Kepler's third law question. HELP PLZ (1 Viewer)

DuhBoyChong · Mar 29, 2015

The orbital radius of Jupiter is five times greater than the Earth’s orbital radius. What is

the period of Jupiter’s orbit in Earth years?

InteGrand · Mar 29, 2015

DuhBoyChong said:
The orbital radius of Jupiter is five times greater than the Earth’s orbital radius. What is

the period of Jupiter’s orbit in Earth years?

Kepler's third law states that $\frac{T_1 ^2}{T_2 ^2}=\frac{R_1 ^3}{R_2 ^3}\Leftrightarrow T_1 = T_2 \left( \frac{R_1}{R_2}\right) ^{\frac{3}{2}}$ (just rearranging), where $T_i$ is the period for planet i and $R_i$ is the orbital radius of planet i, for $i\in \{ 1,2\}$ .

We know the following:

T_Earth = 1 Earth year
R_Jupiter = 5R_Earth ⇔ R_Jupiter/R_Earth = 5

Therefore, substituting these values gives into Kepler's third law equation gives

$T_\text{Jupiter} = T_\text{Earth}\times \left( \frac{R _\text{Jupiter}}{R _\text{Earth}}\right) ^{\frac{3}{2}}$

$=1 \text{ Earth year} \times 5^{\frac{3}{2}}$

$\approx 11.18 \text{ Earth years}$ (to two decimal places).

DuhBoyChong · Mar 29, 2015

thank you bro

Kepler's third law question. HELP PLZ (1 Viewer)

DuhBoyChong

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InteGrand

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DuhBoyChong

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