Kepler's Laws of Planetary Motion/Third Law/Examples
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Examples of Use 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
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.25$: Kepler's Laws and Newton's Law of Gravitation: Problem $2$