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Posts Tagged ‘Escape Velocity’

Last week, I blogged about the theoretical arguments for the Galaxy harbouring a supermassive black hole at its centre, and here I blogged about the observational evidence. The work done by the UCLA and MPE teams, discussed here, has led to a determination that the central black hole has a mass of between 4.4 and 4.5 million solar  masses. I am going to take the upper end  of this range, just for convenience.

astronomers-take-a-major-step-in-understanding-black-hole-at-the-center-of-the-milky-way-galaxy

An artist’s impression of Sgr A*, showing the central supermassive black hole and the accretion disk which surrounds it.

The size of the event horizon

In this blog here I showed that the radius of a blackhole’s event horizon can be calculated by using the equation for the escape velocity v_{esc} when that velocity is equal to the speed of light c. That is

v_{esc} = c = \sqrt{ \frac{2GM }{ R } }

where M is the mass of the blackhole, G is the universal gravitational constant, and R is the size of the object, which in this case is the radius of the event horizon (also known as the Swarzchild radius R_{s}). So, we can write

R_{s} = \frac{ 2GM }{ c^{2} }

Putting in a mass of 4.5 million solar masses, we find

R_{s} = 1.33 \times 10^{10} \text{ metres}

Converting this to AUs, we find the radius of the event horizon is 0.09 AUs, much smaller than the radius of Mercury’s orbit, which is about 0.3 AUs.

At the distance of the Galactic centre, 8 kpc, this would subtend an angle of
\theta = 6.17 \times 10^{-9} \text{ degrees} (remember to double R_{s} to get the diameter of the event horizon). This is the same as

\boxed{ \theta = 22.22 \text{ micro arc seconds} }

Converting this to radians, we get

\theta ( \text{in radians}) = 1.08 \times 10^{-10}

In fact, we do not need to resolve the event horizon itself, but rather the “shadow” of the event horizon, which is about four times the size, so we need to resolve an angle of

\theta ( \text{in radians}) \approx 4 \times 10^{-10}

The resolution of a telescope

There is a very simple formula for the resolving power of a telescope, it is given by

\theta( \text{in radians}) = \frac{ 1.22 \lambda }{ D }

where D is the diameter of the telescope and \lambda is the wavelength of the observation. Let us work out the diameter of a telescope necessary to resolve an object with an angular size of 50 \times 10^{-4} \text{ radians } at various wavelengths.

For visible light, assuming \lambda = 550 \text{ nanometres}

D = \frac{ 1.22 \times 550 \times 10^{-9} }{ 4 \times 10^{-10 } }, \boxed{ D = 1.68 \text { km} }

There is no visible light telescope this large, nor will there ever be. At the moment, visible-light interferometry is still not technically feasible over this kind of a baseline, so imaging the event horizon of the Galaxy’s supermassive blackhole is not currently possible at visible wavelengths.

For 21cm radio radiation (the neutral hydrogen line)

D = \frac{ 1.22 \times 21 \times 10^{-2} }{ 4 \times 10^{-10 } }, \boxed{ D = 640,000 \text { km} }

This is more than the distance to the Moon (which is about 400,000 km away). So, until we have a radio dish in space, we cannot resolve the supermassive blackhole at 21cm either.

For millimetre waves, we have

D = \frac{ 1.22 \times 1 \times 10^{-3} }{ 4 \times 10^{-10 } }, \boxed{ D = 3,100 \text { km} }

which is feasible with very long baseline interferometry (VLBI). So, with current technology, imaging the event horizon of the Milky Way’s supermassive blackhole is only feasible at millimetre wavelengths. Millimetre waves lie in a niche between visible light and radio waves. They are long enough that we can do VLBI, but they are short enough that the baseline to image the supermassive black hole’s event horizon is small enough to be possible with telescope on the Earth.

Next week I will talk about a project to do just that!

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Last weekend was a pretty big sporting weekend. Not only was there the Third and deciding test of the 2013 Lions’ tour of Australia, but there was also the men’s and women’s finals of Wimbledon, and the German Grand Prix. As far as I can tell there is no sport going on this weekend. Before someone comments below that England are playing Australia at cricket in the First Test of the Ashes, I should remind my readers (all 2 of you) that a bunch of overweight men standing around for 5 days not doing much does not constitute sport. So, with this lull in the sporting calendar, I thought I would write this post about black holes, which I have been meaning to do for a while.

Nearly every one has heard of black holes, but what actually are they? And do they actually exist? Well, a black hole gets its name because it is an object from which not even light can escape. The radiation (light and other forms of electromagnetic radiation) which the central part of the black hole gives off is not able to escape the extreme gravitational field the black hole creates.

An artist's impression of the black hole in Cygnus, with matter falling into the black hole.

An artist’s impression of the black hole in Cygnus, with matter falling into the black hole.

Calculating the escape velocity

Newton’s law of gravity allows us to calculate the force of gravitational attraction between two bodies of masses m_{1} and m_{2}. It is simply F = \frac{G m_{1} m_{2} }{ r^{2} }. Let us suppose m_{1} is the mass of the Earth, which we are now going to call M; and m_{2} is the mass of an object on the Earth’s surface, we are going to call this second mass just m. In order for the object on the Earth’s surface to escape the Earth’s gravitational field we have to give it a velocity.

When something is moving it has kinetic energy, and that kinetic energy is given by KE = \frac{1}{2}mv^{2} where v is the object’s velocity. The object also has gravitational potential energy, as it is in a gravitational field. The gravitational potential energy is related to the gravitational force, it is given by PE = - \int \frac{GMm}{r^2}.dr which is - \frac {GMm}{r}

Notice that the gravitational potential energy is negative, whereas the kinetic energy is always positive. The sum of the two, the total energy E = KE + PE is given by E = \frac{1}{2} mv^{2} - \frac{GMm}{r}. In order for an object to escape the gravitational pull of another object, it needs to be able to escape to infinity. If it does not escape to infinity but to a smaller distance then, technically, it has not escaped the gravitational field of the object.

At infinity the PE is PE = 0 as we are dividing by infinity. As the object travels further and further away from its parent body it will slow down (as it is having to do work against the gravitational force), and so the KE will get less and less. At infinity it will be zero. So we can say that, at infinity, E = KE + PE = 0. But E is constant, so it is also going to be zero any distance from the gravitational object, including at the surface of the planet.

Let us suppose the planet has a radius of R, we can then write 0 = \frac{1}{2}mv^{2} - \frac{GMm}{R}. The m on both sides can cancel giving us 0 = \frac{1}{2}v^{2} - \frac{GM}{R} and so, rearranging, we can write \frac{1}{2}v^{2} = \frac{GM}{R}. The escape velocity is then found by writing v^{2} = \frac{2GM}{R} so finally v_{esc} = \sqrt{ \frac{2GM}{R} }.

The escape velocity from Earth, a White Dwarf and a Neutron Star

The equations we have just derived allows us to calculate the escape velocity from any object. We are going to calculate the value for the Earth, a white dwarf and a neutron star.

The escape velocity from the Earth

For the Earth, the mass is M = 5.972 \times 10^{24} \text{ kg} and its radius R = 6375 \text{ km} (note, the Earth is not spherical, it bulges at the equator, so this is an average value). Before we plug these values of M \text{ and } R into the equation above we need to note that the value I have quoted for the Earth’s radius is in kilometres. We cannot put it into the equation in these units, we have to convert it to metres. 6375 \text{ km} = 6.375 \times 10^{6} \text{ m}, so this is the number we can plug into the equation for v_{esc}. When we do this we get that, for the Earth, v_{esc} = 111178.86 \text{ m/s} = 11.2 \text{ km/s}

The escape velocity from a white dwarf

A white dwarf is the stellar remnant of a star like the Sun. They are typically about the size of the Earth, but with about the mass of the Sun. So, for M we shall use the mass of the Sun, which is M = 1.99 \times 10^{30} \text{ kg}, and we shall use the radius of the Earth that we used above, R = 6.375 \times 10^{6} \text{ m}. These numbers give us v_{esc} = 6.45 \times 10^{6} \text{ m/s} = 6.45 \times 10^{3} \text{ km/s} which is 2% of the speed of light.

The escape velocity from a neutron star

A neutron star is the end produce of more massive stars. The Sun is not massive enough to become a neutron star, but a star which is more than about 3 times the mass of the Sun is. In a neutron star all the space that exits in atoms is squeezed out, so it is essentially a pure lump of nuclei. A typical neutron star may have the mass of 2 Suns, but squeezed down into something the size of a city! So, for our calculation, we are going to assume a 2 solar mass neutron star, M = 2 \times (1.99 \times 10^{30} ) = 3.98 \times 10^{30} \text{ kg}. For the radius we will assume 10km, so R = 10 \times 10^{3} \text{ m}. Plugging these values into the equation for the escape velocity gives 2.30 \times 10^{6} \text{ m/s} = 230 \times 10^{3} \text{ km/s} which is 77% of the speed of light.

The event horizon of a black hole

The escape velocity from a neutron star is still below the speed of light. Pulsars are produced by radiation from the surface of a neutron star being beamed past us as the neutron star rotates. So, we have direct observational evidence that we can see radiation from neutron stars.

But, in the same way that a star which is a few times the mass of the Sun will end its life as a neutron star rather than a white dwarf; an even more massive star will not end as a neutron star. This is because of something called the neutron degeneracy pressure. To put it simply, this is a physical law which says that neutrons do not all want to be in the same place. They resist this through a resistive component in the strong nuclear force. But, if a neutron star were to have more than about 3 times the mass of the Sun, the gravity is strong enough to overcome this neutron degeneracy pressure. There is no known force to stop the collapse of the neutron star, and this is what forms a black hole.

We can work out the radius at which the escape velocity becomes equal to the speed of light for an e.g. 2 solar mass black hole. This is the same mass as our neutron star example above. But, as we shall see, it will need to be smaller than the 10km size of a neutron star. The radius at which the escape velocity is equal to the speed of light is what we call the event horizon of black hole.

To do the calculation we just re-arrange our escape velocity equation to find R when v_{esc} = c where c is the speed of light. The re-arrangement is that R = \frac{2GM}{c^{2}}. For M=2 \times (1.99 \times 10^{30}) = 3.98 \times 10^{30} \text{ kg}, and c = 3.0 \times 10^{8} \text{ m/s} we find the radius of the event horizon to be R = \frac{2GM}{c^{2}} = 6.0 \times 10^{3} \text{ m} = 6 \text{km}. Notice how close this is to the actual size of a typical neutron star, just a little over half the size. It shows how little mass has to be added to a neutron star to tip it over the edge into becoming a black hole.

Notice that all of the above calculations have been done assuming Newton’s law of gravity. Newton’s law of gravity is not actually correct, it has been superseded by Einstein’s, which we call the theory of General Relativity. To do the calculations properly we should use this theory, but it is rather complicated. No, it is very complicated. But to illustrate the basic idea, Newton’s laws are fine. It is surprisingly often said that Einstein’s work led to the prediction of black holes. This is not true, they had been suggested by a geologist by the name of John Michell in 1783. But we do need Einstein’s work to do the calculations properly.

Any radiation being emitted from inside of the event horizon will never get to us, the gravitational pull from the black hole stops it from escaping. How do we therefore even know that black holes exist? I will answer that question in a future blog, along with some discussion of what happens as matter crosses the event horizon of a black hole, and what might be right at the centre of a black hole.

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