# A new planet is discovered. Planet X is observed to orbit the sun every 300 years. What is the semi-major axis of Planet X's orbit in AU?

Data yet to be confirmed: Orbit period is about 15000 years, Perihelion = 200 AU and aphelion 600 - 1200 AU. For these tentative (imprecise )data, semi-major axis is half-sum = 400 to 700 AU.

The orbit period for this distant planet from Neptune may be much longer than Neptune's 165 years (see discoverers' articles in Science Magazine).

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To calculate the semi-major axis of Planet X's orbit in astronomical units (AU), you can use Kepler's third law of planetary motion, which states:

[T^2 = \frac{4\pi^2}{G(M_1 + M_2)}a^3]

Where:

- (T) is the orbital period in years,
- (G) is the gravitational constant,
- (M_1) and (M_2) are the masses of the two bodies (in this case, the mass of the Sun and the mass of Planet X),
- (a) is the semi-major axis of the orbit.

Given that (T = 300) years and (M_1 \gg M_2) (the mass of the Sun dominates), we can simplify the equation to:

[a^3 = \frac{G \cdot T^2 \cdot M_1}{4\pi^2}]

[a = \left(\frac{G \cdot T^2 \cdot M_1}{4\pi^2}\right)^{1/3}]

Substituting the known values, including (G = 6.674 \times 10^{-11} , \text{m}^3 , \text{kg}^{-1} , \text{s}^{-2}) and (M_1 = 1.989 \times 10^{30} , \text{kg}):

[a = \left(\frac{(6.674 \times 10^{-11} , \text{m}^3 , \text{kg}^{-1} , \text{s}^{-2}) \cdot (300 , \text{years})^2 \cdot (1.989 \times 10^{30} , \text{kg})}{4\pi^2}\right)^{1/3}]

[a \approx \left(\frac{(6.674 \times 10^{-11}) \cdot (300^2) \cdot (1.989 \times 10^{30})}{4\pi^2}\right)^{1/3} , \text{m}]

[a \approx (8.88 \times 10^{25})^{1/3} , \text{m}]

[a \approx 7.31 \times 10^{8} , \text{m}]

Converting meters to astronomical units (1 AU = (1.496 \times 10^{11}) meters):

[a \approx \frac{7.31 \times 10^{8}}{1.496 \times 10^{11}} , \text{AU}]

[a \approx 4.88 , \text{AU}]

So, the semi-major axis of Planet X's orbit is approximately 4.88 AU.

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