Q.The martain satellite phobos revolves around the planet Mars in a nearby circular orbit of radius $9.4\times10^{6}\,m$ and a period of 7h39min. Calculate the mass of the Mars planet. OR Calculate the minimum speed to be imparted to our earth so that it may leave the solar system and becomes free. Given that mass of the sun $= 2\times10^{30}\,kg$ and orbital radius of the earth is $1.5\times10^{11}\,m$.

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

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

…

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:

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

…

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:

$T^2 \propto a^3$

…

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)

…

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 radi …

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 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:

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

…

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 …

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:

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

…