Kepler's Laws of Planetary Motion/Third Law/Examples

From ProofWiki
Jump to navigation Jump to search

Examples of Kepler's Third Law of Planetary Motion

Let $P$ be a planet orbiting the sun $S$.

Let $P$ be:

$\text{(a)}: \quad$ Twice as far away from $S$ as the Earth
$\text{(b)}: \quad$ $3$ times as far away from $S$ as the Earth
$\text{(c)}: \quad$ $25$ times as far away from $S$ as the Earth.


Then the orbital period of $P$ is:

$\text{(a)}: \quad$ approximately $2.8$ years
$\text{(b)}: \quad$ approximately $5.2$ years
$\text{(c)}: \quad$ $125$ years.


Proof

Let the orbital period of Earth be $T'$ years.

Let the mean distance of Earth from $S$ be $A$.

Let the orbital period of $P$ be $T$ years.

Let the mean distance of $P$ from $S$ be $a$.


By Kepler's Third Law of Planetary Motion:

\(\ds \dfrac {T'^2} {A^3}\) \(=\) \(\ds \dfrac {T^2} {a^3}\)
\(\ds \leadsto \ \ \) \(\ds T^2\) \(=\) \(\ds \dfrac {a^3} {A^3}\) as $T'$ is $1$ year
\(\ds \leadsto \ \ \) \(\ds T\) \(=\) \(\ds \left({\dfrac a A}\right)^{3/2}\)


Thus the required orbital periods are:

$\text{(a)}: \quad 2^{3/2} = 2 \sqrt 2 \approx 2.8$ years
$\text{(b)}: \quad 3^{3/2} = 3 \sqrt 3 \approx 5.2$ years
$\text{(c)}: \quad 25^{3/2} = 125$ years.

$\blacksquare$


Sources