Question

One planet is 6.00 x10^5 m. Another planet is 1.4 x 10^27kg. The radius is 7.14 x 10^7. Find the period.

Answer 1

Answer on Question #57011-Physics-Solid State Physics

One planet is $6.00 \times 10^{5} \mathrm{~m}$. Another planet is $1.4 \times 10^{2} \mathrm{kg}$. The radius is $7.14 \times 10^{7}$. Find the period.

Solution

Newton's second law has on one side, the force of gravity:

$$F = G \frac{m_{1} m_{2}}{r^{2}} = \frac{m_{1} v^{2}}{r}$$ $$G \frac{m_{1} m_{2}}{r^{2}} = \frac{m_{1} (4 \pi^{2} r)}{T^{2}}$$ $$T = \sqrt{\frac{4 \pi^{2} r^{3}}{G m_{2}}} = \sqrt{\frac{4 \pi^{2} (6.00 \cdot 10^{5} + 7.14 \cdot 10^{7})^{3}}{6.67 \cdot 10^{-11} \cdot 1.4 \cdot 10^{27}}} = 1.26 \cdot 10^{4} \, \text{s} = 3.5 \, \text{h}.$$

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