Reading a question on the Worldbuilding SE site a minute ago about a trip to the other side of the galaxy flying by the supermassive black hole in the center of our galaxy, my immediate first thought was: "How silly, how do you plan flying by close and ever again escaping, that's not just some black hole, it's a black hole strong enough to hold together an entire galaxy".

Which led me to say: "Oh, wait a moment...", and here comes the question:
Is it (and if it is, how?), from everyday observation, plausible that there is indeed any such thing as Sgr A* at the center of our galaxy at all?

Earth rotates around itself and around the sun, and the sun orbits the center of the galaxy, in some kinda-elliptical but not alltogether trivial way. This means that I (and every object on Earth) rotate in some kind of semi-chaotic way, facing in every possible direction now and then. Inevitably, while standing on Earth's surface I will face "towards" the center of the galaxy at some point, and "away from" it at other times.

Now, that means one day the black hole's gravitation (which is strong enough to hold together a galaxy, so it is not as trivial as e.g. the Moon's or Mars' gravitation) would add to Earth's gravitation, and on another day it would pull the opposite way.

In other words, the weight of a constant-mass object should be very observably different on different days. That's apparently not the case, or someone would have wondered during the last couple of hundred years. What now?

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    $\begingroup$ Sgr A* does not hold together the Milky Way. The Milky Way is held together by the combined mass of everything in it. Sgr A* weighs about 0.0001% of the mass of the Milky Way so it is far too small to be a dominant force in shaping the galaxy. $\endgroup$ – John Rennie Aug 17 '16 at 9:36
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    $\begingroup$ See also: Is there a black hole in the centre of the Milky Way? $\endgroup$ – John Rennie Aug 17 '16 at 9:38
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    $\begingroup$ Perhaps the gravitational field of the black hole affects the earth and its inhabitants equally. $\endgroup$ – Dawood ibn Kareem Aug 17 '16 at 9:58
  • $\begingroup$ You question is based on a basic misunderstanding of orbits. You are also moving around the black hole and experiencing the corresponding centrifugal/petal force. The only force difference comes from tidal forces. Related: en.wikipedia.org/wiki/Tidal_force $\endgroup$ – Wolpertinger Aug 17 '16 at 10:02
  • $\begingroup$ @DavidWallace or what you said, which is much more concise +1 $\endgroup$ – Wolpertinger Aug 17 '16 at 10:02

There are several flaws in your arguments. I'll try to help you to understand them.

  1. As John Rennie pointed out in comment, the (hypothetical) black hole in the center of the Milky Way does not hold the whole galaxy together. It's the gravitational forces between all the stars, nebula, dark matter, etc. which do. So the fact that stars in the Milky Way do not fly away does not support the hypotheses that there is a black hole in it's center.

  2. Let's imagine a situation when there is a super-massive black hole, and a earth-looking planet which flies around it. Radius of it's orbit is 28,000 light years, period of rotation is 230 millions light years. (These are parameters of Sun orbiting the center of our Galaxy). No other stars, or whatever, only these two objects. Let's calculate to acceleration of this planet. (that would be acceleration of gravity of the back hole on this distance). The formula is: acceleration = distance * (angular speed)^2.

1 year = 86400 * 365 sec = 31,536,000 sec

radius = 28,000 * 300,000,000 * 1 year = 2.64 * 10 ^ 20 meters

period = 230,000,000 * 1 year = 7.25 * 10 ^ 15 sec

angular speed = 2 * pi / period = 8.66 * 10 ^ -16 (1/sec)

acceleration = radius * (angular speed)^2 = 2 * 10 ^ -10 m / sec^2

acceleration of gravity on earth is 10 m / sec^2, this is 50 billions times higher. It would be quite difficult to observe the difference of weight of some object, which is due to gravity of black hole.

  1. The situation is even worse. Because the planet is also "free falling" around the black hole, the difference of weight of some object in different locations of planet would not be m * (black hole acceleration part). You would be able to observe only the difference of black hole gravity in different locations of planet. Which is much-much smaller. It's like trying to measure the earth's gravity when you are on a satellite. Although gravitational forces are almost the same as on earth's surface, you can not feel it because you frame of reference is "free falling" on earth.

Now back to the black hole in the center of Milky Way. There are strong evidences, that there is one, but the fact that Sun is rotation around the center of galaxy is not one of them.

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The supermassive black hole (SMBH) $SgrA^*$ has an estimated mass of $4.1$ million solar masses. This is estimated from the motion of stars nearby. We can look a ratios of acceleration. $SgrA^*$ is $27$ thousand light years away or about $3\times 10^{17}km$ away. The factor $M/r^2$ is then $4.5\times 10^{-29}$ in units of solar mass per square kilometer. Now let us look at the case for the $1AU$ orbital radius of the Earth around the sun, which is $4.4\times 10^{-17}$ solar mass per kilometer square that is $12$ orders of magnitude larger. The influence of the SMBH is then negligible on the orbit of the Earth given by $\vec F~=~-GMm{\vec r}/r^3$.

The SMBH consists of $4.1$ million solar masses and the Milky Way has $300$ million stars. This means the SMBH is $1.4\times 10^{-5}$ times the total mass due to stars. There is then interstellar gas in the galaxy that has a total mass comparable to the mass of these stars. Then one can consider the role of dark matter or halo mass that is around $5$ times the mass of stars. The influence of the SMBH in the Milky Way is not that significant for the dynamics of the galaxy as a whole.

SMBHs have great influence in the center of galaxies. This is particularly for some galaxies with quite spectacularly large SMBHs up to $10^{10}$ solar masses. This is still a minor component of these galaxy masses. The main influence of black holes in general is fairly local.

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  • $\begingroup$ Rule of thumb on mass-energy of the universe is that about $73\%$ is dark energy and $27\%$ is matter. The matter breaks down further to about $22\%$ dark matter and about $5\%$ ordinary or luminous matter. So dark matter is about $5$ times ordinary matter. $\endgroup$ – Lawrence B. Crowell Aug 17 '16 at 11:49

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