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

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## 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

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