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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