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It is a supermassive black hole but its mass is nothing compared to the mass of the whole galaxy. So if you remove the black hole, we will not feel any difference because we are so far away. It is not the central black hole that keeps the galaxy together, our sun is orbiting around the center of mass of the galaxy and the supermassive black hole just happens ...


21

A typical giant galaxy, such as the one you've provided a picture of, has a radius of something like $10\;\rm kpc$ (kiloparsec - $1\;\rm pc \approx 3.2\;ly$). A supermassive black hole hosted in such a galaxy has a mass of something like $10^6-10^9\;\rm M_\odot$ (solar mass, $1\;\rm M_\odot \approx 2\times10^{30}\; kg$). The monstrous billion solar mass ...


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Just to add to John Rennie's answer, the objects where we expect to see the largest frame dragging effects are spinning black holes. There, there is actually a surface called the ergosphere (outside of the event horizon), where it is impossible for observers to stay stationary with respect to observers far from the black hole. In a sense, their reference ...


3

The spacetime outside a spinning mass is described by the Kerr metric. To explain how the Kerr metric produces frame dragging is hard, because it's not something for which there's an easy intuitive model. Frame dragging arises because the spacetime geometry links the angle measured around the spinning object to time, and this means the angle changes with ...


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Depending on the shape of the universe the luminosity distance is given by : \begin{equation} d_L(z) = \left\{ \begin{array}{rl} \frac{(1 + z) c}{H_0 \sqrt{|\Omega_k|}} \sin \left[ \sqrt{|\Omega_k|} \int _0 ^z \frac{dz'}{H(z')/H_0} \right] & \mbox{for $k = 1$} \\ \frac{(1 + z) c}{H_0} \int _0 ^z \frac{dz'}{H(z')/H_0} & \mbox{for $k = 0$} \\ ...



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