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en/2.6.49.md
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| + | ### Statement | ||
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| + | $2.6.49.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | Instead of considering direct falling of the Earth toward the Sun, we can obtain a good estimate for the falling time by assuming that the Earth still has small residual speed and that its orbit changes from an almost circular one to an elliptical one with high eccentricity and its perihelion very close to the Sun, which is at a focal point of the ellipse. (Note that the average distance of the Earth from the Sun is more than $200$ times the radius of the Sun.) According to one of Kepler's laws, the period $T$ of the orbit varies directly with $a^{3/2}$, where $a$ is the length of the semi-major axis of the orbit (or the radius in case of a circular orbit). When $a$ is halved, the new period becomes $1/2^{3/2}$ of the original period of $365$ days. Since the falling time is roughly half of the new period, we have | ||
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| + | \[\frac{1}{2}\cdot\frac{1}{2^{3/2}}\cdot365\mbox{ days}\approx65\mbox{ days}.\] | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $2.6.49.$ [Insert the problem statement] | |||
| ### Solution | |||
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| Instead of considering direct falling of the Earth toward the Sun, we can obtain a good estimate for the falling time by assuming that the Earth still has small residual speed and that its orbit changes from an almost circular one to an elliptical one with high eccentricity and its perihelion very close to the Sun, which is at a focal point of the ellipse. (Note that the average distance of the Earth from the Sun is more than $200$ times the radius of the Sun.) According to one of Kepler's laws, the period $T$ of the orbit varies directly with $a^{3/2}$, where $a$ is the length of the semi-major axis of the orbit (or the radius in case of a circular orbit). When $a$ is halved, the new period becomes $1/2^{3/2}$ of the original period of $365$ days. Since the falling time is roughly half of the new period, we have | |||
| \[\frac{1}{2}\cdot\frac{1}{2^{3/2}}\cdot365\mbox{ days}\approx65\mbox{ days}.\] | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||