Posts Tagged ‘International Space Station’

On Saturday 18 June, as some of you may know, Tim Peake returned from his 6-month stint on the International Space Station (ISS). At the end of January, as a bit of fun, in a blog entitled “Is Tim Peake getting younger or older?”  I worked out whether he was getting younger (due to time dilation in special relativity) or older (due to time running faster due to general relativity). The answer was that the special relativity effect of time slowing down for him was greater than the general relativity effect of time speeding up. But, he would need to stay in space for 100 years to age by 1 second less than if he were on Earth! But, now that he is back on Earth time is running at the same rate for him as for the rest of us. 🙂

Peake held a press conference on Tuesday 21 June, and later that day I was on BBC radio making some comments about his time on the ISS. It was only a short 3-minute interview for the evening news programme (you can listen to it here), but one of the things I was asked was whether Tim Peake’s mission to the ISS had inspired young people (school students).


Astronaut Tim Peake returned to Earth from the International Space Station on 18 June 2016 after a six-month period there.

My answer was that yes, it absolutely had. Peake has captured the public imagination with his trip to the ISS, and has inspired a whole new generation to think about space. As the first person from Britain to go into space at the taxpayers’ expense, he may have had instructions to engage with the public in his time spent there. I don’t know. But, what I do know is that he clearly enjoys communicating science and the wonders of space and the oddities of an astronaut’s life to the public, and has done an excellent job of it.

I just about member the last Apollo mission, Apollo 17, which went to the Moon in December 1971. I’m too young to remember the ones before; even though my mother sat me down in front of the TV to see Neil Armstrong take his historic steps in 1969, I sadly don’t remember it. Seeing astronauts going to the Moon was certainly a factor in igniting my own interest in space and astronomy, but since that time there has been very little to inspire later generations. Going up in the Space Shuttle or going to the ISS are not as exciting as going to the Moon; but thankfully Tim Peake has turned what has become a rather routine activity these days into something very exciting for our younger people.

I don’t know how much it cost the DUK taxpayers to put Peake into space, but I can guarantee you that the money will be recouped dozens of times over. There is no surer way to create wealth than through science and technology, and inspiring a whole new generation of school students into taking an interest in physics, mathematics, engineering and science will, hopefully, see more of them pursue such careers in the future. This can only be a boom for our economy.

Read Full Post »

Last week, as part of the “tech hour” on BBC Radio 5’s “Afternoon Edition” on Wednesdays, I heard one of their reporters do an item on visiting a facility in Australia which is trying to track space debris and locate their positions precisely. After the item a listener asked about geostationary orbits, and it was clear from the reporter’s answer that he did not really understand them. Then again, if he is not a trained physicist or engineer, why should he?

In the light of the wrong answer given to the listener by the reporter, I’ve decided to blog about satellite orbits to clear up any confusion, and also to show that the mathematics of it is not that hard. This is the kind of problem that a student studying physics would be expected to do in their last year of high-school here in Wales. Mark Thompson, one of the astronomers who has appeared on BBC Stargazing Live, did clear up some of the confusion when he came on the programme later in the same hour, but he did not have time to go into all the details (nor would he want to do so on a radio show!).

Low-Earth and geostationary orbits

For those of you not familiar with the terminology, a low-Earth orbit is typically just a few hundred kilometres above the Earth’s surface, which as I will show below leads to an orbit about the Earth which takes a couple of hours at most. These are the kinds of orbits which are used by most spy satellites, and also by the International Space Station (ISS) and, when it used to fly, by the Space Shuttle. The famous Hubble Space Telescope is in a low-Earth orbit. Any satellites in low-Earth orbits will experience drag from the Earth’s atmosphere.

Space “officially” begins at 100km above the Earth’s surface, but the atmosphere doesn’t suddenly stop at any particular altitude,it just gets thinner and thiner. In fact, the “scale height” of the atmosphere is about 1 mile, or about 1.5km. For each 1.5km altitude, the amount of atmosphere roughly halves (based on pressure), so at an altitude of 3km the air is 0.5 \times 0.5 = 0.25 (25\%) as thick, and if you go up to 10.5km which is roughly the altitude at which commercial aeroplanes fly, then the thickness of the atmosphere is about 0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 = 0.5^{7} = 0.0078 as thick, or 0.78 \% of the thickness at the ground. Even though the thickness of the atmosphere a few hundred kilometres up is very very thin, there is still enough of it to slow satellites down. So, satellites in low-Earth orbits cannot maintain their orbits without some level of thrust, otherwise their orbits would just decay and they would burn up in the atmosphere.

Low-Earth orbit satellites can be in any orbit as long as it is centred on the Earth’s centre. So, for example, many spy satellites orbit the Earth’s poles. If you think about it, if they are taking about 90 minutes to orbit (some take less, they will be even closer to the Earth), then each time they orbit they will see a different part of the Earth, which is often what you want for spy satellites. The ISS orbits the Earth in an orbit which is titled to the Earth’s equator, so it passes over different parts of the Earth on each orbit, but it is not a polar orbit. A satellite cannot, for example, orbit, say, along an orbit which is directly above the tropic of Cancer, and satellites do not orbit the Earth in the opposite direction to the Earth’s rotation, they all orbit the Earth in the same direction as we are rotating.


Geostationary orbits, on the other hand, are a particular orbit where the satellite takes 24 hours to orbit the Earth. So, as seen by a person on the rotating Earth, they appear fixed relative to that person’s horizons. Geostationary orbits are used for communication satellites so that, for example, your TV satellite dish can be fixed to the side of your house and point in the same direction, as the satellite from which it’s getting the TV signals stays fixed in your sky. They are also used by Earth-monitoring satellites such as weather satellites and satellites which measure sea and land temperatures. A geostationary satellite has to be at a particular altitude above the Earth’s surface, and it has to be in orbit about the Earth’s equator (so it sits above the Earth’s equator), and of course orbits in the same direction as the Earth’s rotation.

How do we calculate the altitude of these respective orbits, the low-Earth and the geostationary ones? It is not too difficult, as I will show below.

The altitude of a satellite which takes 90 minutes to orbit the Earth

To make the mathematics simpler, I will do these calculations assuming a circular orbit. In reality, some satellites have elliptical orbits where they are closer to the Earth’s surface at some times than at others, but that makes the maths a little more complicated, and it’s not necessary to understand the basic ideas.

As I showed in this blog, when an object is moving in a circle it must have a force acting on it, because its velocity (in this case its direction) is constantly changing. This force is called the centripetal force, and as I showed in the blog this is given by

F_{centripetal} = \frac{ mv^{2} }{ r }

where m is the mass of the object moving in the circle, v is its speed, and r is the radius of the circle.

For a satellite orbiting the Earth, the centripetal force is provided by gravity, and the force of gravity is given by

F_{gravity} = \frac{ G M m}{ r^{2} }

where M is the mass of the Earth (in this case), m is the mass of the satellite, r is the radius of the orbit as measured from the centre of the Earth, and G is a physical constant, known as Big G or the universal gravitational constant.

Because the centripetal force is being provided by gravity, we can set these two equal to each other and so we have

\frac{ mv^{2} }{ r } = \frac{ G M m }{ r^{2} } \rightarrow v^{2} = \frac{ GM }{ r } \text{ (Equ. 1) }

The speed of orbit, v is related to how long the satellite takes to orbit via the definition of speed. Speed is defined as distance travelled per unit time, and as we are assuming circular orbits the distance travelled in an orbit is the circumference of a circle. So, we can write

v = \frac{ 2 \pi r }{ T }

where T is the time it takes to make the orbit. Substituting this expression for v in Equation (1) we can write

\left( \frac{ 2 \pi r }{ T } \right)^{2} = \frac{ GM }{ r }

Remember we are trying to calculate r for a satellite which takes 90 minutes to orbit, so we re-arrange this equation to give

r^{3} = \frac{ GM T^{2} }{ 4 \pi^{2} } \text{ (Equ. 2)}

Notice that the only two variables in this equation are r \text{ and } T, everything else is a constant (for orbiting the Earth, obviously if the object were orbiting e.g. the Sun we’d replace the mass M by the mass of the Sun). Equation (2) is, in fact, just Kepler’s Third law (which I blogged about here), which is that

\boxed{ T^{2} \propto r^{3} }

Now, to calculate r for a satellite taking 90 minutes to orbit, we just need to plug in the values for the mass of the Earth (5.972 \times 10^{24} kg), the value of G=6.67 \times 10^{-11}, and convert 90 minutes to seconds which is T=5400 \text{ s}. Doing this, we get

r^{3} = 2.942 \times 10^{20} \rightarrow r = 6.65 \times 10^{6} \text{ m}

But, remember, this is the radius from the Earth’s centre, so we need to subtract off the Earth’s radius to get the height above the Earth’s surface. The average radius of the Earth is R = 6.37 \times 10^{6} \text{ m} so the height above the Earth’s surface of a satellite which takes 90 minutes to orbit is

6.65 \times 10^{6} - 6.37 \times 10^{6} = 281 \times 10^{3} \text{ m  } \boxed{ =  281 \text{ km} }

which is roughly 300km.

The altitude of a geostationary satellite

To calculate the altitude of a geostationary satellite we just need to replace the time T in equation (2) with 24 hours (but put into seconds), instead of using 90 minutes. 24 hours is 24 \times 3600 = 8.64 \times 10^{4} \text{ s}, and so we now have

r^{3} = 7.53 \times 10^{22} \rightarrow r = 4.22 \times 10^{7} \text{ m}

Again, subtracting off the Earth’s radius, we get the altitude (height above the Earth’s surface) to be

4.22 \times 10^{7} - 6.37 \times 10^{6} = 3.59 \times 10^{7} \text{ m  } \boxed{ = 35,900 \text{ km} }

which is roughly 36,000 km.


It surprises most people that geostationary satellites are so far above the Earth’s surface, some 36,000 km. There is no atmosphere at that altitude, so geostationary satellites really pose no threat in terms of man-made space debris. Their orbits will not decay, as there is no atmospheric drag. It is the low-Earth orbit satellites which are the source of our man-made space debris, and if their orbits are not maintained the orbits will decay and they will burn up in the Earth’s atmosphere. Of course, most satellites entering the thicker parts of the atmosphere burn up completely, but larger satellites have been known to have parts of them survive to hit the Earth’s surface. For example this is what happened to the Russian space station MIR when its orbit was decayed by switching off its thrusters.

The time it takes a satellite to orbit the Earth is entirely dependent on how far from the centre of the Earth the satellite is. Low-Earth orbit satellites orbit the Earth in a matter of hours, many less than two hours. Geostationary satellites have to be about 36,000 km from the surface of the Earth in order to take 24 hours to orbit. All satellites orbit the Earth in the same direction as the Earth is rotating, except for satellites in a polar orbit which essentially allow the Earth to rotate below them as they go around the poles. Geostationary satellites have to orbit above the Earth’s equator, but other satellites can be in an orbit which is inclined to the equator, as long as the centre of their orbit is the centre of the Earth.

Finally, coming back to the issue of space debris; one of the reasons that space debris is so dangerous to other satellites and astronauts is that objects in Earth-orbit are moving very quickly. For example, for an object taking 90 minutes to orbit the Earth, such as the orbit the ISS is in, we can work out how quickly (in metres per second) it is moving quite easily.

If T=90 \text{ minutes } = 5400 \text{ s} and the radius of the orbit (as we showed above) is r = 6.65 \times 10^{6} \text{ m } then

v = \frac{ 2 \pi \cdot 6.65 \times 10^{6} }{ 5400 } \; \; \boxed{=  7737.6 \text{ m/s } }

so, nearly 7,740 m/s or 7.7 km/s or nearly 28 thousand km/h, which by anyone’s standards is FAST!

The damage a piece of space debris can make is dependent on its kinetic energy (KE), which is given by

\text{Kinetic energy } = \frac{ 1 }{ 2 } mv^{2}

An object with a mass of 1kg moving at 7,740 m/s will thus have a KE of

KE = \frac{ 1 }{ 2 } 1 \cdot (7740)^{2} = 30.0 \times 10^{6} \text{ Joules}

(or 30 Mega Joules). This is a huge amount of energy, because the object is moving so fast. By comparison, a car, which we will assume to have a mass of 1,000 kg, moving at 100 km/h will have a kinetic energy of 0.77 \times 10^{6} \text{ Joules}, or nearly 0.8 Mega Joules, so about a factor of 40 (forty!) less than a 1kg object in low-Earth orbit. This is why such small objects can cause such damage!

Read Full Post »