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
hint: use keplers 3rd law to find the commets new orbital period
-
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.
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.
-
Hint: this is a homework question.... Do your own homework!