Stairway to Heaven
Group 4
July 11, 2020
Problem 1: 1D Beanstalk
Figure: 1D beanstalk from earth
I imagine 1-dimensional beanstalk from earth and a crawlerstarting with some initial velocity v0
Problem 1: 1D Beanstalk
forces acting upon the crawler:
I gravitational force: gmR2
r2
I centrifugal force: mω2r
I hence:
r = ω2r − gR2
r2
r(0) = R
r(0) = v0
Problem 1: 1D Beanstalk
I approximate solution
r(t + ∆t) ≈ r(t) + r(t)∆t
r(t + ∆t) ≈ r(t) + r(t)∆t
Problem 1: 1D Beanstalk
Figure: initial speed 7 kms
Problem 1: 1D Beanstalk
I add air drag and friction between crawler and beanstalk
I air drag:
FD =1
2ρv2CDA
ρ =pM
RgT
I friction:FR = −µ · 2mωv
I e.g. µsteel = 0.1
Problem 1: 1D Beanstalk
Figure: initial speed 7 kms with µ = 0, 0.1, 1, 10, 1000, m = 100kg ,
A = 1m2.
Problem 1: 1D Beanstalk
Figure: escape velocities: 9.85 kms without air drag/friction, 10.03 km
s with
air drag/friction. Actual escape velocity of earth: 11.2 kms .
Problem 2: 2D Dynamics
Problem 2: 2D Dynamics
Shuttle initial velocity: vs = ωe · rs + veVis viva equation:
v =
√GM ·
(2
r− 1
a
)For the orbit:
as =rapo + rperi
2=
rapo + re2
bs =√rapo · re
Problem 2: 2D Dynamics
Figure: Parameters: rs = 6.4 · 107 m
Problem 2: 2D Dynamics
Dynamical simulation:
I radial coordinates (r(t), ϕ(t))
I velocities (vr (t), vϕ(t))
with respect to basis er =
(cosϕsinϕ
), eϕ =
(− sinϕcosϕ
)I gravitation v = −GMr−2er
r = vr ,
vr = −GMr−2 + r−1v2ϕ,
vϕ = −r−1vrvϕ.
Problem 2: 2D Dynamics
Dynamical simulation:
I radial coordinates (r(t), ϕ(t))
I velocities (vr (t), vϕ(t))
with respect to basis er =
(cosϕsinϕ
), eϕ =
(− sinϕcosϕ
)
I gravitation v = −GMr−2err = vr ,
vr = −GMr−2 + r−1v2ϕ,
vϕ = −r−1vrvϕ.
Problem 2: 2D Dynamics
Dynamical simulation:
I radial coordinates (r(t), ϕ(t))
I velocities (vr (t), vϕ(t))
with respect to basis er =
(cosϕsinϕ
), eϕ =
(− sinϕcosϕ
)I gravitation v = −GMr−2er
r = vr ,
vr = −GMr−2 + r−1v2ϕ,
vϕ = −r−1vrvϕ.
Problem 2: 2D Dynamics
Dynamical simulation:
I radial coordinates (r(t), ϕ(t))
I velocities (vr (t), vϕ(t))
with respect to basis er =
(cosϕsinϕ
), eϕ =
(− sinϕcosϕ
)I gravitation v = −GMr−2er
r = vr ,
vr = −GMr−2 + r−1v2ϕ,
vϕ = −r−1vrvϕ.
Problem 2: 2D Dynamics
Dynamical simulation:
I radial coordinates (r(t), ϕ(t))
I velocities (vr (t), vϕ(t))
with respect to basis er =
(cosϕsinϕ
), eϕ =
(− sinϕcosϕ
)I gravitation v = −GMr−2er
r = vr ,
vr = −GMr−2 + r−1v2ϕ,
vϕ = −r−1vrvϕ.
Problem 2: 2D Dynamics
Figure: Parameters: rs = 6.4 · 107 m
Problem 3: 3D Dynamics
Figure: Source: http://planetfacts.org/ecliptic/
Problem 3: 3D Dynamics
Figure: Parameters: rs = 6.4 · 107 m
Problem 3: 3D Dynamics
To reach mars, best case scenario, travelling time:
t =1
2
√4π2
GM· a3s ≈ 7.8 months
Beanstalk length:
rs =
√GM ·
(2re− 1
as
)− ve
ωe≈ 3.65 · 107 m
Problem 3: 3D Dynamics
Figure: Parameters: rs = 3.65 · 107 m
Problem A1: Longitudinal displacement of beanstalk
Elongation / compression caused by centrifugal force andgravitation.
d
dx(EA
du
dx) = −c1Ar + c2Ar
−2
where c1 = ρω2, c2 = ρgR2.
I A: area of cross-section
I r : distance to earth centre
I u: displacement
I dudx : strain
I EAdudx : tension (internal force)
Boundary conditions:{u(0) = 0 (hinged support)dudx (L) = 0 (free end)
Problem A1: Longitudinal displacement of beanstalk
Elongation / compression caused by centrifugal force andgravitation.
d
dx(EA
du
dx) = −c1Ar + c2Ar
−2
where c1 = ρω2, c2 = ρgR2.
I A: area of cross-section
I r : distance to earth centre
I u: displacement
I dudx : strain
I EAdudx : tension (internal force)
Boundary conditions:{u(0) = 0 (hinged support)dudx (L) = 0 (free end)
Problem A1: Longitudinal displacement of beanstalk
Elongation / compression caused by centrifugal force andgravitation.
d
dx(EA
du
dx) = −c1Ar + c2Ar
−2
where c1 = ρω2, c2 = ρgR2.
I A: area of cross-section
I r : distance to earth centre
I u: displacement
I dudx : strain
I EAdudx : tension (internal force)
Boundary conditions:{u(0) = 0 (hinged support)dudx (L) = 0 (free end)
Problem A1: Longitudinal displacement of beanstalk
Elongation / compression caused by centrifugal force andgravitation.
d
dx(EA
du
dx) = −c1Ar + c2Ar
−2
where c1 = ρω2, c2 = ρgR2.
I A: area of cross-section
I r : distance to earth centre
I u: displacement
I dudx : strain
I EAdudx : tension (internal force)
Boundary conditions:{u(0) = 0 (hinged support)dudx (L) = 0 (free end)
Problem A1: Longitudinal displacement of beanstalk
Elongation / compression caused by centrifugal force andgravitation.
d
dx(EA
du
dx) = −c1Ar + c2Ar
−2
where c1 = ρω2, c2 = ρgR2.
I A: area of cross-section
I r : distance to earth centre
I u: displacement
I dudx : strain
I EAdudx : tension (internal force)
Boundary conditions:{u(0) = 0 (hinged support)dudx (L) = 0 (free end)
Problem A1: Longitudinal displacement of beanstalk
Elongation / compression caused by centrifugal force andgravitation.
d
dx(EA
du
dx) = −c1Ar + c2Ar
−2
where c1 = ρω2, c2 = ρgR2.
I A: area of cross-section
I r : distance to earth centre
I u: displacement
I dudx : strain
I EAdudx : tension (internal force)
Boundary conditions:{u(0) = 0 (hinged support)dudx (L) = 0 (free end)
Problem A1: Longitudinal displacement of beanstalk
Elongation / compression caused by centrifugal force andgravitation.
d
dx(EA
du
dx) = −c1Ar + c2Ar
−2
where c1 = ρω2, c2 = ρgR2.
I A: area of cross-section
I r : distance to earth centre
I u: displacement
I dudx : strain
I EAdudx : tension (internal force)
Boundary conditions:{u(0) = 0 (hinged support)dudx (L) = 0 (free end)
Problem A1: Longitudinal displacement of beanstalk
Figure: Steel E = 2e+11,ρ = 8e+3, length 3.6e+7 vs. 8e+7
Problem A1: Longitudinal displacement of beanstalk
Figure: Graphene E = 1e+12,ρ = 2.3e+3, length 3.6e+7 vs. 8e+7
Problem A2: Will the Beanstalk break?
Constant cross-section
Figure: strain dudx (graphene, length 5e7)
Problem A2: Will the Beanstalk break?
Cross-section A(x) = c(L− x) for x > 2.85e7.(Note: Higher power may be necessary for other materials.)
Figure: strain dudx (graphene, length 5e7)
Advanced problem 3: What will happen to Earth?
Without beanstalk
Center of mass:(0 0 0
)TMoment of inertia:
Im =2
5·me · r2e ≈ 9.696 · 1037 kg.m2
With BeanstalkCenter of mass:
x =mb ·
(hb−re
2 + re)
mb + me≈ 9.005 · 10−6 m
Moment of inertia:
Ib =1
3· mb · h2b −
1
3· (mb −mb) · r2e ≈ 1.293 · 1026 kg.m2
I = Ie + Ib ≈ Ie