A passing star knocks a comet loose from its orbit at the very edge of the Oort Cloud (50,000 A.U. away)
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A passing star knocks a comet loose from its orbit at the very edge of the Oort Cloud (50,000 A.U. away)

[From: ] [author: ] [Date: 12-02-27] [Hit: ]
Say that the comet is body x and Earth is body y (AUs and years are Earth-based units,Tx = (1 year) * √((25,Tx = (1 year) * √(1.Tx = (1 year) * √(1.Tx = (1 year) * (3.Tx = 3.......
and ends in the Sun. How many years will it take the comet to reach the inner Solar System from the edge of the Oort Cloud?

hint: use keplers 3rd law to find the commets new orbital period

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Kepler's Third Law states that the square of an orbit's period is proportional to the cube of its semi-major axis.

The semi-major axis is half the greatest distance of an ellipse, or perhaps the "big radius." If you had a planet whose orbit was 24 AU from end to end one way and 5 AU from end to end the other way, its semi-major axis would be 12 AU (and its semi-minor axis is 2.5 AU). The comet's new orbit goes from 50,000 AU to 0 AU, so half this distance is its semi-major axis: 25,000 AU.

A mathematical version of Kepler's Third Law is

Tx^2 / ax^3 = Ty^2 / ay^3 (imagine the x's and y's being subscripted, and calling them "T-sub-x" or "T of x")

where Tx and Ty are the periods of the orbits of the planetary bodies x and y and ax and ay are their respective semi-major axes. x and y must orbit the same mass.

Say that the comet is body x and Earth is body y (AUs and years are Earth-based units, so Ty = 1 year and ay = 1 AU)

Solving for Tx:

Tx^2 / ax^3 = Ty^2 / ay^3

Tx^2 = Ty^2 * (ax^3 / ay^3)

Tx = Ty * √(ax^3 / ay^3)

Tx = (1 year) * √((25,000 AU)^3 / 1 AU^3)

Tx = (1 year) * √(1.56 x 10^13 AU^3 / 1 AU^3)

Tx = (1 year) * √(1.56 x 10^13)

Tx = (1 year) * (3.95 x 10^6)

Tx = 3.95 x 10^6 years

or about 4 million years to make a complete orbit.

The answer, however, is half of that, since you're finding the time it takes to go half an orbit (from 50,000 AU to 0 AU), so the answer is close to 2 million years.

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Hint: this is a homework question.... Do your own homework!
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