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If the speed of light is about 3x10^5 km/sec, what is the greatest distance an object cn be seen if the hubble constant is 20 km/sec/10^6 l.y.

2007-11-02 04:29:03 · 2 answers · asked by ted 1 in Science & Mathematics Astronomy & Space

2 answers

hubble bubbles law is v=Hd

you know v... look up H (hubbles constant)

divide and conquer, baby!

2007-11-02 04:46:17 · answer #1 · answered by Faesson 7 · 0 0

Hubble's law says the universe is expanding and Hubble's constant gives the rate at which it is expanding. (Note that since a light year is a unit of length, the figure of 20 km/sec/10^ l.y. is is dimensionally a ratio/time or a percentage increase per unit time.)

Given any two different initial positions, as time goes by they will become arbitrarily far apart. Furthermore, since the expansion rate is constant, the distance between the two positions will not only continue to grow, it will grow at an accelerating rate.

(i.e if the universe doubles its linear dimension every N years, and it started out L across, then after N years it is 2L across and has grown L/N per year. After another N years, it has grown to 4N across, and has been growing at 2L/N per year, etc.)

This means two things:

1. There are points in the universe that started out far enough from our initial position that their light hasn't reached us yet.

(i.e. as the light is travelling from point A to point B, the distance from A to B is increasing so the speed of light as a fraction of the total distance per unit time keeps decreasing even as its absolute velocity stays constant. So the effective rate at which light covers the distance is less than the nominal speed of light.)

2. There are points in the universe that started out far enough from our initial position that we will never see their light.

It isn't clear whether the "can be seen" refers to objects that we haven't seen because the light hasn't reached us yet, or refers to objects we'll never be able to see because they are receding from us too quickly as a result of being too far to start with. Either way, you can compute a visibility horizon as a function of time given the data you have.

2007-11-03 19:18:29 · answer #2 · answered by simplicitus 7 · 0 0

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