3
$\begingroup$

We know that life is almost 4 billion years old on Earth.

We also know that time contracts as we approach the event horizon of the massive black hole at the center of the Milky Way galaxy.

Now the question is that how can we compute the radius of the orbit close to that event horizon for which the whole four billion years long story of life on Earth is contracted into just "one second"? Is its order of magnitude around meters, kilometers, or light hours or light years?

$\endgroup$
0
$\begingroup$

You seem to be asking for a distance above the event horizon at which time dilation due to the gravitational acceleration of the black hole would convert 4 billion years to 1 second. As all time is relative, what you really are asking for is the distance above the event horizon at which it would APPEAR to someone on Earth that an event which takes 4 billion years to transpire on Earth, occurs in 1 second.

The Schwarzschild radius marks the event horizon around a black hole. It's the radius of a sphere into which enough mass is compressed, such that escape velocity would be the speed of light. Any object with radius smaller than its Schwarzchild radius (based on the amount of mass in the object) is a black hole. The Schwarzschild radius of our black hole at galactic center is about 13.3 million kilometers. The Earth is about 25 to 28 thousand lightyears above the event horizon of that black hole.

Time dilution due to the gravity of the black hole can be approximated as follows:

Tr/T = (1-Rs/r)^1/2, where: "Tr" is elapsed time of an observer at radial coordinate r above the event horizon (1 second), "T" is elapsed time for an observer on Earth (4 billion years), "Rs" is the Schwarzschild radius of the black hole at galactic center (13.3 kilometers), "r" is the radial coordinate for which we are solving.

Solve for "r" in the time dilution equation, and you have the answer. It will be very, very small, and not very accurate!

See ACuriousJim's comment below. He found "r" to be 8.3 x 10^-25 meters. However, the equation above is for an observer hovering above the event horizon, rather than for one who orbits the black hole, as John Rennie points out in his comment, below. Also, it would be impossible for an observer to achieve a stable orbit so close to the event horizon, as Hypnosifl says in his comment, below.

$\endgroup$
  • $\begingroup$ @ACuriousJim: You're right! I misplaced a decimal. Earth is about 8 kilo parsecs, or about 26 thousand lightyears from galactic center. Thank you for this correction. $\endgroup$ – Ernie May 11 '15 at 15:19
  • 1
    $\begingroup$ I'm not sure if you want to include the number in your answer, but I did the math using your numbers and you'd have to be around $8.3\times10^{-25}m$ from the event horizon in order for one second for you to be equivalent to 4 billion years for a distant observer $\endgroup$ – Jim May 11 '15 at 15:28
  • $\begingroup$ For what it's worth, the time dilation from the matter between SgrA* and Earth is going to be a lot larger than that due to SgrA* itself. $\endgroup$ – Jerry Schirmer May 11 '15 at 16:29
  • $\begingroup$ The formula given for the time dilation is true only for an observer hovering above the event horizon. For an object in a circular orbit there is an additional time dilation due to the orbital speed. $\endgroup$ – John Rennie May 11 '15 at 16:36
  • $\begingroup$ The question does ask about "the radius of the orbit", and as John Rennie says your answer is about hovering, not orbiting; I'm not sure if VictorM cares about the difference, but if so, it should be noted that for a non-rotating black hole it's impossible to have a stable orbit any closer than 3 times the Schwarzschild radius, though this can be decreased for a rotating black hole (the innermost stable orbit gets closer to the event horizon the closer the black hole's rotation rate is to a certain critical value). $\endgroup$ – Hypnosifl May 11 '15 at 19:27
1
$\begingroup$

A few things:

1) time dilation is only a relative effect. A clock on earth would tick differently than one far from the milky way, but people on Earth would notice no abnormal effect.

2) To first order, time dialation effects are governed by the gravitational potential energy at a point. It turns out that the time dilation due to the mass of the ordinary stars in teh galaxy is much larger than that of the central black hole, which only accounts for something like 1/120 of the mass of the galaxy (this is still a huge number).

3) we are orbiting around the center of mass of the milky way, but we are not falling into the center, either. Just like how the Earth is not falling into the sun.

I may come back and turn this into a proper answer.

$\endgroup$
0
$\begingroup$

I think you can't do it with this data. because time dilation of our so-called "UNIVERSAL TIME" is lagging behind w.r.to a body in deep void because of mainly due to earth's gravitation, not sun or the center black hole of milky way

proof:

$$R=\frac{CT}{2\pi}\sqrt[2]{\frac{\Delta t}{t}}$$ where $T$ : time period of revolution of milky way.
$\Delta t / t $: 1 second per 4 billion years then we get $R \approx .112 Lyrs$ which means that the center of the galaxy would be nearer than the nearest star ( $proxima Centauri$ )!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy