An advanced civilization lives on a solid spherical planet ofuniform density. Ru

Question

An advanced civilization lives on a solid spherical planet ofuniform density. Running out of room for their expandingpopulation, the civilization's government calls an engineering firmspecializing in planetary reconfiguration. Without adding anymaterial or angular momentum, the engineers reshape the planet intoa hollow shell whose thickness is one-fifth of its outerradius. Find the ratio of the new to the old A) Surface Area B) Length of Day An advanced civilization lives on a solid spherical planet ofuniform density. Running out of room for their expandingpopulation, the civilization's government calls an engineering firmspecializing in planetary reconfiguration. Without adding anymaterial or angular momentum, the engineers reshape the planet intoa hollow shell whose thickness is one-fifth of its outerradius. Find the ratio of the new to the old A) Surface Area B) Length of Day

Explanation / Answer

An advanced civilization lives on a solid spherical planet ofuniform density. Running out of
room for their expanding population, the civilization's governmentcalls an engineering firm
specializing in planetary reconfiguration. Without adding anymaterial or angular momentum,
the engineers reshape the planet into a hollow shell whosethickness is one-fifth of its outer radius. Find the ratio of the new to the old A) Surface Area B) Length of Day
answer:
A) let radius of the solid sphere be r and density be .
original surface area = 4r2
in the hollow planet, if the thickness of the shell is d,
outer radius = 5d
inner radius = 4d
volume = (4/3)(125 -64)d3 =(4/3)r3 >>> since volume is conserved
61 d3 = r3
d = .254 r
outer radius = 1.270 r
inner radius = 1.016 r
surface area = 4(1.27 r)2
ratio of new area to old = 1.272 = 1.61
so surface area increases by 61%.
B)
Let I be the MOI of the solid sphere and be the angularmomentum.
I = (2/5)Mr2
For the spherical shell of same mass,
MOI = I' = (2/5)M[(1.27r)5 -(1.016r)5]/[(1.27r)3 -(1.016r)3]
= (2.222)(2/5)Mr2
since angular momentum is conserved,
I = I''
new angular momentum = ' = /2.222
length of day = period of one revolution
= 2/'= 2.222 (2/)
which means length of day increase by a factor 2.222.

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