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## Do GPS satellites move in the sky?

Following on from my blog “Is Tim Peake getting younger or older?” , a bit of fun to work out whether time was passing more slowly or more quickly for Tim Peake in the ISS than it is for us on the ground, it got me thinking about the global positioning system (GPS) that so many of us use on a daily basis. Whether it is using a SATNAV in our car, or a GPS-enabled watch to measure how far and fast we have run, or using maps on a smartphone, GPS must be one of the most-used satellite developments of the last few decades.

As I blogged about here, communication satellites need to be at a particular height above the Earth’s surface so that they orbit the Earth in the same time that it takes the Earth to rotate. In addition to their altitude, communication satellites can only orbit the Earth about the equator, no other orientation will allow the satellite to hover in the same place relative to a location on Earth.

But what about the satellites used in GPS? What kind of an orbit are they in?

## The GPS satellites’ orbits

It turns out that the GPS satellites are not in a geo-stationary orbit, but are in fact in an orbit which leads to their orbiting the Earth exactly twice in each sidereal day (for a definition of sidereal day see my blog here).

The GPS system consists of 31 satellites in orbit around the Earth

We can work out what radius from the Earth’s centre this needs to be by remembering that the speed of orbit is given by

$v = \sqrt { \frac{ GM }{ r } } \text{ (1) }$
where $v$ is the speed of orbit, $G$ is the universal gravitational constant, $M$ is the mass of the Earth and $r$ is the radius of orbit from the centre of the Earth (not from its surface).

A sideral day is 23 hours and 56 minutes, which in seconds is $8.6160 \times 10^{4}$ seconds. So, half a sidereal day is $4.308 \times 10^{4}$ seconds. We will call this the period $T$. The speed of orbit, $v$ is related to the period via the equation
$v = \frac{ 2 \pi r }{ T }$
where $r$ is the radius of the orbit, the same $r$ as in equation (1), and $2 \pi r$ is just the circumference of a circle. So, squaring Equation (1), we can write
$v^{2} = \frac{ GM }{ r } = \left( \frac{ 2 \pi r }{ T } \right)^{2}$
So, in terms of $r$ we can write
$r^{3} = \frac{ G M T^{2} }{ 4 \pi^{2} } \rightarrow r = \sqrt[3]{ \frac{ G M T^{2} }{ 4 \pi^{2} } }, \; \text{ so } r = 26.555 \times 10^{6} \text{ m}$
In terms of height above the Earth’s surface, we need to subtract off the radius of the Earth, so the altitude, which I will call $a_{gps}$, is going to be
$a_{gps} = 26.555 \times 10^{6} - 6.371 \times 10^{6} = 20.184 \times 10^{6} \text{ m } \text{ or } \boxed{ 20.2 \text{ thousand kilometeres} }$

## Why are GPS satellites in this kind of an orbit?

As I didn’t know what kind of an orbit GPS satellites were in before I wrote this blog, the next obvious question is – why are they in an orbit which is exactly half a sidereal day? It is clearly not coincidental! To answer this question, we need to first of all discuss how GPS works.

GPS locates your position by measuring the time a signal takes to get to your GPS device from at least four satellites. Your device can identify from which satellites it gets a signal, and the system knows precisely the position of these satellites. By measuring the time the signals take to you reach you from each of the satellites, it is able to calculate how far each one is from you, and then by using triangulation it can work our your location. There are currently 31 satellites in the system, so often there are more than four visible to your GPS device. The current 31 satellites have all been launched since 1997, the original suite of 38 satellites launched between 1978 and 1997 are no longer in operation.

As I mentioned in my blog about geostationary satellites, a satellite in a geostationary orbit can only orbit above the Earth’s equator. This would clearly be no good for a GPS system, as all the satellites would lie to the south of someone in e.g. Europe or North America. As I said above, there are currently 31 operational satellites; the 31 are divided into 6 orbital planes. If there were 30 satellites this would be 5 in each orbit. The orbits are inclined at $55^{\circ}$ to the Earth’s equator. Each orbit is separated from the other one by 4 hours (equivalent to $60^{\circ}$) in longitude.

As one can see approximately 6 hours in right ascension to both the east and west of one’s location, this means that there will be at least 3 of the orbits above the horizon, and sometimes more. If there were 5 satellites in each orbit this would mean that each one would pass a particular latitude 4 hours before the next one. So, at any particular time there should be some satellites further north than one’s location and some further south, as well as some further east and some further west. This configuration allows for the necessary triangulation to obtain one’s location.

The orbits are inclined at $55^{\circ}$ to the equator and separated by 4 hours (equivalent to $60^{\circ}$) in right ascension, as this diagram attempts to show

## Is the time-dilation effect due to SR or GR more important for these satellites?

We already showed in this blog that, for the International Space Station, the time-dilation due to Special Relativity (SR) has a greater effect on the passage of time than the time-dilation due to General Relativity (GR). What about for the GPS satellites?
The speed of orbit for the GPS satellites at a radius of $26.555 \times 10^{6}$ from the Earth’s centre is, using Equation (1),
$v = \sqrt{ \frac{ GM }{ r } } = 3.873 \times 10^{3} \text{ m/s}$
As we showed in my blog about Tim Peake, the speed of someone on the Earth’s surface relative to the centre of the Earth is $v_{se} = 463.35 \text{ m/s}$, so the relative speed between a GPS satellite and someone on the Earth’s surface is given by
$v = 3.873 \times 10^{3} - 463.35 = 3.410 \times 10^{3} \text{ m/s}$
Compare this to the value for the ISS, which was $7.4437 \times 10^{3}$, it is less than half the speed.

This value of $v$ leads to a time dilation factor $\gamma$ in SR of
$\gamma = \frac{ 1 }{ \sqrt{ 0.9999999999} } \approx 1$
which means that the time dilation due to SR is negligible.
The time dilation due to GR is given by (see my blog here on how to calculate this)
$\left( 1 - \frac{ gh }{ c^{2} } \right) = (1 - 2.2 \times 10{-9}) = 0.9999999978$, or 22 parts in $10^{10}$. Compare this to the ISS, where it was about 1 part in $10^{11}$. Clearly the GR effect for GPS satellites is greater, by about a factor of 5, than it was for the ISS. But, conversely, the SR time-dilation effect has become negligible.

To conclude, the time dilation for GPS satellites is nearly entirely due to General Relativity, and not due to Special Relativity. Time is passing more quickly for the clocks on the GPS satellites than it is for us on Earth, the converse of what we found for the ISS, which is in a much lower orbit.

Because the timings required for GPS to work are so precise, the time dilation effect due to GR needs to be taken into account, and is one of the best pieces of evidence we have that time dilation in GR actually does happen.

## Tim Peake is home from space

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.

## Is Tim Peake getting younger or older?

As anyone who hasn’t been living under a rock knows, the International Space Station (ISS) orbits the Earth with (typically? always?) six astronauts on board. It has been doing this for something like the last fifteen years. One of the astronauts currently on board is the Disunited Kingdom’s first Government-funded astronaut, Tim Peake.

The first British person to go into space was Helen Sharman, but she went into space in a privately funded arrangement with the Russian Space Programme in 1991. Other British-born astronauts have gone into space through having become naturalised Americans, and going into space with NASA. But, Tim Peake has gone to the ISS as part of ESA’s space programme, and his place is due to Britain’s contribution to ESA’s astronaut programme. So, he is the first UK Government-funded astronaut, which is why there has been so much fuss about it in these lands.

Official NASA portrait of British astronaut Timothy Peake. Photo Date: August 28, 2013.

Anyway, I digress. This blog is not about Tim Peake per se, or about the ISS really. I wanted this blog to be about whether Tim Peake is getting older or younger whilst in orbit. Of course everyone is getting older, including Tim Peake, as ‘time waits for no man’ as the saying goes. What I really mean is whether time is passing more or less quickly for Tim Peake (and the other astronauts) in the ISS compared to those of us on Earth.

As some of you might now, when an astronaut is in orbit he is in a weaker gravitational field, as the Earth’s gravitational field drops off with distance (actually as the square of the distance) from the centre of the Earth. Time will therefore pass more quickly for Tim Peake than for someone on the Earth’s surface due to this effect. This is time dilation due to gravity, a general relativity (GR) effect.

But, there is also another time dilation, the time dilation due to one’s motion relative to another observer, the time dilation in special relativity (SR). Because Tim Peake is in orbit, and hence moving relative to someone on the surface of the Earth, this means that time will appear to move more slowly for him as observed by someone on Earth. Interestingly (at least for me!), the SR effect works in the opposite sense to the GR effect.

Which effect is greater? And, how big is the effect?

## Time dilation due to SR – slowing it down for Tim Peake

As I showed in this blog, the time dilation due to SR can be calculated using the equation

$t^{\prime} = \gamma t \text{ where } \gamma = \frac{ 1 }{ \sqrt{ ( 1 -v^{2}/c^{2} )} }$

If he is in orbit at an altitude of 500km (I guessed at this amount, according to Wikipedia it is 400km, but it does not alter the argument which ensues) then his distance from the centre of the Earth (assuming a spherical Earth) is $6.371 \times 10^{6} + 500 = 6.3715 \times 10^{6}$ metres. The centripetal force keeping him in orbit is provided by the force of gravity, and in this blog I showed that the centripetal force $F_{c}$ is given by

$F_{c} = \frac{mv^{2} }{r}$

where $m$ is the mass of the object in orbit, $v$ is its velocity and $r$ is the radius of its orbit.This centripetal force is being provided by gravity, which we know is

$F_{g} = \frac{ GMm }{ r^{2} }$

where $G$ is the universal gravitational constant, and $M$ is the mass of the Earth. Putting these two equal to each other

$\frac{ mv^{2} }{ r } = \frac{ GMm }{ r^{2} } \rightarrow v^{2} = \frac{GM}{r}$

Putting in the values we have for the ISS, where $r=6.3715 \times 10^{6}, G=6.67 \times 10^{-11}$ and $M= 5.97237 \times 10^{24}$, we find that

$v^{2} = 6.2522 \times 10^{7} \rightarrow v = 7.907(067129) \times 10^{3} \text { m/s} = \boxed{ 7.907(067129) \text{ km/s} }$

But, this is the motion relative to the centre of the Earth. People on the surface of the Earth are also moving about the centre, as the Earth is spinning on its axis. But, we cannot calculate this speed as we have done above; people on the surface are not in orbit, but on the Earth’s surface. For something to stay e.g. 1 metre above the Earth’s surface in orbit it would have to move considerably quicker than the rotation rate of the Earth.

The Earth turns once every 24 hours, so for someone on the equator they are moving at

$v_{se} = \frac{ 2 \pi \times 6.3715 \times 10^{6} }{ 24 \times 3600 } = 463.348(5554) \text { m/s}$

where $v_{se}$ refers to the speed of someone on the surface of the Earth. Someone at other latitudes is moving less quickly, at the poles they are not moving at all relative to the centre of the Earth. The speed of someone on the surface will go as $v_{se} \cos (\theta)$ where $\theta$ is the latitude. This is why we launch satellites as close to the Earth’s equator as is feasible; we maximise $v_{se}$ and thus get the benefit of the speed of rotation of the Earth at the launch site to boost the rocket’s speed in an easterly direction.

The difference in speeds between the ISS and someone at the equator on the surface of Earth is therefore

$7.907(067129) \times 10^{3} - 463.348(5554) = \boxed { 7.443(718574) \times 10^{3} \text { m/s} }$

Referring back to my blog on time dilation in special relativity that I mentioned at the start of this section, this means that the time dilation factor $\gamma$, using this value of $v$, is

$\gamma = \frac{ 1 }{ \sqrt{(1 - (v/c)^{2})} } = \frac{ 1 }{ 0.9999999997 }$
(where $c$ is, of course, the speed of light).
This value of $\gamma$ is equal to unity to 3 parts in $10^{10}$, so it would require Tim Peake to orbit for about $3 \times 10^{9}$ seconds for the time dilation factor to amount to 1 second. $3 \times 10^{9}$ seconds is just over 96 years, let us say 100 years.

## The time dilation due to GR – speeding it up for Tim Peake

For GR, the time dilation works in the other sense, it will run more slowly for those of us on the Earth’s surface; we experience gravitational time dilation which is greater than that experienced by Tim Peake. In this blog here, I derived from the principle of equivalence the time dilation due to GR, and found

$\Delta T_{B} = \Delta T_{A} \left( 1 - \frac{ gh }{ c^{2} } \right)$

where, in this case, $\Delta T_{B}$ would be the rate of time passing on the Earth’s surface, $\Delta T_{A}$ the rate of time passing on the ISS,  $g = 9.81$ (the acceleration due to gravity at the Earth’s surface) and $h$ is the height of the orbit, which we have assumed (see above) to be 500 km = $500 \times 10^{3}$.

Plugging in these values we get that

$\frac{ \Delta T_{B} }{ \Delta T_{A} } = 1 - 5.45 \times 10^{-11}$

So, the GR effect is about one part in $10^{11}$ (100 billion). In six months, the number of seconds that Tim Peake will be in orbit is about $1.6 \times 10^{7}$ seconds, so a factor of about 10,000 less than for the GR effect to amount to 1 second. Tim Peake would need to be in orbit for about 5,000 years for the GR effect to amount to 1 second of difference!

## Conclusions

In conclusion, the SR effect on how quickly time is passing for Tim Peake is about 3 parts in 10 billion, in the sense that it passes more slowly for Tim Peake. The GR effect is even smaller, about one part in 100 billion, but in the sense that time is passing more quickly for him. The SR ‘time slowing down’ effect is greater than the GR time ‘passing more quickly effect’, by roughly a factor of 300.

Tim Peake is therefore actually ageing more slowly by being in orbit than if he were on Earth. But, he would need to orbit for nearly 100 years for this difference to amount to just 1 second! And, none of this of course takes into account the detrimental biological effects of being in orbit, which are probably not good to anyone’s longevity!

## Space debris and geostationary satellites

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.

## Summary

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!

## Ground control to Major Tim

Last week, in a big media event at London’s Science Museum, it was announced that Major Tim Peake would become Britain’s first “government backed” astronaut. He is scheduled to go to the International Space Station in late 2015.

He will not be the first British person in space, that was Helen Sharman who, in 1991, went into space through a joint venture between a number of private UK companies and the Russian government. There are also several British born astronauts who have gone into space through NASA, but to do so they have had to become US citizens. Michael Foale is the most well-travelled British born astronaut.

In my opinion it is wonderful news that the Disunited Kingdom’s government is now officially getting involved in sending humans into space. Although we learn a great deal about the Universe beyond Earth when we send space probes, there is nothing which captures the public imagination more than sending humans into space. One only had to see the press coverage this announcement received to see the proof of that, and today’s teenagers will hopefully be inspired to take up a career in science or engineering through being inspired by people like Tim Peake.

Of course the title of my blog is a play on the lines in David Bowie’s Space Oddity, so it also gives me a good excuse to add a YouTube clip of that wonderful song.