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2 years ago

What does the geocentric model of the solar system look like

1 answer:

hjlf2 years ago

5 0

In the geocentric system, the Earth is considered to be the center of the solar system. The Moon, the planets, the Sun, and the stars all rotate around the Earth (which stays still), with uniform circular motion.

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52.38 m/s

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3 years ago

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A burnt paper on the road has a picture, which shows a speed boat runs fast on the lake and produces V-like water waves. This re

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Moessbauer Effect = eggy eggs

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3 years ago

A mouse is walking at 2.45 m/s. He sees a cat and accelerates at 4.92 m/s/s. How much time does it take him to reach a speed of

alexandr1967 [171]

Speed= (4.92-2.45) 2.47 m/s
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You didn't mention about the distance?

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3 years ago

oe finds that the temperature of a substance is 12 degrees Celsius. What does this tell Zoe about the substance? Its internal en

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Its internal energy is less than 12 degrees

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3 years ago

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

Tju [1.3M]

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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4 years ago