation providing the centripetal force for a circular orbit: $\frac{GMm}{R^2} = \frac{m(2\pi/T)^2 R}{1}$, which rearranges to $T^2 = \frac{4\pi^2}{GM}R^3$, confirming $T^2 \propto R^3$.

Kepler's third law relates the orbital period of a planet to the size of its orbit: $T^2 \propto a^3$.

Kepler's law of periods (Kepler's third law) states that the square of the time period of revolution of a planet around the Sun is directly proportional to the cube of the semi-major axis of its elliptical orbit:

$T^2 \propto a^3$

Equivalently, $\dfrac{T^2}{a^3} = \text{constant}$ for all planets orbiting the Sun (or, more generally, for all satellites orbiting the same central body).

In the two-body approximation where the central body's mass (Earth) dominates, the orbiting body's own mass drops out of the period formula entirely - so changing the Moon's mass doesn't change its orbital period.

For a body of mass m orbiting a much more massive central body of mass M (here, Moon orbiting Earth), equating gravitational force to the required centripetal force gives:

GMm/r^2 = m (4 pi^2 / T^2) r

Notice the orbiting body's mass m appears on BOTH sides and cancels out completely:

GM/r^2 = 4 pi^2 r / T^2

T^2 = 4 pi^2 r^3 / (GM)

This is Kepler's third law: the period T depends only on the central mass M (Earth's mass) and the orbital radius r - not on the mass of the orbiting body at all. So even if the Moon's mass were doubled (with Earth's mass and the orbital radius unchanged), the period of revolution would remain essentially the same, about 29 days.

Kepler's third law: T^2 ∝ r^3.

Kepler's law of periods states that the square of the orbital period T of a planet is directly proportional to the cube of the semi-major axis (mean orbital radius) r of its orbit:

T^2 ∝ r^3, or T^2 = (4π^2/GM) r^3.

Kepler's three laws each have a name: the first is the law of orbits, the second is the law of areas, and the third (T^2 proportional to a^3) is the law of periods.

Kepler's first law (law of orbits): planets move in elliptical orbits with the sun at one focus.

Kepler's second law (law of areas): the radius vector from sun to planet sweeps equal areas in equal time intervals.

Kepler's third law (law of periods): the square of the orbital period T is proportional to the cube of the semi-major axis a of the orbit, T^2 proportional to a^3.

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

Kepler's third law, also called the Law of Periods, states: the square of the time period of revolution of a planet around the Sun is directly proportional to the cube of the semi-major axis of its elliptical orbit.

Mathematically:

$T^2 \propto r^3 \quad \Rightarrow \quad \frac{T^2}{r^3} = \text{constant (same for all planets orbiting the Sun)}$

This can be derived from Newton's law of gravitation for a circular orbit: the gravitational force provides the centripetal force,

$\frac{GMm}{r^2} = \frac{m(2\pi/T)^2 r}{1} \Rightarrow T^2 = \frac{4\pi^2}{GM}r^3$

which confirms $T^2 \propto r^3$.

Setting gravitational force equal to centripetal force for an orbiting satellite, the satellite's mass cancels out algebraically, so the orbital time period depends only on orbital radius and the mass of the central body (planet), never on the satellite's own mass.

For a satellite of mass m orbiting a planet of mass M in a circular orbit of radius r, gravity supplies the centripetal force:

GMm/r^2 = m v^2/r = m (4 pi^2 r / T^2)

The satellite's mass m appears on both sides and cancels:

GM/r^2 = 4 pi^2 r / T^2

T^2 = 4 pi^2 r^3 / (GM)

T = 2 pi sqrt(r^3 / GM)

This clearly shows T depends only on the orbital radius r and the mass M of the planet being orbited — it is completely independent of the satellite's own mass.

Kepler's third law: T² ∝ r³, which on taking a square root becomes T ∝ r^{3/2}.

Kepler's third law of planetary motion states that the square of the time period T of a planet's revolution around the Sun is directly proportional to the cube of the semi-major axis (mean orbital radius) r of its orbit:

$T^2 \propto r^3$

Taking the square root of both sides:

$T \propto r^{3/2}$

This law follows from Newton's law of gravitation applied to circular orbits: equating gravitational force to the centripetal force requirement, $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r} = \dfrac{m}{r}\left(\dfrac{2\pi r}{T}\right)^2$, which rearranges to $T^2 = \dfrac{4\pi^2}{GM}r^3$.