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Posts Tagged ‘mars’

## How to calculate your horizon distance

While on a recent trip to the remote South Atlantic island of St Helena (exile place of Napoleon, and location of Edmond Halley’s observatory) [blog post to follow!] I ascended the highest mountain on the island, Diana’s Peak.

At 823m above sea level it commanded splendid views of the island, but the most striking thing was the unbroken 360° view of the horizon. I did a quick calculation in my head of how far I could see, and that forms the basis of this blog post: how do you calculate your horizon distance?
It turns out it’s pretty straight forward if you know a little simple maths. It helps to start by drawing a picture, so I did:

Definitely not to scale

For an observer of height h above sea level, the horizon distance is D. The Rs in this diagram are the radius of the planet you’re standing on, in this case the Earth. The only real assumption here is that you’re seeing a sea level horizon.As you can see you can draw a right-angled triangle where one side is D, the other is R, and the hypotenuse (the side opposite the right angle) is R + h.

Using Pythagoras’s Theorem, discovered around 2500 years ago, the square of the hypotenuse is equal to the sum of the squares of the other two sides. So we can say that:

(R + h)2 = R2 + D2

If you expand the part to the left of the bracket you get (R + h)2 = R2 + 2Rh + h2 so that:

R2 + 2Rh + h= R2 + D2

There’s an R2 term on both sides of the calculation so you can cancel them out, leaving:

2Rh + h= D2

Therefore the horizon distance, D, is:

D = √(2Rh+h2)

Here’s where you can make life much simpler for yourself. In almost every case R is much, much larger than h, which means that 2Rh is much, much larger than h2 so you can just ignore h2 and your equation simplifies to:

D ≈ √2Rh

(the ≈ sign here means “almost equals”. Honestly.)

So if you know R and h you can calculate D. To make this calculation easily you can carry round the value of √2R in your head meaning you only have to calculate √h and multiply those two numbers together.

So for the Earth, R is 6371000m, so √2R is 3569.6. Multiplying this by √h in metres would give you D in metres, so lets convert that into km to make things easier. This means dividing this number by 1000, giving an answer of 3.5696 which is ≈ 3.5.

So as a rough rule of thumb, your horizon distance on Earth,

D = 3.5 x √h

where D is measured in km and h in metres.

On Diana’s Peak, at 823m high, √h = 28.687… which multiplied by 3.5 gives a horizon distance of almost exactly 100km!

This is pretty cool, and is true of anywhere you can see the sea from a heigh of 823m.

One final calculation which sprung to mind on the mountain top was the area of sea I could see, which is easy to work out using the fact that the area of a circle is πr2, where r in this case is D, or 100km.

π is 3.14159 which means that the area of sea I could see was 31415.9 km2. Just a tad larger than Belgium, at 30528 km2.

And in that Belgium-sized circle of ocean was only one ship, the RMS St Helena that was taking me home the following day.

If you’re on Mars your horizon distance is shorter, at 2.6√h. On Mercury it’s smaller still at 2.2√h. This is due to Mars and Mercury being much smaller than the Earth, and so their surfaces curve away from you quicker. Venus is almost exactly the same size as the Earth (only a fraction smaller) so there you’d have to use the same calculation as here on Earth, 3.5√h.

Hovering above the surface of Jupiter your horizon would stretch to 11.8√h and on Saturn to 10.8√h. Uranus and Neptune are about the same size, giving a horizon distance of 7.1√h.

Mercury 2.2√h
Venus 3.5√h
Earth 3.5√h
Mars 2.6√h
Jupiter 11.8√h
Saturn 10.8√h
Uranus 7.1√h
Neptune 7.1√h

What about the dwarf planets? Being so small their surfaces will curve away from you very quickly, shortening your horizon distance. One of the smallest spherical objects in the solar system is the dwarf planet Ceres (as in cereal), which is the largest object amongst the fragments of rock in the asteroid belt. Your horizon distance on Ceres is almost exactly √h, making that a pretty simple horizon calculation!

Categories: Other stuff

## The Night Sky in May 2014

This month sees a glut of amazing stargazing sights in the night sky, even as the days lengthen towards summer.

Saturn as it might look through a large telescope, image by Kenneth Crawford and Michael A. Mayda

Saturn is coming to opposition this month (10 May) meaning it shines in the sky all night long throughout the month. A small telescope (even a pair of binoculars on a tripod) will show Saturn’s beautiful rings and one of its moons.

Mars is even brighter than Saturn, shining a soft orange colour in the constellation of Virgo, near the bright star Spica.

Jupiter is still an evening object although it sets in the west around 1am.

There’s the possibility of a spectacular new meteor shower on 23/34 May as the Earth passes through the dust trail of comet 209P/Linear.

And May sees the start of the noctilucent cloud season, where these elusive high-altitude begin to shine in deep twilight.

Full moon this month is on 14 May, when the Moon will sit near Saturn.

Categories: Stargazing

## Mars at Opposition 2014

This week the red planet Mars reached opposition, making it best placed for observing. Opposition is, as the name suggests, the point where a planet is directly opposite the Sun in our sky.

This means that Mars is up all night long at the moment, rising as the Sun sets and setting as the Sun rises, and so you should be able to spot it whatever time of night you’re out.

Mars reached opposition on 8 April, but on 14 April it will reach its closest approach to Earth, at a mere 57.4 million miles!

On that night – and on nights near that date – the planet Mars will shine very brightly at magnitude -1.5, brighter than anything else in the night sky except the Moon (which is Full on 14 April, and sits near Mars) and Jupiter.

Mars in the sky at midnight on 9 April

Mars also looms larger than normal when seen through a telescope, at a whopping 15″ (15 arcseconds = 0.25 arcminutes = 1/240 of a degree!). Stargazers with a decent sized telescope, good observing skills, and good observing conditions should be able to make out the north polar cap of Mars which is tilted towards us at the moment.

Through a small scope you might catch it looking like this:

Mars as seen through a small telescope

Categories: Stargazing