How to weigh objects in space

10th April 2015

Size comparison between the Sun (which is 1 pixel) and a massive star (which is too large to fully fit in the frame).

Size comparison between the Sun and the massive red hypergiant star VY Canis Majoris.
Image credit: Helen Klus/CC-NC-SA.

1. Gravity

Everything with mass produces a gravitational field, which causes things to accelerate towards it. The gravity on the surface of the Earth is about 9.8 ms-2, which is often referred to as 1 g. This is just high enough to keep us from falling off the surface and drifting into space, but low enough to allow us to jump up and down and walk around.

Newton showed that the gravity of an object can be found using[1a]:

Gravity of object =
Mass of object × G/Distance from object2
,

where G is a number known as the gravitational constant, which is equal to 6.67384×10-11 m3kg-1s-2.

The Earth weighs about 5.972×1024 kg, and so the gravity on the surface of the Earth, 6,372,000 m from its centre, is:

Gravity on the surface of the Earth =
5.972×1024 × 6.67384×10-11/6,372,0002
,
= 9.8 ms-2 = 1 g.
= 9.8 ms-2 = 1 g.

This number gets rapidly lower the further you get from the surface of the Earth, by the time you are about 400 km away (which is roughly the distance from London to Paris), it falls by about 10%. At 4000 km, it falls by about 60%, and by the time you get to the Moon, which is 384,400 km away, it has fallen by over 99.9%.

This number gets rapidly higher the closer you get to the centre of the Earth. The fact that it's infinite at a radius of 0 is indicative of our lack of understanding about how physics works at very small lengths.

The force of gravity 10 cm from the centre of the Earth is:

Gravity 10 cm from the centre of the Earth =
5.972×1024 × 6.67384×10-11/0.12
= 4×1016 ms-2 = 4×1015 g
= 4×1016 ms-2 = 4×1015 g

If you could tunnel to the centre of the Earth, however, then you would not feel this force. At the centre of the Earth, the gravity of the Earth would be accelerating you equally in all directions, and so you would feel weightless[2].

All objects with strong enough gravitational fields become approximately spherical. This is because the surface is pulled inwards equally in all directions, and so any imperfections are smoothed out. The mass that an object needs to become spherical depends on what it is made of. It is easier to smooth out water than rock, for example. Objects made of rock tend to be spherical if they have a mass over about 1020 kg, which is the mass of some asteroids[3]. This is about 100,000 times less massive than the Earth, and produces objects with a radius of about 300 km[4].

Gravity on an asteroid =
1020 × 6.67384×10-11/300,0002
= 0.25 ms-2 = 0.026 g
= 0.25 ms-2 = 0.026 g

The gravity produced by objects much less massive than this is usually negligible. The gravity produced by a 60 kg person from 10 cm away, for example, is over 25 million times smaller than the gravity produced by the Earth on its surface.

Gravity produced by a person =
60 × 6.67384×10-11/0.12
= 0.0000004 ms-2 = 0.00000004 g
= 0.0000004 ms-2 = 0.00000004 g

2. Weight

The International Space Station orbits the Earth from about 400 km away[5], and so:

Gravity on the International Space Station =
5.972×1024 × 6.67384×10-11/(6,372,000 + 400,000)2
= 8.7 ms-2 = 0.9 g
= 8.7 ms-2 = 0.9 g

This is about 90% of the gravity on the surface of the Earth.

This means that people on the International Space Station do not appear weightless because of their distance from the Earth. Instead, they appear weightless because the Space Station accelerates towards the Earth at about 8.7 ms-2, this means that it is in free-fall.

Skydivers would feel weightless when they are in free-fall if it were not for the drag of the air. You do not feel this drag if you are inside an enclosed space, like an aeroplane, and so you can experience weightlessness on a zero-G flight. Here an airplane accelerates towards the Earth at about 9.8 ms-2. You feel weightless on a zero-G flight because the floor falls away from you at the same rate as you are pulled towards it.

Credit: via Space Affairs.

The most important difference between someone on a zero G flight and someone on the International Space Station is that the International Space Station is in orbit around the Earth. This means that it falls to Earth at the same rate as the spherical surface of the Earth curves away, and so it is able to fall continuously while remaining at a constant distance from the Earth's surface.

Diagram of Newton's cannon thought experiment, this shows how gravity describes the motion of objects on Earth and in space.

Objects in orbit are always falling. Image credit: Brian Brondel/CC-SA.

You may feel weightless when in free-fall, but you still have the same mass.

While the terms 'mass' and 'weight' are commonly used interchangeably, astronomers need the terms to mean two different things.

In physics, mass is a fundamental property that particles have, like spin, or charge. A person's mass is the sum of the mass of all the particles in their body, and so their body mass only changes if they add or remove some of these particles.

Weight is the force you feel due to gravity, and 'apparent weight' is the force you feel due to your total acceleration. This means it is your weight that will decrease if you move away from the Earth's gravitational field.

Newton's 2nd law shows that[1b]:

Force = Mass × Acceleration.

The force of weight is due to acceleration caused by a gravitational field[6], and so:

Weight = Mass × Gravity.

Where Mass refers to the mass of the object you want to weigh (m) and Gravity includes all the non-negligible gravitational fields felt by that object. On Earth, everything is negligible but the gravity produced by the Earth itself, and so the weight of a 60 kg person on the surface of the Earth is:

Weight = m ×
Mass of Earth × G/Distance from centre of Earth2
= 60 kg × 9.8 ms-2
= 60 kg × 9.8 ms-2
= 589 kg ms-2 = 589 N
= 589 kg ms-2 = 589 N

This means that the Earth's gravitation exerts a 589 N force on a 60 kg person due to its gravitational field.

If the two masses were reversed, and you calculated the force on the Earth caused by the gravitational field of a 60 kg person, then the equation would remain the same, and so a 60 kg person also exerts a 589 N force on the Earth. This means that the weight of the Earth caused by its acceleration under the gravitational field of a 60 kg person is 589 N.

Astronomers tend to refer to weight as the gravitational force, and so Newton's law of gravitation states that the force exerted by an object of mass (m) on to an object of mass (M) is[1c]:

Gravitational force =
mMG/r2
,

where r is distance between the two objects.

An object, like a satellite, will orbit the Earth in a circle when the force of gravitation between the object and the Earth is equal to the object's centripetal force. The centripetal force is caused by the rotation of the Earth; it pulls objects away from the Earth in the same way that an object on a string will move away from the centre if the string is swung in a circle[7a].

Centripetal force =
mV2/r
,

where V is the velocity of the object.

If  
mMG/r2
=
mV2/r
, then the required velocity for a circular orbit is V =
MG/r
and so in order to remain in a circular orbit, an object like the International Space Station, which is about 400 km from the surface of the Earth, must travel at:
V =
5.972×1024 × 6.67384×10-11/6,372,000 + 400,000
= 8 km/s = 28,000 km/h
= 8 km/s = 28,000 km/h.

The circumference of the Earth is 40,075 km, and so the International Space Station orbits the Earth about once every 90 minutes.

2.1 What is 1 kg?

1 kilogram (kg) is a unit of mass.

The gram was defined in 1795 as the mass of one cubic centimetre of water at the melting point of water[8].

The General Conference on Weights and Measures (CGPM) formed in 1875 and representatives from different countries met in Paris in order to develop a common international measuring system, known as the metric system[9].

By 1900, member nations included 14 countries in Europe, as well as Russia, Argentina, the United States, Venezuela, Japan, and Mexico. There are currently over 50 member states, which meet in Paris every four to six years[10].

In 1889, the CGPM created an object out of a platinum-iridium alloy that was about 1000 times the mass of a gram; this is a kilogram, where kilo refers to 1000[11a].

The CGPM declared that from then on, the weight of a kilogram in the metric system would refer to the weight of this object, and copies were taken to each member country.

The terms 'mass' and 'weight' were still often used interchangeably when referring to objects on Earth. The CGPM clarified this in 1901, when they declared that the kilogram was a unit of mass, not weight[11b].

3. Weighing objects on Earth

On Earth, people can weigh themselves using balancing scales.

These work using the law of torques, which state that two objects on a pivot, like a seesaw, will balance when the torque is the same on each side, where[7b]:

Torque = Force × Distance from pivot.

The downwards force that an object exerts on a balancing scale is equal to its weight. This means that you can weigh an object by comparing it to an object with a known weight, like the CGPM's prototype kilogram, which has a weight on the surface of the Earth that is equal to 1 kg × 9.8 ms-2 = 9.8 N[11c]. If you place the CGPM's prototype kilogram on one side of a balancing scale, then you can work out how much another object weighs by placing it on the other side and moving it until the scale is balanced.

Diagram showing how torques balance on balancing scales or a seesaw.

Balancing scales. Image credit: Helen Klus/CC-NC-SA.

Photograph of balancing scales.

Image credit: Nikodem Nijaki/CC-SA.

An object with twice the weight of the prototype kilogram will need to be twice as close to the pivot than the prototype kilogram in order to make the scales balance.

We also often weigh things on Earth using spring scales, like those found in bathroom scales. These use the law of torques combined with Hooke's law[7c].

Diagram of the spring in spring scales.

Spring scales. Image credit: Svjo/CC-SA.

Photograph of bathroom scales.

Bathroom scales. Image credit: Angelsharum/CC-SA.

Hooke's law shows that the force needed to stretch a spring is proportional to the distance it's stretched. The proportionality constant depends on how stiff the spring is. This is known as the stiffness constant, where stiffer springs have higher stiffness constants.

Hooke showed that:[7d]

Force = -Stiffness of spring × Distance the spring stretches.

When you weigh an object using a spring scale, the force on the spring is due to the object's weight, and so if you take an object with a known weight, like the CGPM's prototype kilogram, then you can measure how much the spring stretches and work out the stiffness constant.

You can then measure the mass of anything using the spring, as long as it retains its elasticity.

In bathroom scales, a person stands on a lever, which is like a seesaw with only one side. Because they stand very close to the pivot, their torque is small, and so the lever does not need as much space to move as a real seesaw. The edge of the lever is attached to a spring, and so when the level goes down, the spring is stretched by the force due to the person's weight. The movement of the spring moves the scale.

The more force on the scale due to the person's weight, the more the lever moves, the more the spring stretches, and the more the dial turns. Mass can be determined from weight since the gravitational field is known, and so the dial is calibrated to show mass.

In digital scales, the force of a person due to their weight causes a strain on a conductive material, like foil. This increases its electrical resistance.

Ohm's law shows that:

Voltage = Current × Resistance[12].

This means that a heavier object will cause more strain on the conductive material, this causes it to have a higher electrical resistance. The mass needed to produce that voltage is then calculated and displayed.

4. The mass of objects on the International Space Station

None of these methods for measuring the mass of objects will work if the object is either in free-fall or far away from any strong gravitational fields. This is because if your apparent weight is zero, like it would be on the International Space Station, then the scales will be moving towards the Earth at the same rate that you are being pulled towards them by gravity.

To measure the mass of a person in space, the mass must be derived from something other than the force of gravity. On the International Space Station, the mass of an astronaut is measured by applying a force to them using a spring. Two devices are generally used: these are NASA's SLAMMD (Space Linear Acceleration Mass Measurement Device) and Russia's BMMD (Body Mass Measurement Device).

NASA's SLAMMD determines a person's mass by pushing them with a small force and then measuring their acceleration using a camera.

Force = Mass × Acceleration. The force is generated by a spring so that Force = -Stiffness of spring × Distance the spring stretches, and so:

Mass =
-Stiffness of spring × Distance the spring stretches/Acceleration
.

Credit: Kowch737.

Russia's BMMD also determines a person's mass by applying a force using a spring. Instead of measuring a person's acceleration, however, BMMD measures their frequency. The force of the spring makes them oscillate in simple harmonic motion. Simple harmonic motion refers to repetitive motion around a central position, like the swing of a pendulum, or the up and down motion of an object attached to a spring, like a pogo stick.

The number of times a simple harmonic oscillator oscillates per second is known as its frequency, and this depends on its mass. The more mass they have, the slower they oscillate, and so the lower their frequency.

For objects undergoing simple harmonic motion from the force of a spring:[7e]

Mass =
-Stiffness of spring/Frequency2
.

Credit: Kowch737.

5. The mass of objects in the Solar System

The mass of an object like the Earth or the Sun cannot be measured using scales or the force of springs.

The mass ratio between a planet, like the Earth, and a star, like the Sun, can be determined, however, if you know how far the planet is from the star, and how long it takes for the planet to make one complete orbit.

If a planet orbits in a circle, then the gravitation force, pulling it towards the star, and the centripetal force, pushing it away, balance so that:

V =
MG /r
,

where M is the mass of the star and r the distance between the star and the planet.

Velocity is also equal to distance/time, where the planet travels a distance of 2πr, which is the circumference of a circle, over a time equal to one orbital period (P), and so:

V =
r/P
.

If

MG/r
=
r/P
,

then

P2 =
2r3/MG
,

and so

M =
2r3/GP2
.

This equation assumes that the planet has a very small mass compared to the star. This is because a planet does not orbit the exact centre of a star; instead, it orbits the system's common centre of mass.

Diagram showing that binary stars balance around a centre of mass in the same way that objects balance on balancing scales.

Binary stars balance like objects on a seesaw. Image credit: Helen Klus/CC-NC-SA.

The centre of mass between two massive objects is like the pivot of a balancing scale. The system will balance if m1r1 = m2r2, where m1 and m2 are the masses of the two objects, r1 is the distance between m1 and the centre of mass, and r2 is the distance between m2 and the centre of mass.

This means that the centre of mass is closest to the most massive object. In the case of satellites orbiting the Earth, the mass difference between the two objects is so large that the centre of mass is within the radius of the Earth, and so we can assume that this is where the centripetal force originates.

For objects like stars and planets, the masses are similar enough that the centre of mass can be outside of the most massive object, and this is where the centripetal force actually originates from.

Here,

Centripetal force =
m1V12/r1
,

where V1 is the velocity of the object with mass m1.

r1 can be determined using m1r1 = m2r2 and r=r1+r2.

These equations can be rearranged to show that:

r1 =
rm2/m1+m2
,

and so

M =
4π2r3/GP2

becomes

m1+m2 =
2r3/GP2
.

This is known as Newton's derivation of Kepler's 3rd Law[1d].

All of this means that if you know the period of a planet, and the distance between the planet and the star, then you can work out the total sum of the mass of the planet and the star.

Once the mass of the Earth was known, the mass of the Sun could also be determined, as could the mass of any planet in the Solar System with a known distance and period.

5.1 The mass of the Earth

g =
Mass of Earth × G/r2
,

where r is the radius of the Earth, and so:

Mass of Earth =
g × r2/G
.

The radius of the Earth can be derived from the circumference, which was first measured by Ancient Greek astronomer Eratosthenes using the angle of shadows in different locations during the solstice, when the Sun is highest in the sky[13].

Galileo showed how you could derive g in the 1500s. He did this by simply measuring the acceleration of falling objects[14].

Cavendish showed how G could be determined in the 1790s[15]. Cavendish did this by measuring the gravitational force between two objects of known masses in a laboratory using a torsion balance.

The Earth was found to be 5.972×1024 kg.

6. The mass of objects outside of the Solar System

For objects outside of the Solar System, such as stars in binary systems or planets orbiting stars other than the Sun, it's often only possible to measure the object's individual velocities, rather than their distance from each other.

In that case, the individual velocities are derived from r.

r = r1+r2, and using

Distance = Velocity × Time,

r = (V1+V2) × P.

Putting this back into

m1+m2 =
2r3/GP2
,

gives

m1+m2 =
P(v1+V2)3/G
.

This is further complicated because usually we can't measure the actual velocity of stars and planets, we can only measure the velocity that stars move towards and away from us. This is known as the radial velocity, and can be determined from the Doppler shift of spectral lines.

Diagram showing how the radial velocity changes depending on the angle.

Image credit: Helen Klus/CC-NC-SA.

We can only measure the correct velocity of stars that we view exactly from the side, so that they are sometimes moving directly towards us and are sometimes moving directly away. The velocity we observe becomes more and more incorrect the more the system is turned, so that we do not observe any velocity if we view the system from the top.

The angle we view the system from is known as its inclination (i), where a system has an inclination of 0° if it is viewed from the top, and an inclination of 90° if it is viewed from the side.

Once this is taken into account:

m1+m2 =
P/G
× (
V1(radial) + V2(radial)/sin(i)
)3.

We cannot easily measure the inclination of star systems, but we do know that we must be viewing the system from the side if we see an eclipse, which is caused by one object passing in front of the other.

Eclipsing systems are also useful to astronomers because the change in brightness during an eclipse can be used to determine the radius of the two objects[16].

Before the eclipse, the total brightness we observe is the sum of the brightness of the two objects. As the small object obscures the large object, some of the large object's light is blocked and so the brightness goes down. It then stays the same until the small object has moved past the large one. If you time how long it takes for the brightness to go down to a constant level (t2-t1 in the image below), then you can work out the radius of the smaller object, since you know its velocity. You can then work out the radius of the large object by timing how long it takes for the brightness to return to the same level (t4-t2).

Diagram showing how the total brightness of two stars in a binary system changes as one star passes in front of another.

The radius of objects that eclipse each other can be determined by measuring their brightness. Image credit: Helen Klus/CC-NC-SA.

You can see what different plots of brightness vs time look like for systems with different inclinations, and containing objects of different sizes, using the eclipsing binary simulator from the University of Nebraska-Lincoln.

The brightness of a star is related to its luminosity, where brightness refers to how bright the star appears from Earth, and luminosity refers to how bright it actually is[17]. This means that two stars of the same luminosity can have different levels of brightness, where the brighter star is closer to the Earth.

The surface temperature of a star can be determined from its colour[18]. This is because it behaves like a blackbody, and so cooler stars are redder and hotter stars are bluer.

Finally, you can determine what a star is made of from its spectra[19].

There are only about 200 well-studied eclipsing binary star systems but when astronomers plotted these parameters, on a plot known as the H-R diagram, they noticed a pattern[20][21].

The H-R diagram - a plot of colour against luminosity for stars. Colour is directly related to temperature and spectral type.

H-R diagram. Image credit: Richard Powell/penubag/CC-SA.

For the majority of stars, known as main sequence stars, there's a link between mass and luminosity, where more massive stars are generally more luminous. More massive stars are also generally larger and have a higher temperature, which makes them bluer. There's also a link between spectral type and mass, where more massive stars are generally made of more massive atoms, like carbon and oxygen, rather than less massive atoms like hydrogen.

More massive stars are therefore generally bigger, hotter, bluer, more luminous, and made of heavier atoms.

Less massive stars are therefore generally smaller, cooler, redder, less luminous, and made of lighter atoms.

Non-main sequence stars do not follow these rules. Red giant stars, for example, are very luminous despite being red because of their unusually large size. The relationship between mass and luminosity for non-main sequence stars is more complicated, but they can sometimes be ssive objects in the known u out by looking at their spectra[22].

This means that the mass of almost any star can be approximated if you look at their spectra, where two stars with similar spectra will have similar masses and sizes.

The mass of galaxies, and more massive objects, like galaxy clusters, can be determined from their velocities.

The mass of galaxies can be determined using V =
MG /r
, which gives M =
rV2/G
.

The mass of galaxy clusters can be determined using the virial theorem[23], which states that the total kinetic energy of a stable, self-gravitating system is approximately equal to minus half of the gravitational potential energy of the system.

Average kinetic energy = -
1/2
Average gravitational potential energy,

which gives:

M
2rV2/G
.

7. The most massive objects in the known universe

As objects become more massive, they go from being bumpy, like asteroids, to spherical, like moons and planets. If spherical objects become too massive, then nuclear reactions will begin at their centre, and they will become stars. The most massive stars are hundreds of times as massive as the Sun[24], and can be thousands of times as large[25].

The relative sizes of stars. Larger stars tend to be more massive. Credit: Michael Hasted.

The most massive stars become black holes. Black holes can gain more mass if other objects fall into them. The most massive black holes are known as supermassive black holes, and so supermassive black holes could be considered the most massive objects in the observable universe.

Supermassive black holes are thought to be at the centre of most galaxies, including our own[26]. The most massive are billions of times as massive as the Sun, although they are only tens to hundreds of times the size[27].

Stars are almost never isolated, however, and billions of stars can be gravitationally bound together in galaxies[28]. Thousands of galaxies can be gravitationally bound in galaxy clusters[29], and hundreds of thousands of galaxies are bound in galaxy superclusters[30]. Superclusters form structures known as galaxy filaments, and these are the largest structures in the observable universe.

The largest structure in the observable universe may be the Hercules-Corona Borealis Great Wall, a filament of superclusters that is about 10 billion light-years wide[31].

Assuming a person is 1.7 metres tall (5 foot 6 inches), then the Hercules-Corona Borealis Great Wall is 5.6×1025 times larger. This is the difference in size between a person, and something 5.6×1025 times smaller - objects this size are millions of billions of times smaller than atoms. The scale of objects in the universe can be explored on this interactive animation created by Cary Huang.

Diagram showing the position of the Laniakea Supercluster relative to other superclusters.

Map showing our position in the universe, we are in the Solar System, orbiting the Sun, which is within the Milky Way galaxy, the Local Group galaxy cluster, and the Laniakea Supercluster. Image credit: Klaus Dolag/CC-SA.

8. References

  1. (a, b, c, d) Newton, I. and Motte, A. (trans), 1846 (1687), 'The Mathematical Principles of Natural Philosophy', Daniel Adee.

  2. University of California, Santa Barbara: Science Line, 'If you traveled through the center of the earth', last accessed 15-02-16.

  3. NASA, 'Asteroid Fact Sheet', last accessed 15-02-16.

  4. Hughes, D. W. and Cole, G. H., 1995, 'The asteroidal sphericity limit', Monthly Notices of the Royal Astronomical Society, 277, pp.99-105.

  5. ESA, 'Where is the International Space Station?', last accessed 15-02-16.

  6. Einstein, A., 1916, 'The foundation of the generalised theory of relativity', Annalen der Physik, 354, pp.769-822, reprinted in 'The Principle of Relativity: A Collection of Original Memoirs on the Special and General Theory of Relativity', 1920, Courier Corporation.

  7. (a, b, c, d, e) Giordano, N., 2012, 'College Physics: Reasoning and Relationships', Cengage Learning.

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  9. Fertell, R., 2009, 'A Day at the BIPM', Cal Lab, 16, pp.50-51.

  10. International Bureau of Weights and Measures, 'Member States', last accessed 15-02-16.

  11. (a, b, c) National Institute of Standards and Technology, 'Historical context of the SI', last accessed 15-02-16.

  12. Ohm, G. S., 1827, 'Die galvanische Kette, mathematisch bearbeitet' ('The galvanic chain, processed mathematically'), Riemann.

  13. Eratosthenes and Roller, D. W. (trans), 2010 (c240 BCE), 'Geography', Princeton University Press.

  14. Nicodemi, O., 2010, 'Galileo and Oresme: Who Is Modern? Who Is Medieval?', Mathematics Magazine, 83, pp.24-32.

  15. Cavendish, H., 1798, 'Experiments to determine the Density of the Earth', Philosophical Transactions of the Royal Society of London, 88, pp.469-526.

  16. NASA, 'Kepler's eclipsing binary stars', last accessed 15-02-16.

  17. Fischer-Cripps, A. C., 2014, 'The Physics Companion', CRC Press.

  18. Wien, W., 1893, 'Die obere Grenze der Wellenlängen, welche in der Wärmestrahlung fester Körper vorkommen können; Folgerungen aus dem zweiten Hauptsatz der Wärmetheorie' ('The upper limit of the wavelengths which can occur in a solid body of the heat radiation; Consequences of the second law of thermodynamics'), Annalen der Physik, 285, pp.633-641.

  19. Kirchhoff, G. and Bunsen, R., 1860, 'Chemical Analysis by Observation of Spectra', Annalen der Physik und der Chemie, 110, pp.161-189.

  20. Hertzsprung, E., 1911, 'Publikationen des Astrophysikalischen Observatorium zu Potsdam' ('Publications of the Astrophysical Observatory at Potsdam'), 63, pp.21.

  21. Russell, H. N., 1913, 'Giant and dwarf stars', The Observatory, 36, pp.324-329.

  22. Reimers, D., 1975, 'Circumstellar absorption lines and mass loss from red giants', Memoires of the Societe Royale des Sciences de Liege, 8, pp.369-382.

  23. Clausius, R., 1870, 'On a mechanical theorem applicable to heat', The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, 40, pp.122-127.

  24. Crowther, P. A., et al, 2010, 'The R136 star cluster hosts several stars whose individual masses greatly exceed the accepted 150 M(sun) stellar mass limit', Monthly Notices of the Royal Astronomical Society, 408, pp.731-751.

  25. Levesque, E. M., Massey, P., Plez, B., and Olsen, K. A., 2009, 'The Physical Properties of the Red Supergiant WOH G64: The Largest Star Known?', The Astronomical Journal, 137, pp.4744.

  26. Schödel, R., et al, 2002, 'A star in a 15.2-year orbit around the supermassive black hole at the centre of the Milky Way', Nature, 419, pp.694-696.

  27. Ghisellini, G., et al, 2010, 'Chasing the heaviest black holes of jetted active galactic nuclei', Monthly Notices of the Royal Astronomical Society, 405, pp.387-400.

  28. ESA, 'Structure of Milky Way', last accessed 15-02-16.

  29. NASA, 'Clusters of Galaxies', last accessed 15-02-16.

  30. NASA, 'Massive Merger of Galaxies is the Most Powerful on Record', last accessed 15-02-16.

  31. Steinicke, W. and Jakiel, R., 2007, 'Pairs, Groups, and Clusters of Galaxies' in 'Galaxies and How to Observe Them', Springer.

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