A Numerical Error
in Hawking’s book, ”Brief answers to big questions”
Last time I had an erratum of mine to share. This time: it’s
somebody else’s!
On p.115 of Stephen Hawking’s book, “Brief answers to big questions”
[1], it is written:
“The speed at which we
can send a rocket is governed by two things, the speed of the exhaust and the
fraction of its mass that the rocket loses as it accelerates. The exhaust speed
of chemical rockets, like the ones we have used so far, is about three kilometres
per second. By jettisoning 30 per cent of their mass, they can achieve a speed
of about half a kilometre per second and then slow down again.” (emphasis mine).
This looks just like
momentum conservation, so I tried to get the answer: ‘about ½ km/sec’. The only
hassle is that the mass keeps changing as the fuel is expended and exhausted. So
I wrote a simple BASIC program, but I got a different number: 1.07 km/sec. Then
I realized you can get the same result by integration:
Vexh dm = m
dV
Rearranging:
dV = - Vexh dm/m
Integration of the equation from V = Vi (initial
velocity) & mass M = Mi at t = ti to V = Vf
& M = Mf at t = tf
yields:
DV = -
Vexh ln (Mi/Mf), where DV = Vf – Vi.
Assuming 30% of the fuel is used, as mentioned, and Vexh
= 3 km/s, we get:
DV = -
3 ln(1/0.7) @ 1.07
km/s.
After I did all this I belatedly remembered that there was
something called a rocket equation that was derived by Konstantin Tsiolkovsky
in 1903 [2].
At this point I decided to write to the publisher of the
book but I got no reply, mainly because I was not sure if it was published by
Hodder & Stoughton, Hachette or whoever. I even mailed some website claiming to be 'connected with Lucy Hawking'. Finally I got a reply from an editor at John
Murray, (to the mail to Hachette!), who thanked me for contacting them and said they would have to send
it to an expert. After a reminder, about 45 days, the editor wrote that the expert
felt that the writing ‘could be clearer’ and that they might just delete the
passage since they were aiming at a non-specialist audience. (This sounds
plausible, since the book was published in the memory of Stephen Hawking, who
died in March 2018).
I wrote that they should just correct the error, and add a
footnote crediting the rocket equation to Tsiolkovsky [2]. No reply, to date.
I then sent the question to the author F.Todd Baker [3] (at
his website Ask the Physicist) - who agreed with the number that I had calculated.
He also speculated as to why the discrepancy occurred:
“I do not know
how Hawking came up with his answer, but it might hinge on what he means by
"30% of its fuel"; you and I have interpreted that to mean the
exhausted fuel has a mass equal to 30% of the mass of the rocket before the
fuel was burned, hence M/(M+m)=0.7.
Also, maybe he was thinking about a launch from earth which
would require some of the energy from the rocket to overcome the gravitational
force and the energy loss from air drag.
Or maybe he was guesstimating the effects of efficiency of
the engines.”
My explanation is simpler: whoever did the calculation, used
the rocket equation, but calculated log10 instead of ln. The factor
of 2.303 accounts for the discrepancy, resulting in a number of 0.46 km/sec –
which was then approximated as ‘about ½ km/sec’.
However, Baker made another point: that the above equation
neglects the mass of the rocket itself (and the mass of the payload). So what
are the numbers for the ‘dry mass’?
“For a typical rocket, the total mass of the vehicle might
be distributed in the following way: Of the total mass, 90% is
the propellants; 6% is the structure (tanks, engines, fins, etc.); and 4%
can be the payload ” [4].
According to NASA [5], the percentage varies, depending upon
the propellant used, but it varies between roughly
85-95% :
|
Propellant
|
Rocket
Percent Propellant for Earth Orbit
|
|
|
|
|
Solid Rocket
|
96
|
|
|
|
|
Kerosene-Oxygen
|
94
|
|
|
|
|
Hypergols
|
93
|
|
|
|
|
Methane-Oxygen
|
90
|
|
|
|
|
Hydrogen-Oxygen
|
83
|
I found an easy way out: an online calculator from Wolfram
[6]. (Probably Hawking’s assistant should have used it too!). The calculator
requires that you input both the initial and final mass, but the DV calculated depends only
on their ratio (as above). But I checked it independently as well, accounting separately for the rocket mass and the fuel in the equation for initial and
final masses.
The results are tabulated below for ratios of (rocket
mass)/(initial mass) varying from 0 to 15%. For the case of 30% fuel used, and
zero rocket mass, the number 1.07 km/sec is obtained from the online calculator
(see below). As expected, the DV
decreases as the percentage of rocket mass increases – but not drastically.
This takes care of F.Todd Baker’s suggestion (one of them, rather). The other
point that he made: there is a difference between 30% of the initially available
fuel being consumed vs 30% of the initial mass being burnt as fuel. For the case
of low dry mass fractions (<15 a="" air="" application="" are="" as="" but="" can="" checked="" difference="" drag="" efficiency="" engine="" for="" gravitation="" here="" is="" it="" not="" o:p="" occam="" of="" others="" out.="" razor="" results="" rule="" s="" shown="" small.="" the="" them="" there="">
|
Rocket mass/initial mass
|
30% of fuel used
|
40% of fuel used
|
50% of fuel used
|
70% of fuel used
|
80% of fuel used
|
100% of fuel used
|
|
0
|
1.07
|
1.532
|
2.079
|
3.6119
|
4.8283
|
|
|
0.05
|
1.006
|
1.434
|
1.933
|
3.28
|
4.2813
|
8.987
|
|
0.1
|
0.9441
|
1.3388
|
1.794
|
2.9827
|
3.819
|
6.908
|
|
0.15
|
0.8831
|
1.246
|
1.66
|
2.7116
|
3.418
|
5.691
|
The tabulated data are plotted in a graphical form above. It
is interesting that there seems to be greater deviation from linearity for a
larger fraction of fuel use.
At the end of the day, the whole calculation is rather
trivial – especially if you know that you should use Tsiolkovsky’s rocket
equation [2] – and even more so if you use the online calculator [6] from
Wolfram. Nevertheless, it is important to get the numbers right.
Addendum:
An interesting application of the rocket equation was done
by Hippke [7] in a recent paper. He argues that – assuming intelligent life
develops on a super-Earth (up to 10X of Earth’s mass ME), it would
need to overcome the planet’s higher gravity and thus higher escape velocity Vesc
(which is proportional to M1/2, where M is the mass of the planet). This
becomes difficult as he explains:
“Many rocky exoplanets are heavier and larger
than the Earth, and have higher surface gravity.
This makes spaceflight
on these worlds very challenging, because the required fuel mass for a given
payload is an
exponential function of planetary surface gravity. We find that chemical
rockets still
allow for escape
velocities on Super-Earths up to 10X ME . More massive rocky worlds,
if they exist, would require other means to leave the planet, such as nuclear
propulsion.”
The mass ratio of the
vehicle becomes:
Mi/Mf =
exp(Vesc/Vth)
which translates to a
mass ratio of ~26 on
Earth, and~2,700
on the super-Earth planet Kepler-20b (9.7X ME). This does not make
it impossible – just that a humongous amount of fuel would be needed to reach
escape velocity of the super-Earth. Hippke gives some arguments about why
super-Earths would be suitable for life (at least they have more surface area
than Earth!), but the jury is out on that one.
References:
1 1. “Brief Answers to Big Questions”
Stephen Hawking (John Murray, 2018) p.114
7 7. Michael Hippke,” Spaceflight from Super-Earths
is difficult” arXiv 18th May 2018 arXiv: 1804.04727v2
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