[amsat-bb] BY70-1

Phil Karn karn at ka9q.net
Sat Feb 25 05:41:01 UTC 2017

On 2/24/17 07:55, Howie DeFelice wrote:
> Very interesting Phil, it seems to make sense that this calculation
> could be used in reverse to calculate the energy needed raise the
> perigee height of a GTO orbit. Assuming a flight to GTO was available to
> a 1U or 3U cubesat, if the perigee is not raised the satellite will not
> stay in orbit very long, if I understand it correctly. Given the limited
> size of the spacecraft and the prohibition on volatile propellants this
> poses a difficult challenge. It would be interesting to determine if
> enough thrust can be generated by electrical thrusters to accomplish this ?
> - Howie AB2S

It would be most relevant if you can use a tether to form an electric
motor with the earth's magnetic field to raise your orbit.

Otherwise, things are much more complicated with a chemical or
electrical rocket because you have to carry your reaction mass with you
and then put energy into it to blow it out the nozzle at high speed.

There's a fundamental tradeoff in rocketry between rocket power and
propellant mass flow rate. You can produce a given amount of thrust with
high power and a low propellant mass flow rate, or with low power and a
high propellant mass flow rate.

E.g., to produce a thrust of 1 N with a mass flow rate of 1 kg/s, you
have to eject it at 1 N / 1 kg/s = 1 meter/sec. Ignoring relativity, the
kinetic energy in 1 second of exhaust (1 kg) will therefore be

1/2 mv^2 = 1/2 * 1 kg * (1 m/s)^2 = 0.5 joules

and since you need 0.5 joules every second, the required power will be
0.5 watts (assuming 100% efficiency).

If you double the exhaust velocity to 2 m/s, you can drop the mass flow
rate to only 1/2 kg/s and still get 1 N of thrust (1/2 kg/s * 2 m/s = 1
N). But you'll now need a power of

0.5 * 0.5 kg/s * (2 m/s)^2 = 1 watt

i.e., twice as much power for that same 1 newton of thrust.

So, which do you have more of, propellant mass or energy? In a chemical
rocket the energy is stored in the unburned propellant, so the energy
per unit mass is set by the propellant chemistry. That's why every
propellant combination has a theoretical specific impulse, e.g. 455
seconds for hydrogen/oxygen in vacuum. Specific impulse is just
effective exhaust velocity divided by g = 9.8 m/s^2, so the theoretical
exhaust velocity for hydrogen/oxygen is 4,462 m/s.

But in an electric rocket the energy source is external to the
propellant mass, so the energy/mass ratio can vary; you decide how fast
to eject it. If mass is cheaper than energy, then you want a low exhaust
velocity. If energy is cheaper than mass, then you want a high exhaust

Since the rocket is free to move, the kinetic energy it produces will be
split between the payload/rocket itself (which you want) and the exhaust
(which is effectively wasted). The only way to get 100% of the energy
into the payload/rocket and none into the exhaust is to set the exhaust
velocity equal to the current velocity of the rocket so that the exhaust
comes out stationary. Of course, velocity is relative so you measure it
relative to the reference frame in which the rocket is initially
stationary. So to minimize energy consumption you want to increase the
exhaust velocity as the rocket accelerates. That's the exact idea behind
the VASIMR (Variable Specific Impulse Magnetoplasma Rocket).

The recent "EM drive" hype notwithstanding, I know of only one way to
produce thrust in vacuum without some kind of propellant: the photon
rocket. Even a flashlight will work, but let's do the numbers. The
momentum of a photon is equal to its energy divided by the speed of
light, so to get 1 newton of thrust from a 100% efficient photon rocket
requires a power input of 1 N * c = 300 megawatts!

That kind of power in space requires either a very large solar array or
a very big nuclear reactor (which still needs a very large radiator to
reject waste heat).

But there's a simpler way to power a photon rocket with the sun. Instead
of turning solar photons into electricity and back into photons, why not
use solar photons directly? Voila -- that's what a solar sail does. The
thrust produced by a solar sail per unit area is equal to the incident
solar power per unit area divided by the speed of light. At 1 AU that's
about 1361 W/m^2, so the thrust will be 1361 W/m^2 / c = 4.54 micro
newton/m^2. That's actually units of pressure, so solar radiation
pressure at 1 AU is 4.54 micropascal on a sail normal to the sun that
simply absorbs solar photons. If you reflect them back, you'll get twice
as much, 9.08 micropascal. Doesn't seem like much, but you'll get it
continuously, no local power source or propellant mass needed.

The one big problem with solar sails is that you can't use them in low
orbits because they'll generate far more drag than thrust.

73, Phil

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