Astronomy Basics

How Do they Know That...Part 1

Part 1: The Solar System Part 2: The Stars and Beyond

You've probably seen in books and magazines facts and figures they give for different things, like "Sirius is 8.7 light years away..." and "Each second the sun lets out enough energy to...". If you're anything like me you'll wonder: how on Earth do they know that?

Always take claims with a pinch of salt- an astronomer's viewpoint is limited to one point in time and space in the universe, hence, we can't really be sure of anything. For instance I find it difficult to accept that we know exactly what happened 0.000000000001 seconds after the 'big bang', when we know only that the age of the universe is somewhere between 10 and 20 billion years old.

However, being so limited makes deducing information about the universe a challenge, and the ingenuity of some of the methods fascinates me.

Part 1: The Solar System

How big is the Earth?
How heavy is the Earth?
How far away is the Moon?
How heavy is the sun?
How far away is the sun?

How far from the sun are the Planets?
How heavy is Jupiter?

Part 2: The Stars and Beyond>>

Radius of the Earth

Two lines coming out fom the centre of the Earth with an acute angle θ between them. The suns rays come in parallel so the difference in shadow is due to (and equivalent to) the angle the distance represents of the Earth circumference.

The Greek mathematician and astronomer Eratosthenes calculated this in 200 B.C.. He knew that on a particular day in Syene, the sun would be directly overhead. He set up a stick in Alexandria 5,000 stadia away (800-900 kilometres) on this day, and measured the angle of the shadow that the sun formed at midday. He measured this to be 7.2°. He deduced that this is the same angle from the Earth's centre that the distance between the two cities represents. Since there are 360° in a circle, he worked out that 5,000 stadia is one-fiftieth of the circumference of the Earth. He therefore suggested the value of 250,000 stadia (40,000-46,000km). Astronomers recently measured it using satellite equipment and submitted the figure of 40,008 km. From this, using the geometry of circles, he could calculate the Earth's radius:

C equals 2 Pi R; 40000 over 2 Pi equals r; r is approximately equal to 6,370 km

Pretty good for a man who lived 2,200 years ago who only used only a stick!

Distance to the Moon

A triangle with one vertex at the centre of the Earth, one on the Earth's surface, and the other at the centre of the Moon. The line from the surface of the Earth to the moon is a tangent to the Earth's surface.

The distance to the moon can be determined by considering a very large imaginary triangle, and was first put forward by the Greek Hipparchus about 100 years after Eratosthenes suggested the radius of the Earth.

The moon is known to be directly overhead in one place, so if the place can be found where the moon is on the horizon at this time and the distance measured, it gives us the base for a very long, thin triangle (see diagram). The angle at A is a right angle, so:

tan θ equals d over r

Where θ is the angle at the Earth's centre, which can be calculated by:

θ equals 360 a over 2 Pi r

Where a is the measured arc length. Using this method Hipparchus estimated the distance to the moon to be 59 times the radius of the Earth (397,000km), which is not far off the modern accepted figure (382,000km).

How far away is the sun?

How far are the planets from the sun?

The first simple and very important question is actually quite difficult to answer, and an accurate figure eluded science until the seventeenth century. The same method as for the moon was tried, but failed due to the massive distance involved. To find the answer to the first question, we have to go the long way round and address the latter question first.

Johannes Kepler studied the orbits of the planets and found that the time period of a planet's orbit squared in years is equal to the distance from the sun cubed in AU (that is, astronomical units, or the distance from the Earth to the sun):

T squared equals R cubed

So Earth has an orbital period of one year, so is 1 AU from the sun. Mars has an orbital period of 1.88 years (these orbital periods had been known since ancient times by observation) and so it's distance from the sun is:

R cubed equals T squared; R cubed equals 1.88 squared; R equals the cube root of 1.88 squared which is approximately 1.524 AU

This could likewise be repeated for all the planets, so that the distance from the sun for all the planets was known in terms of the distance from the Earth to the Sun (the AU):

PlanetDistance from the sun
Mercury0.38 AU
Venus0.723 AU
Earth1.000 AU
Mars1.524 AU
Jupiter5.204 AU
Saturn9.582 AU

Astronomers only had to work out the actual distance to just one planet to find the distances to all the rest, including the Earth. This was achieved in 1672 when Giovanni Cassini measured the distance to Mars using Parallax. This is a similar method to the one for finding the distance to the moon, forming a tall thin triangle. The situation is more complex than the measurement of the distance to the moon however, as the two observers couldn't be on opposite sides of the world because they wouldn't both be able to see Mars at the same time and the triangle isn't necessarily a neat right-angled or isosceles triangle, but the principle is the same. (Click here if you want to see the mathematics in more detail)

Cassini sent a colleague to French Guiana while he remained in Paris. At an agreed time the two measured the position of Mars against background stars (assumed to be at an infinite distance). When they met back again, they compared the two positions and measured the tiny angle of parallax. Knowing this, and the distance (through the Earth) between the two locations (which can be calculated as the radius of the Earth is known), they could form a large triangle and calculate the length of one of the sides (which gives the distance between Mars and Paris).

The sun, Earth and Mars in a straight line. the distance from the sun to Earth is E, to Mars M, and from earth to Mars EM

The figure they obtained was 93% accurate compared with the modern figure (78,000,000 km minimum between Mars and Earth).

The night they chose to do this was when Mars was at opposition to Earth, i.e. it was near enough exactly on the other side from the sun, giving the positions shown on the diagram on the left. The distance that they measured, EM on the diagram, is clearly equal to M-E. Since earlier we worked out that M=1.524 E from Kepler's laws, we can use simultaneous equations to get E, the distance from the Earth to the sun in kilometers:

78,000,000 equals M take E and M equals 1.524 E so 78,000,000 equals 1.524 E take E equals 0.524 E; E equals 78,000,000 over 0.524 equals 150,000,000 km

This is an extremely important result as it means that we can now work out the geography of the entire solar system, as we now know that 1 AU = 150,000,000 km:

PlanetDistance from the sun
Mercury58,000,000 km
Venus108,000,000 km
Earth150,000,000 km
Mars228,000,000 km
Jupiter779,000,000 km
Saturn1,433,000,000 km

(When planets were discovered later their distances were calculated likewise). More accurate estimates of the AU were made in 1771 using the same method on Venus during a transit (when it passes in front of the sun). Nowadays we can use this method but instead of measuring the distance to Venus with parallax, we can measure it directly and extremely accurately with radar. A beam of radiation is aimed at the planet and the time it takes to receive the echo is measured using an atomic clock. Because the speed of light is known exactly the distance to Venus is half the time (the signal travels there and back) divided by the speed of light.

How heavy is the Sun?

A few years after Kepler died, Isaac Newton was born. In 1687 he published Philosophae Naturalis Principia Mathematica which was one of most influential books in physics. In it he published his theory of gravity. His contemporaries suspected that Kepler's laws could be a result of a centre-seeking force inversely proportional to the object's distance, but no-one could prove it. Isaac Newton presented the correct mathematical proof that this was the case, and, that it was also proportional to the two masses involved (If you're mathematically-minded I've written out such a proof- that an inverse square force produces elliptical orbits amongst others- on the gravity pages). This lead to Newton's Universal Law of Gravitation:

F equals G M m over r squared

Where G is a constant known as the Universal Gravitational Constant, which we will learn more about later.

We can now apply this to work out the masses of celestial objects. If we assume that the planet moves in a circle (which is a good assumption for many orbits... the same calculation can be applied to eliptical orbits- the mathematics just becomes a bit more complex when you allow for eccentricity) then there is a constant, centre-seeking (centripetal) force. For any object moving in a circle it is known that the centripetal force acting on it is given by:

F equals m r (2 Pi over T) squared

Where m is the mass of the object, r is the radius of it's motion and T the time for 1 full revolution. In our case of objects in orbit, this centripetal force is gravity, so we can equate the two forces:

G M m over r squared equals m r (2 Pi over T) squared; G M over r squared equals 4 Pi squard r over T squared; T squared equals (4 Pi squared over G M) times r cubed

Notice the similarity between this and Kepler's law above. This was the proof of Kepler's third law. Now we have a formula relating time of orbit, distance from sun (in real measurements, not AU), and mass. Rearranging the formula to make mass the subject gives:

M equals 4 Pi squared r cubed over T squared G

So we can find an object's mass if we know the time period of and distance to an orbiting object.

But how do we know what the value of 'G' is? That is a very good question, to which Newton did not really know the answer. It was not until 1798 that a reliable figure was established by an experiment conducted by Henry Cavendish. He set up two masses on either end of a rod hung by a string, free to turn. He moved two very large masses near the two smaller masses and measured the twist in the string that this produced (for more detail about the experiment see this page by the Physics Classroom). His experiment resulted in a value of 6.75x10-11 for G, which is not far from it's modern value, 6.67x10-11. This is an extremely small number, which explains why we only appreciate gravitational force when it comes to very large masses.

We are now free to calculate the mass of any object which has a satellite (if we know it's period and distance) This includes the Sun (by looking at any of the planets) Earth (by looking at the moon) or Jupiter (by looking at it's moons).

The Earth has a period of 365.24x24x60x60 seconds (31,556,296 seconds) and orbits at a distance of about 150,000,000 km (150,000,000,000 metres) so the mass of the sun is:

M equals 4 Pi squared r cubed over T squared G; M equals (4 Pi squared  times (150 times 10 to the nine) cubed) over (31,556,296 squared times 6.67 times 10 to the minus 11); M equals 2 times 10 to the 30 kilograms

How heavy is the Earth?

The Moon has a period of 2,551,443 seconds and orbits at a distance of about 384,403,000 metres (its orbit may be assumed to be roughly cicrular) so the mass of the Earth is:

M equals 4 Pi squared r cubed over T squared G; M equals (4 Pi squared  times 384,403,000 cubed) over (2,551,443 squared times 6.67 times 10 to the minus 11); M equals 6 times 10 to the 24 kilograms

However, there is a more accurate way to determine the Earth's mass that doesn't depend on the accuracy of our knowledge of the distance to the moon.

An object's weight on Earth is equal to the Earths gravitational pull on it, using Newton's second law that F=ma (force = mass x acceleration):

m a equals G M m over r squared; a equals G M over r squared

The 'm' cancel's out and so the acceleration, (which we will call g), of any object toward Earth at a distance 'r' (the radius of the Earth) is:

g equals G M over r squared

The value of g can be accurately be determined by experiment to be 9.8ms-1. Rearranging then, to get M, the Mass of the Earth is:

g r squared over G equals M; M equals 9.8 times 6,378,000 squared over 6.67 times 10 to the minus 11 which is approximately equal to 5.97 times 10 to the 24

How heavy is Jupiter?

A planet at a distance with the angle its size represents from Earth marked, and a planet at a smaller distance with the larger angle the planets size represents to us drawn in.

A similar method can be applied to find the mass of Jupiter, Saturn, or any planet with an observable moon. First, though, we need to find out how to work out an object's size from observation, if we know how far away it is.

Things look bigger close up than they do further away. This obvious truth is because an further object has a smaller angular diameter than a close one (see diagram). We can measure directly an objects angular diameter from where we are, so if we know how far away it is we can work out how big it is. The actual size of an object is calculated by:

r over d equals tan θ; r equals d tan θ
d is the distance to the planet, θ is the angular radius of the planet and r is the actual radius of the planet.

Where θ is the object's angular radius, d is the distance to the object, and s is the actual radius of the object (see diagram on the right). You could think of it as parallax in reverse- the triangle is upside down.

Now, we know how far away Jupiter is, so by measuring the angular size of the radius of the orbit of Jupiter's moons, we can work out it's actual radius by using the method above. Let's look at one of Jupiter's four large moons, Io. We can calculate the radius of it's orbit to be 421,600 km using this method. We observe it to take 152,841 seconds to orbit the planet. Using the same calculation for the mass of the sun, then, we can calculate Jupiter's mass:

M equals 4 pi squared times 421,600,000 cubed over (6.67 times 10 to the minus 11 times 152,841 squared)

Note that the principle that you can find the size of an object if you know how far away it is applies to anything; whether it be objects here on Earth, the Moon, the Sun, Neptune, The Orion Nebula, or even whole Galaxies. It can even be used to find the size of craters on the moon, mountains on Mars or the size of Jupiter's great red spot. Bear in mind as well that now you know the size and mass of objects, you can easily calculate their volume and average density.

Have a go yourself, try this question (you'll need a calculator, and maybe a piece of paper and pen)

Saturn and Titan, at maximum angular separation (194 arcseconds)

QWhen Saturn is at opposition to Earth (when the Sun, Earth, and Saturn are in a line) the radius of orbit of Titan, one of Saturn's moons, is measured. It has an angular size of 194 arcseconds (that's 194/3600ths of a degree). It's time period is observed to be 1,377,648 seconds. You know that Saturn is 1,443,500,000,000m from the sun and Earth is 149,600,000,000m from the sun (G is 6.67x10-11). What is the mass of Saturn?

All the methods needed are listed above. Think- what do you need to know to work out an central objects mass?

Part 2: The Stars and Beyond >>