Rockets and Satellites | | DJS

Rockets and Satellites

It has struck me for some time that the ways in which we attempt to reach orbit are remarkably expensive, even ineffective. I began to wonder why.

A rocket functions by blowing exhaust at considerable speed, u. This reduces the mass of the rocket and raises the momentum. The change of mass is ∆m, exhaust ejected and the change of velocity is ∂m, so the law of conservation of momentum says that
 m∂v = -u∂m, where the negative shows that the mass ejected is going the other way from the rocket. This means that ∂v = -u∂m/m  and we can integrate this with intital conditions v₀ and m₀, so that

 v-v₀ = u ln (m₀/m)                       See Tsiolkovsky and his equation.

Rocket engineers like to express the exhaust velocity as 'specific impulse' x gravity, which makes specific impulse a unit in seconds and using a standard gravity, g. Rocket engineers also refer to 'delta-v', written ∆v, meaning the impulse (not velocity?)  required to perform a manoeuvre, and so is likely to be scalar not vector.

Typical rocket exhaust speeds are 2000 to 4500 m/s, though Jove says the practical limit for conventional rockets is 2500 m/s.

Centripetal force on a satellite is F = m v²/r  and gravitational force is F = G Mm/ r², so for a sateliite (mass m, the planet is mass M) in orbit, v² = GM/r. The gravitational constant is G= 6.6743x10⁻¹¹  Nm²/kg² and the mass of Earth is 5.9722x10²⁴ kg. So GM=3.9860x10¹⁴ [units, 5 sig fig.]

Low earth orbits [LEOs] lie below 2000km and are necessarily elliptical, even if nearly circular. At 600km a satellite will have atmospheric drag (and solar wind) and so its orbit will soon decay. Decay is rapid from 200km, under a day; at 300km typical orbit lifetime is a month, at 400km a year at 500km 10 years and at 700m 100 years. The international space station is at 340km, the Hubble telescope at 540km; sun synchronous satellites are between 600-800km and polar satellites between 700 and 1700km. Thus various things in orbit under 600km need periodic boosts, such as the Hubble telescope. This will be part of the 'maintenance' cycle. Medium earth orbits {MEOs] run from 2000km to 36000km and above that is the high earth orbit, HEO. A geosynchronous {GEO] or geostationary [GSO] satellite rotates at the same speed as the planet.

The planet rotates once every 23:56:04 = 86164 seconds, a sidereal day. The mean radius of the earth is 6371km. 

Q1. Work out the velocity of the surface of the earth in m/s, moving one whole circumference in a day.

Q2. Use  v² = GM/r. and vt=2πr to find the orbit (from the centre of the earth) for a geosynchronous satellite. Hence find the orbital altitude.

To cause a rocket to reach orbit we have a collection of problems not yet considered. the rocket sets off vertically but the objective is to be travelling in orbit. One assumes that there is a clever balancing act that optimises the curve travelled so as to best use the available fuel.

Q3.  Using G = 6.6743x10⁻¹¹  Nm²/kg²,  M= 5.9722x10²⁴ kg and r=6371km, produce a value for g at the planetary surface.
Repeat this for a LEO of 629 km. (Hint: why did I pick this number?)

Q4. Draw a graph showing orbital lifteimes and predict or discuss the fequency of visits to the international space station [ISS] and the  Hubble telescope so as to maintain their orbits, assuming that the mean orbital altitude for the ISS is 340-385km and that   [interesting paper and another]. The Hubble went up in 1990 and has had its orbit adjusted once since. 

The ISS has an orbital period of 91-94 minutes, 15.75 orbits per day at a height of, usually, between 340 and 350km (data). The ISS is visited and its orbit adjusted as shown below. Comment on the fit with your answer.

Q5. What is the orbital speed of the ISS?

------ I need to come back to this ---------

 Q6.  I read that for low earth orbit, a typical speed will be 7800m/s, which hopefully matches your last answer.  

If the exhaust speed is 2000m/s find the ratio of m₀/m.  Too hard?

thrust to weight ratio

angled flight

modelling software availble

A1. Circumference is 2π.6371000 = 4.003x10⁷m. this distance in 86164 seconds is 464.6m/s.

A2.  v² = GM/r  = 4 π² r² / t² => r³ = GMt²/4π² => r= 4.21642x10⁷m (4 s.f. preferred). So the orbital height is less by one earth radius, 35,790 km. Wikipedia says 42,164 km and 35786km, making the planet radius 6378 km which is the maximum value; the minimum is 6357 km.

A3.  F=mg = GM/r² => g = 9.8203 [m/s²], where 4 s.f. is appropriate. If we use the maximum planetary radius 6378, g reduces to 9.7965. If we use 6357km, g is 9.861.
For low earth orbit, r increases to 7000km. The new value of gravity, g₇ = 8.133 m/s². Using  v-v₀ = u ln (m₀/m)  and therefore  (v-v₀)/u = ln (m₀/m)        7800/2000 = 3.9 = ln 49.4. This doesn't seem anywhere close to right, since I already know that 96% of a typical rocket's weight is fuel.

A4.  The Hubble went up in 1990. Between 1993 and 2009 there were five visits,  [link] none of which admit to adjusting the orbit. A graph of orbital height against log time (I used log₁₀ (days) is close to a straight  (blue; y = 105x+165) line.  At 540km, and x= (y-165)/105, x= 3.57 where x is log (days) so days is 3727, still about ten years.  I got a fair fit (R=0.995)  with a power curve, y=199.84x^(0.1168) which suggests 13.6 years (4970 days, call that 5000 days to decay from 540km. So any value approximating 5000 days, around 13 to 14 years, is valid for 540km.

For the ISS at 340km, anywhere around three months looks like a fair figure and 120 days for 350km. So if each trip moves the station out 10km from 340 to 350km, that makes it possible for the next trip to fail, just.

A5. The ISS has an orbital period of 91-94 minutes, 15.75 orbits per day at a height of, usually, between 340 and 350km., which is 6371+345 (ish) = say about 6715km. 

Using v² = GM/r = 3.9860x10¹⁴ / 6.715x10⁵ => v=7705m/s. 

Using vt=2πr, the day is still 86164 seconds, roughly  91.2*60*15.75 seconds.  In a day the ISS travels nearly 16 orbits, so v = 2πr/t =15.752*πr/t =15.75*2π*6.715x10⁶/86164 = 7712 m/s, which is in close enough agreement to the previous 7705 and we'll call the answer 7700 m/s, or 7.7 km/s. We can declare two sig fig with some confidence.

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