Question

The mean distance of mars from the earth is 0.5A.U. and its orbital period is 687 days . Calculate the orbital period of jupitar given that its mean distance the earth is 4A.U.

Answer 1

Answer on Question #45874 – Physics – Astronomy | Astrophysics

For solving this problem we need $3^{\text{rd}}$ reduced Kepler’s law: if there is two planets are rotating around the same star, we can express ratio between mean distances planet-star and its’ orbital period for them

$$\frac{D^{3}_{m1}}{D^{3}_{m2}} = \frac{T^{2}_{1}}{T^{2}_{2}} \qquad (1)$$

Where T – period, $D_{m}$ – mean distance planet-star. For Mars we get

$$D_{M} = 1 + D_{\text{Mars-Earth}} = 1 + 0.5 = 1.5 \, a.u.$$

For Jupiter

$$D_{J} = 1 + D_{\text{Jupiter-Earth}} = 1 + 4 = 5 \, a.u.$$

Transforming formula (1)

$$T_{j} = T_{j} \sqrt{\frac{D^{3}_{j}}{D^{3}_{M}}} = 687 \times \sqrt{\frac{125}{3.375}} = 687 \times 6.086 = 4181.08 \, \text{days}$$

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