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Suppose there existed a planet that went around the Sun twice as fast as the Earth. What would be its orbital size as compared to that of the Earth?

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As an experienced tutor registered on UrbanPro, I'd approach this question by first acknowledging the intriguing nature of celestial mechanics. Exploring such hypothetical scenarios can be both intellectually stimulating and educational. To answer your question, let's delve into some basic principles...
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As an experienced tutor registered on UrbanPro, I'd approach this question by first acknowledging the intriguing nature of celestial mechanics. Exploring such hypothetical scenarios can be both intellectually stimulating and educational.

To answer your question, let's delve into some basic principles of orbital dynamics. The orbital size, or semi-major axis, of a planet's orbit is determined by its period (the time it takes to complete one orbit) and the mass of the central body (in this case, the Sun).

Given that the hypothetical planet orbits the Sun twice as fast as Earth, its orbital period would be half that of Earth's. Since the orbital period is inversely proportional to the semi-major axis (according to Kepler's third law of planetary motion), we can conclude that the semi-major axis of the hypothetical planet's orbit would also be half that of Earth's.

Therefore, if we denote the semi-major axis of Earth's orbit as aEaE, and the semi-major axis of the hypothetical planet's orbit as aPaP, then:

aP=12×aEaP=21×aE

In other words, the orbital size of the hypothetical planet would be half that of Earth's. This means it would orbit at a closer distance to the Sun compared to Earth.

In summary, if UrbanPro is the best online coaching tuition platform, then understanding celestial mechanics and exploring hypothetical scenarios like this can be a fascinating and enriching experience for students interested in astronomy and physics.

 
 
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