Determine the compression of the spring

[Westin]

Homework Statement

Russian aviator Vsevolod Mikhailovich Abramovich invented the Abramovich Flyer based on the design of the Wright brothers' first plane. After this first success, Abramovich became obsessed with deep space travel designing a spring based launcher to fire a probe of mass 90kg from Earth (mass 6.00×10^24kg, radius 6.40×10^6m) into deep space.

Determine the minimum speed to launch this probe into deep space such that it never returns.

vesc= 11183.1346 m/s

Determine the compression of the spring, having spring constant 5.50×105N[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char3D.pngm, needed to launch this probe using Abramovich's design.

s=___________________

Homework Equations

vesc=sqrt(2GM/R)

F⃗ spring=−kŝ

Us=∫(dUs/ds)ds=∫ksds=1/2ks^2−Es[/B]

The Attempt at a Solution

First part I just plugged it into the V escape equation.

Second part attempt: 5.50×10^5N[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char3D.pngm * 6.40×10^6m = 3.52E12 Newtons

3.52E12 =

U = mc^2
U = 6E24*(3E8)^2 = 5.41E41

Integrate (1/2)(5.50×10^5N[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char3D.pngm )^2 - 5.41E41

Maybe use Youngs Module to find compression?

I think I'm somewhere on the right track but I'm kinda lost..[/B]

 

[gneill]

Have you considered a straight forward conservation of energy approach? What's the KE required for the probe to escape?

 

[Westin]

KE = (1/2)mv^2
KE = (1/2) (90kg) (11183.1346)^2

Would I use that answer to then multiply it by the given stiffness?

 

[gneill]

KE = (1/2)mv^2
KE = (1/2) (90kg) (11183.1346)^2

Would I use that answer to then multiply it by the given stiffness?

Not quite. You'd want to make sue that your spring, when compressed, is storing at least that much energy so that when it relaxes it imparts that much energy to the probe. What's the expression for the PE stored in a compressed spring?

You might also want to convince yourself that the gravitational PE change for the probe through the spring's relaxation distance is not a significant contributor to the calculation.

 

[Westin]

Fs=−ks=−dUs/ds
dUs/ds=ks

Integrating this equation once would give us the Potential Energy that was obtained.

My second attempt:
Set KE=(1/2)mv^2 and -W=(1/2)kx^2 equal to each other
kx^2=mv^2

Solve for x
x=sqrt((mv^2)/k)
x=sqrt((90kg*11183.1346^2)/5.5e5)

x=143.055m

 

[gneill]

That result looks good.