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Andrej [43]
3 years ago
11

Si un planeta tuviese un periodo de traslación de 65 años terrestres a que distancia se encontraría del sol

Physics
1 answer:
Tju [1.3M]3 years ago
4 0

Answer:

R \approx 2.418\times 10^{9}\,km

Explanation:

(The following exercise is written in Spanish and for that reason explanation will be held in Spanish)

Supóngase que el planeta tiene una órbita circular, el período de rotación del planeta es:

T = \frac{2\pi}{\omega}

Asimismo, la rapidez angular se describe como función de la aceleración centrípeta:

\omega = \sqrt{\frac{a_{r}}{R} }

Ahora se reemplaza en la ecuación de período:

T = 2\pi \cdot \sqrt{\frac{R}{a_{r}} }

La aceleración experimentada por el planeta es:

a_{r} = G\cdot \frac{M_{sun}}{R^{2}}

Se reemplaza en la ecuación de período:

T = 2\pi \cdot \sqrt{\frac{R^{3}}{G\cdot M_{sun}} }

La distancia del planeta con respecto al sol es finalmente despejada:

R^{3} = G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}

R = \sqrt[3]{G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}}

Finalmente, se sustituyen las variables y se determina la distancia:

R = \sqrt[3]{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (1.989\times 10^{30}\,kg)\cdot \left[\frac{(65\,a)\cdot \left(365\,\frac{d}{a} \right)\cdot \left(86400\,\frac{s}{d} \right)}{2\pi} \right]^{2}}

R \approx 2.418\times 10^{12}\,m

R \approx 2.418\times 10^{9}\,km

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Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

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given Radius between earth and sun to be = 1.50 × 10¹¹m

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substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

<em>P </em>sun = <em>I </em><em>sun-earth </em><em>A </em><em>sun-earth</em>

where A = 2.863 10²³×m², and <em>I </em> is 1360 W/m²

<em>P </em>sun =  2.863 10²³ × 1360

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(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

<em>I </em><em>sun-mars = </em><em>P </em>sun / A sun-mars

where <em>P </em>sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

<em>I </em><em>sun-mars =  </em>3.85×10²⁶W / 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>589.6 ≈ 590 W/m²

<em>I </em><em>sun-mars = </em>590 W/m²

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