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19

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 ...

3

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 ...

1

Depending on the shape of the universe the luminosity distance is given by : 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 $k = 0$} \\ ...

-4

Galaxies stay together because gravity is about A BILLION TIMES faster than the speed of light.* *This is pure heresy to Einsteinians of course (disciples of Einsteins general and special relativity theories), since they firmly believe that nothing travels faster than light. Not even gravity. If this were true however,it would make galaxies like Hercules A ...

2

Not all galaxies are disk-shaped, but some certainly are. (Some others are spiral, etc.) For one thing, we see a lot of galaxies, and several of them look exactly like they would if they were disk-shaped and we were just seeing them from different angles. Some seem circular because we are seeing them head-on, while others seem more linear or elliptical ...

0

Assuming that the radiation source is a blackbody then redshift stretches all the wavelengths by a factor $(1+z)$ as per your formula. You are then correct to say that an application of Wien's law will yield a temperature which is lower by a factor of $(1+z)$. Hence the microwave background appears now to have a temperature of 2.73K, but it was emitted ...

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