Effect of 2% radius decrease on Earth's surface gravity and rotational kinetic energy

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baldbrain
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Homework Statement


Let g be the acceleration due to gravity at the Earth's surface and K be the rotational kinetic energy of the Earth. Suppose the Earth's radius decreases by 2%. Keeping all other quantities constant,
(a) g increases by 2% and K increases by 2%
(b) g increases by 4% and K increases by 4%
(c) g decreases by 4% and K decreases by 2%
(d) g decreases by 2% and K decreases by 4%

Homework Equations


g=GM/R2 , where M & R is the mass and the radius of the Earth respectively
K=(1/2)Iω2 , where I is the moment of inertia of the Earth about its axis of rotation and ω is it's angular velocity about the same axis

The Attempt at a Solution


(dR/R)100 = 2% ...(decrease)
Since all other quantities are constant,
i) g=GM/R2 ⇔ g ∝ R-2
⇒ dg/g = 2(dR/R)
⇒ (dg/g)100 = 2((dR/R)100)
= 2(2) = 4% ....(increase)
Since there is inverse proportionality, g increases by 4%

ii) K=(1/2)Iω2
Now, assuming the Earth to be a homogeneous sphere of uniform mass density, its moment of inertia about the diameter is
I=(2/5)MR2
Therefore K= (1/2)(2/5)MR2ω2 = (1/5)MR2ω2
Keeping all other quantities constant,
K ∝ R2
⇒dK/K = 2(dR/R)
⇒(dK/K)100 = 2((dR/R)100)
= 2(2) = 4% .... (decrease)
Since there is direct proportionality, K decreases by 4%Hence, g increases by 4% and K decreases by 4%
So, I think option (b) should've been - g increases by 4% and K decreases by 4%, instead of K increases by 4%
Thoughts?
 
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Discussion

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Are you claiming that ω is unaffected by the change in radius? What conservation law is involved?
 
gneill said:
Are you claiming that ω is unaffected by the change in radius? What conservation law is involved?
I understand ω will get affected. But they've said all other quantities are kept constant. By the way, if u substitute ω2=v2/R2, then the expression becomes independent of R altogether, which would then be meaningless.
 
I think that the phrase "Keeping all other quantities constant" should not be applied too broadly, otherwise one could argue that nothing should change despite the change in radius. In particular I believe that you must take into account how ω will change and thus affect the KE.
 
gneill said:
I think that the phrase "Keeping all other quantities constant" should not be applied too broadly, otherwise one could argue that nothing should change despite the change in radius. In particular I believe that you must take into account how ω will change and thus affect the KE.
Yeah, but as you can see, the expression is becoming independent of R if I substitute ω2 = v2/R2
 
HPPAS said:
Yeah, but as you can see, the expression is becoming independent of R if I substitute ω2 = v2/R2
I don't see how that substitution gets you anywhere. v is not a constant: it will change with both ω and R.

Take a look at conservation of angular momentum to see how ω changes with R.
 
gneill said:
I don't see how that substitution gets you anywhere. v is not a constant: it will change with both ω and R.

Take a look at conservation of angular momentum to see how ω changes with R.
Well, even my professor said that. But he too agrees that it should be 4% decrease.
Anyways, I'm getting the same answer even after considering angular momentum (we have to assume either of ω or v to be constant, or else it isn't possible)
 
Find the ratio of the new ω to the original ω in terms of the new radius and original radius. Use conservation of angular momentum. So,

##\frac{ω}{ω_o} = ?##
 
gneill said:
Find the ratio of the new ω to the original ω in terms of the new radius and original radius. Use conservation of angular momentum. So,

##\frac{ω}{ω_o} = ?##
(Ro/R)2
 
gneill said:
Keeping all other quantities constant" should not be applied too broadly
It's worse than that. The question is unanswerable.
If the radius changes then either the mass changes or the density changes. Which one are we to take as constant?
Same question with spin rate versus angular momentum.
 
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HPPAS said:
(Ro/R)2
Okay. Now apply that information to working out the ratio of the new KE to the original KE.
 
haruspex said:
If the radius changes then either the mass changes or the density changes. Which one are we to take as constant?
Same question with spin rate versus angular momentum.
No, mass is constant. Distribution of mass changes and hence Moment of Inertia (MI) changes.
I'm also thinking whether our assumption of the Earth as 'being a homogeneous sphere of uniform mass density' is valid. If not, I don't know anything about the MI of a geoid.
 
gneill said:
Okay. Now apply that information to working out the ratio of the new KE to the original KE.
I'm still getting the same answer
4% decrease
I think that sufficiently cross verifies that our answer is right, now that we've seen 3 different methods??
 
HPPAS said:
I'm still getting the same answer
4% decrease
I think that sufficiently cross verifies that our answer is right, now that we've seen 3 different methods??
Can you show your work?
 
HPPAS said:
No, mass is constant.
How do you know? Is the wording in post #1 not exactly what you were given?
But to get one of the given answers, you do have to suppose mass and angular momentum are constant.
 
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I realized the hitch. Our approach wasn't wrong, it just wasn't the same as theirs.
If we substitute
K= (1/2)Iω2=(1/2)(Iω)*ω
K= (1/2)Lω=(1/2)L2/I
... since L=Iω & again ω=L/I
Then, we get K∝R-2
Which yields us their answer of 4% Increase