Bezos says 24X less liftoff energy on the Moon –
but is it 22.09?
“As Bezos laid out in his May 9th 2019 address, in a
presentation at the Air and Space Museum, utilizing resources from the moon
instead of bringing equivalents to space from the earth’s surface provides an
extraordinary advantage: it takes 24 times less energy to lift one pound off
the moon compared to lifting one pound off Earth.” [1]
The Moon’s lighter gravity means that it takes "24
times less energy to lift a pound off the moon than it does Earth." [2]
Jeff Bezos said that it takes 24X less energy to get an
object out of Moon’s gravity than to get it out of Earth’s – as per the above
quotes [1,2].
The Moon’s gravity is 6X less than Earth’s, so how does this
24X smaller energy happen?
See the table below [3]:
|
|
Earth
|
Moon
|
Ratio
|
|
Escape velocity (km/sec)
|
11.186
|
2.38
|
4.7
|
|
g ( m/sec2)
|
9.807
|
1.62
|
6.054
|
|
R (kms)
|
6,371
|
1,737
|
3.66
|
The ratio that Bezos mentioned is: (11.186/2.38)2
= 22.09.
Not 24X, but close.
Also: 4.7 @
(6.054 x 3.66)1/2 = 4.707.
Actually, this should be obvious from the definition of
escape velocity: the minimum velocity you have to give an object to take it
infinitely far from the astronomical body it is escaping from.
The reason? The escape velocity is given by:
vesc = [GM/R]1/2
But :
g = [GM/(R)2]1/2
So that: vesc = [gR]1/2
The radius of Earth at the equator is 6,378 kilometers [4],
according to NASA's Goddard Space
Flight Center. However, Earth is not quite a sphere. The planet's rotation
causes it to bulge at the equator. Earth's polar radius is 6,356 kms — a
difference of 13 miles (22 km). This amounts to 0.3% ‘error’.
Some more data [5] is necessary to do some simple checks:
|
|
|
Moon
|
Earth
|
|
Equatorial radius
|
1738.1
|
6378.1
|
|
Polar radius
|
1736.0
|
6356.8
|
|
Density (kg/m3)
|
3344
|
5514
|
Since the mass M is proportional to the density, and assuming
both bodies are spheres:
g µ
[rR]1/2
The ratio of the surface gravity g on the Earth and the Moon
is thus: (5514/3344)(6378/1737) = (1.649)(3.672) = 6.054
Which matches the ratio in the table above.
The variation in g on Earth is (6378.1/6356.8) = 1.0033.
This is a variation of 0.33%.
For a mass of 1000 kgs, the total energy required to escape
Earth is: (1000)(11,200)2/2 @
6.2x1010 J
According to this gpoles = 9.832 m/s2
and gequator = 9.780 m/s2 - which is a difference of
0.53% [6].
This is explained in detail in the quote below [6]:
“The variation in apparent gravitational acceleration (g) at
different locations on Earth is caused by two things. First, the Earth is not a
perfect sphere—it's slightly flattened at the poles and bulges
out near the equator, so points near the equator are farther from the
center of mass. The distance between the centers of mass of two objects affects
the gravitational force between them, so the force of gravity on an object is
smaller at the equator compared to the poles. This effect alone causes the
gravitational acceleration to be about 0.18% less at the equator than at the
poles.
Second, the rotation of the Earth causes an
apparent centrifugal force which points away from the axis of rotation, and
this force can reduce the apparent gravitational force (although it doesn't
actually affect the attraction between two masses). The centrifugal force
points directly opposite the gravitational force at the equator, and is zero at
the poles.
Together, the centrifugal effect and the center
of mass distance reduce g by about 0.53% at the equator
compared to the poles.
You can use the following
equation to calculate g at a certain latitude,
accounting for both of these effects:
g = g45 – [(gpoles –
gequator)/2] cos (2pl/180)
where l = latitude in degrees.
- gpoles
= 9.832 m/s2
- g45 =
9.806 m/ s2
- geq =
9.780 m/s2 "
That equation assumes you're at sea level, but if you
want to account for the effect of altitude when you go up in a plane you can
use this additional equation:
gh = g0 – [re/(re
+ h)]2
where re is the Earth's mean radius
(6,371.0088 km) and g0 is the standard gravitational
acceleration (9.80665 m/s2).”
The deviation from a spherical shape is greater for the
Earth (@0.33%)
than for the Moon (@0.11%),
but these can only partly explain the discrepancy between 24X and 22X.
The Lagrange L1 point:
The Lagrange L1 point for the Earth-Moon system is the point
where the Moon’s gravitational force is exactly equal and opposite to the
Earth’s gravitational force.
Lagrange L1 point of Earth-Moon system [7]:
The question could be asked: instead of taking a given mass
infinitely far from both the Earth and the Moon, how much energy is needed to
reach the L1 point from the Earth and the Moon and what is their ratio?
The energy needed to lift a mass from the Earth’s surface to
the L1 point is:
EEL1 = (GMEm/RE) - (GMEm/(RE + REL1))
= GMEm REL1/[ RE(RE + REL1)]
@ (GMEm/RE),
since REL1>>RE (323,050 kms vs
6,378 kms).
Similarly,
EML1 = (GMMm/RE) - (GMMm/(RM + RML1))
= GMMm RML1/[ RM(RM + RML1)]
@ (GMMm/RM),
since RML1>>RM (61,350 kms vs
1,737 kms).
So, to a good approximation, this (energy to L1) is the same as the escape
velocity.
Air Drag:
An obvious difference between the Earth and the Moon is that
only the former has an atmosphere, and the added air drag will make it more
difficult for a rocket to escape Earth – but how much more difficult?
This question has been discussed [8] in the context of
whether it would help to launch a rocket from (say) Tibet, to reduce the effect
of air drag in the thickest part of the atmosphere. The calculation has been
done by equating the change in the kinetic energy to the effect of drag force
on a rocket as it traverses the atmosphere:
D(MV2/2)
= (ACdV2/2) òr dh = (ACdV2/2)m
MVDVdr
= m Cd A V2/2
i.e. DVdr
= m Cd A V/(2M)
where it is assumed that the atmospheric density is the
strongest variable with height (exponential dependence), and the dependence of
drag coefficient Cd and velocity V is less significant. This
estimate has been done [8] for a Falcon rocket with a diameter of 3.66 metres
and a mass of 333.4 tonnes. The value of m
is estimated as 10 tonnes/m2, and the drag coefficient is taken as
0.5, and V is 300 m/s. This gives:
DVdr
= (10)(0.5)[(p)(1.88)2](300)/[(2)(333.4)]
@ 23.7 m/s
This value is negligible with respect to the escape velocity
(@0.21%).
This is a good back-of-the-envelope calculation, but it
neglects the fact that the velocity V changes with altitude (because of
acceleration and because the fuel is being burned, decreasing the total mass),
and thus so does the drag. The main reason why the effect of the drag is so low
is that , when the atmosphere is dense, the velocity is low; and by the time
the velocity is high, the atmosphere is thin. All these effects are best
incorporated by numerical integration. This would start with the equation of
motion:
a = (Fth – Fdr – mg)/m
where the drag force is given by the v2-law (in
terms of the drag coefficient as above), the density of the atmosphere decays
exponentially:
r =
1.2exp(- h/8000)
with a scale height of 8,000 metres. (This is a simplified
form, and not what any self-respecting meteorologist would want to be seen
associating with – but good enough for practical purposes).
In addition, the weakening of gravity with increasing
altitude h is given by the square-law dependence as above [6]. The dry mass of
the rocket is assumed to be 10% of the total initial mass. The parameters for
the rocket (for a nano-satellite) are taken from a paper by A.Uranga et al [9]:
900 kgs fuel and 100 kgs dry mass, cross-sectional area 0.3 m2 with
a thrust of 20 kilo-newtons. The fuel burn time is 90 secs and the rocket
reaches an altitude of hf @
1170 kms according to Uranga [9].
|
tbo (secs)
|
hbo (kms)
|
hf (kms)
|
drt
|
|
90
|
88
|
-
|
-
|
|
100
|
110
|
304
|
9x107
|
|
110
|
133
|
871
|
8.7x107
|
|
120
|
160
|
1407
|
8.3x107
|
|
130
|
190
|
1920
|
8x107
|
The first plot is acceleration a vs altitude h:
The acceleration keeps increasing with altitude because of
the constant thrust and weakening gravity, and then drops to close to zero
after all the fuel burns out.
The second plot is velocity v vs altitude h:
The velocity of the rocket keeps increasing under
acceleration, until burn out after which it keeps decreasing due to the action
of gravitational force.
The third plot is drag force Fdr vs altitude h:
As pointed out earlier, the drag force keeps increasing
because of the increasing speed of the rocket – until the point where the
exponential decay of the atmospheric density (not shown), takes over and the
drag force drops to zero by the time 100 kms is reached.
The fourth plot is energy needed to overcome drag Edr
vs altitude h:
The drag energy initially keeps increasing with altitude,
but saturates when atmospheric density goes to zero. This pattern is followed
not only by Uranga’s nanosatellite rocket, but also by the larger Super Strypi
and by the even larger Falcon9 (1st stage) rocket (that are
discussed in more detail later).
The saturated value of the drag energy for the Uranga rocket
is @108 J maximum.
This is a small fraction (0.16%) of the total energy required to escape:
(1/2)(1000)(11.2x103)2 = 6.27x1010
J = (6.674x10-11)(5.972x1024)(1000)/(6.378x106)
It does not suffice to explain the gap between Bezos’s
number of 24X and the escape velocity ratio of 22.1. Even the estimate for the
heavier Falcon rocket showed DVdr
as a small fraction (0.21%) of the escape velocity.
Other effects? Mostly magnetic …
Are there any other possible effects? Solar wind does exist
in the space between the Earth and the Moon, but its density is very low (<100 a="" against="" aligned="" always="" an="" and="" be="" but="" cc="" effect.="" even="" going="" having="" high="" insignificant="" is="" it="" its="" kms="" moon.="" not="" o:p="" protons="" sec="" spacecraft="" speed="" the="" though="" thus="" to="" towards="">
The Earth’s magnetic field (30-60 mT micro-Teslas) could also have
some effect. But the Lorentz force cannot do work on the rocket/satellite; it
can only deflect or rotate it. That could add to the total energy needed if
extra thrust is required to re-direct the rocket.
There are two Van Allen belts: the inner one from 1,000 to
6,000 kms above the surface of the Earth, and the outer one 13,000 to 60,000
kms above it [11]. And solar storm(s) could also contribute a magnetic field
but it is very small, up to 10 nT (nano-Tesla) [10]. Magnetic drag is known to
slow down satellites in low Earth orbit (LEO), but how much effect it would have
on rockets is not very clear – but it is discussed further below.
“Atmospheric drag on the station (the ISS, 400 kms orbit) will
be about 0.3 to 1.1 Nt (depending on the time of year).
Pressure due to solar photons is about 5 Nt . “[12]
According to Rawashdeh [13] there are 4 regions of differing
torques on the angular motion of satellites:
|
|
Altitude range (kms)
|
Environmental effects
|
|
Region I
|
<350 o:p="">
|
Aerodynamic torques dominate
|
|
Region II
|
350-650
|
Aerodynamic and gravitational torques are comparable
|
|
Region III
|
650-1,000
|
Aerodynamic, gravitational and solar torques are comparable
|
|
Region IV
|
>1,000
|
Solar and gravitational torques are comparable
|
The various torques have been plotted vs altitude by
Zagorski[14]:
According to Zagorski, the gravity torque varies as R-3
(R = the distance to the centre of the Earth), and a similar dependence is
observed for the magnetic torque (although the latter also depends on the
inclination of the orbit). The aerodynamic torque is proportional to
atmospheric density, and thus decreases exponentially with altitude, while the
solar torque is practically constant. It seems that the magnetic torque is
about two orders of magnitude smaller than the aerodynamic torque, and though
it does extend much further, it has an R-3 dependence, so its effect
will also decrease at high altitudes. Its magnitude also seems to be about 10%
of that of the gravity torque (and it drops below even the solar torque at ~5,000 kms altitude). One
could argue that the effect of magnetic drag is then 10% of the effect of the
gravitational force, with the same functional dependence on R [14]. This would
then account for the discrepancy between 22 and 24. But this assumes that
magnetic torque has the same dependence as magnetic force. Note also that the
aerodynamic drag is stronger than the magnetic drag until the altitude reaches
about 500 kms (since the former has an exponential dependence, and the latter
an R-3 dependence).
However, Beard & Johnson [15] have this to say:
“Satellite motion across the magnetic lines of the earth will induce a voltage
on the satellite of as much as 0.2 volt per meter of satellite size... The
magnetic drag resulting from the induced currents is proportional to the cube
of the satellite dimensions and may exceed the mass drag for satellites larger
than 50 meters in diameter; this can occur only above 1200‐km altitude, where
the charge density exceeds the neutral density. Thus the magnetically induced
current is an insignificant cause of drag.” This means that this effect is
weaker than the mass drag calculated above by a factor of about 6,500X (A =
0.3m2 above).
This would imply that magnetic drag is not even close to
aerodynamic drag (V2 term). However, this may refer to a satellite
and not to a rocket. Most discussions of rocket drag [13,16,17,18] do not refer
to magnetic drag, other than Zagorski [14]. Rawashdeh [13] mentions magnetic
coils in the satellite used to prevent angular motion or tumbling of the
satellite.
Kawashima et al [19] have calculated the effect of magnetic
drag on a satellite with a magnetic torquer as about 105 nano-newtons assuming
the magnetic moment of the torquer is 10 Amps-m2. Just to get an
idea of this effect, we can multiply this force by the Earth-Moon distance:
(1.05x10-7)(3.84x108) @
40J, which is negligible compared with the energy required to attain escape
velocity (6.27x1010 J). In comparison, the magnetic torque at the
Earth’s surface in Zagorski’s plot [14] above is 10-4 Nt-metre.
Since the Earth’s magnetic field at its surface is at least 30 micro-Tesla, the
magnetic moment that Zagorski has assumes is m = (10-4)/(3x10-5)
@ 3 Amps-m2. This is of the same order
of magnitude as that assumed above by Kawashima [19]. Certainly it is true that
Kawashima [19] is talking about a satellite, not a rocket – but, then, so are
Zagorski [14] and Beard & Johnson [15]. The calculation done by Beard &
Johnson (that I was unable to access) seems to hinge upon the definition of the
magnetic moment [20]:
m = (BrV)/m0
where V is the volume, m0
is the magnetic permeability of vacuum and Br is the residual flux
density (in Teslas) of the magnetic material. This explains why Beard &
Johnson say that the magnetic drag is proportional to “the cube of the
satellite dimensions”. Even if we assume the volume of a rocket is 100X that of
a satellite (the lowest payload fraction is listed as 1.4% [21]), it still will
not increase the magnetic drag energy to even a tiny fraction of the escape
energy.
Kawashima [19] points out that the magnetic drag force F is
proportional to the 2/3 power of the magnetic moment: F µ m2/3. A formula for
the magnetic moment is given by Sinha [22] for a thin-walled cylinder:
Ke = psr3Lt [1 –
(2t/L)tanh(L/2t)]
where the cylinder
has length L, radius r, thickness t and conductivity s.
And the magnetic moment is given by:
M = KewB
Assuming that the cylinder is rotating with an angular
velocity w in
the magnetic field B. The data from the super Strypi rocket [23] show that it
is 18 metres in length, 1.32 metres in diameter and rotates at 2.5 revolutions
per sec. Since L/t >>1, tanh(L/2t) = 1 and Ke = psr3Lt.
Other parameters of super Strypi rocket: launch mass 25
metric tons, thrust = 136,000 kgf ( 1 kgf = 9.80665 newtons, so: thrust =
1.33x106 Nt), burn time 73 secs and it launches a 275 kgs payload to
LEO.
Assuming the skin of the rocket is made of aluminium [24], s = 38x106
siemens/metre. The thickness t is taken as 5 mm [24]:
Ke = (3.14)(38x106)(0.66)3
(18)(0.005) = 3.09x106
M = KewB
= (3.09x106)(6.28)(2.5)(3x10-5) @ 1,455 Amp-m2
This magnetic moment is much larger than the range of values quoted
for a satellite 1-10 Amp-m2 [19] or about 3 Amp-m2,
inferred from Zagorski [14]. Assuming the 2/3 power dependence mentioned by
Kawashima [19], the magnetic drag on the super Strypi rocket should be about
(105)(145.5)2/3 @
2900 nN. If we repeat the calculation, multiplying this by even a large
distance does not result in a huge magnetic drag energy: (2.9x10-6)(3.84x108)
@ 1100 J, which is still
negligible compared with 6.27x1010 J.
Sinha [22] also mentions that magnetic torque can occur due
to hysteresis for a soft magnetic material. Since aluminium exhibits magnetic
effects only under strong magnetic fields [25], one can neglect this
contribution to magnetic torque.
Assume satellite dia is 1 m, then Zagorski’s figure [14]
shows, at zero altitude, magnetic torque t
= F(l/2) = F (0.5) = 1x10-4 so: F = 200 mN. Over a distance to the Moon,
the total energy is (2x10-4)(3.38x108) = 6.7x104
J. So it is negligible.
The above arguments still do not explain why Zagorski’s plot
shows that the magnetic torque is 10% of the gravitational torque. This can be
done by looking at the definition of gravitational torque used by Zagorski
[14]:
t = (F1
– F2) (l/2) sina,
Where F1 – F2 = GMm[(1/R)2
– (1/(R + l)2] @
GMm[2l/R3] = [GMm/R2](2l/R)
That is, the gravitational force is effectively reduced by a
factor (2l/R). So, the magnetic torque being 10% of gravitational torque
translates to a magnetic drag that is about 107 times smaller than
the gravitational potential energy. This is consistent with the above
calculations of magnetic drag based on Beard & Johnson [15], Kawashima [19]
and Sinha [22]. Thus magnetic drag flatters to deceive: it is unable to explain
the discrepancy between 24 and 22.09.
|
|
Uranga nano sat [9]
|
Super Strypi [23]
|
Falcon 9 (1st stage) [24]
|
|
Dia, Area
|
0.3 m2
|
1.32
|
3.7 m
|
|
Length (m)
|
-
|
18
|
68
|
|
Launch Mass(kgs)
|
1,000
|
25,000
|
549,000
|
|
Thrust
|
20,000 Nt
|
136,000*9.80 Nt
|
7.67X106 Nt
|
|
Burnout time (secs)
|
90
|
73
|
162
|
The parameters of the Super Strypi[23] and the Falcon 9 (1st
stage)[24] rocket are given in the above table. It is assumed that 10% of the
launch mass remains after all the fuel is burnt and that the aerodynamic drag
coefficient Cd is 0.25. The maximum height that these rockets reach,
the total drag energy and the total work done to reach hmax are
tabulated below:
|
|
Uranga nano sat [9]
|
Super Strypi [23]
|
Falcon 9 (1st stage) [24]
|
|
Hmax (kms)
|
720
|
18,323
|
1246.6
|
|
Edr (J)
|
1x108
|
1.44x109
|
1.9x109
|
|
Wh-max (J)
|
1.17x09
|
1.63x1011
|
6.85x1011
|
|
Wesc (J)
|
6.25x1010
|
1.56x1012
|
3.43x1013
|
For the heavier rockets, the total atmospheric drag energy
is more than 10X greater, but the total work done Wh-max is more
than 100X larger, so it is still not a significant contributor. 5.27x109
J is the energy required for the 1,000 kgs (Uranga) rocket to reach Bezos’s 24X
ratio to the Moon’s gravitational energy – and this is much larger than the
atmospheric energy drag Edr (by 50X).
Conclusion:
Since atmospheric drag contributes less than 1% to the total
energy required to escape Earth’s gravity, it cannot explain the discrepancy.
The effect of magnetic drag would seem to add on 10% to the
gravitational force, thus pushing the ratio from 22.1 to the ratio 24 mentioned
by Bezos – if one goes by Zagorski [14] – but this argument does not work
because of the definition of gravitational torque which is ~ (2l/R) of the
gravitational energy (as discussed above). When we calculate the magnetic drag
force based on Kawashima’s paper [19] for the super Strypi rocket (based on
magnetic moment formula of Sinha [22]), we end up with micro-newtons, which is
too small to make a difference. Even with a much bigger rocket, at most the
magnetic drag would increase by a factor of 10-100X. Thus, the discrepancy is
still unaccounted for.
It might well be argued that escape velocity is the wrong
parameter to consider. After all, it is just a number related to the energy
required to get a given mass infinitely far from an astronomical body (GMm/R).
It does not address the specifics of which rocket is needed, with how much
thrust and how many stages, to actually accomplish the task. For example, Wesc
in the table above is calculated by multiplying the launch mass by the square
of the escape velocity – but 90% of the launch mass is actually shed by burnout of the fuel (for a first stage rocket). But we cannot just reduce it by 10X, and call it
quits, because a single stage rocket will not do the job! And a similar 90% fuel consumption may occur for the 2nd stage. And are two stages enough?
Formulated like that
the problem is more realistic – and even further out of my pay grade. Will that
complete answer give 24X? Probably not, because it is unlikely to result in a
single number…
References:
9 9. A.Uranga et al. “Rocket performance analysis
using electrodynamic launch assist,”43rd AIAA Conference, Reno,
Nevada, 10-13th Jan.2005 doi: 10.2514/6.2005-1449.
1 13.
S.A.Rawashdeh “CubeSat aerodynamic stability at
ISS altitude and inclination” SSC12-VIII-6 26th Annual AIAA-USU
Conference on Small Satellites
1 14.
P.Zagorski “Modeling disturbances influencing an
Earth-orbiting satellite” Pomiary Automatyka Robotka (2012) 98-103
1 16.
J.Peraire and S.Widnall Lecture 14
MIT6_07F09_Lec14 Fall 2008
1 18.
Y.I.Khlopkov et al Int’l J. of Aeronautical
& Space Sci. 14(3)(2013) 215-221
DOI:10.5139/IJASS.2013.14.3.215
(2013)
1 19.
R.Kawashima et al “Particle simulation of plasma
drag force generation in the magnetic plasma de-orbit” arXiv 16May2018
1805.06123v1
2 22.
N.K.Sinha
nptel https://nptel.ac.in/courses/101106046/Lecture23.pdf
