Question

Orbital period of satellite is proportional to?

Answer 1

hey there

Kepler's 3rd Law of Periods: This law is known as the law of Periods. The square of the time period of the planet is directly proportional to the cube of the semimajor axis of its orbit. That means the time ' t ' is directly proportional to the cube of the major axis..

hope it help

{MARK BRAINLIEST}

Answer 2

The square of the orbital period of a satellite is directly proportional to the cube of the orbit's semi-major axis.

Explanation:

• The time taken by a satellite to complete one single orbit is called its orbital period. It is denoted by $T$.

• According to Kepler's Third Law, the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its elliptical orbit i.e., $\[{T^2} \propto {a^3}\]$.
• Using Newton's Laws and assuming the satellite's orbit to be circular, the orbital period of the satellite is given as: $\[T = \frac{{2\pi }}{{\sqrt {GM} }}{a^{\frac{3}{2}}}\]$ where $a$ is the radius of the satellite's orbit, $G$ is the gravitational constant, and $M$ is the mass of the planet it is orbiting.
• Squaring the equation $\[T = \frac{{2\pi }}{{\sqrt {GM} }}{a^{\frac{3}{2}}}\]$ on both sides we get: $\[{T^2} = \frac{{4{\pi ^2}}}{{GM}}{a^3} \Rightarrow {T^2} \propto {a^3}\]$ which is Kepler's Third Law.

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