Question from a child that I was unable to answer

Question from a child that I was unable to answer

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RobM77

35,349 posts

233 months

Tuesday 29th November 2016
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V8LM said:
RobM77 said:
wsurfa said:
RobM77 said:
wsurfa said:
Hugo a Gogo said:
What is the force of gravity one metre from the centre of the earth then?
probably about 0.01% ish of that acting on you at the surface

assuming you're 100kg then c 0.1N 1km from the centre vs c 1000n at the surface

ish - very much ish
yes The earth's average radius is 6371km, so at 1 metre from the centre, your weight (i.e. the gravitational force pushing downwards due to your mass) would be 0.016% of what it is at the surface. This assumes a uniform density, which as we know is not the case. In actual fact, the earth gets denser as you go inwards, so the real figure would be slightly higher than this.
Using 12gcm3 I think its close to 0.03% (ish)
Thanks - I was about to work that out having looked up the density of the inner core! spin

ETA: I'd weigh just a smidge over 2kg biggrin
Aren't we a few orders of magnitude out here? (radius is 6 371 000 m).
Ah, yes, probably. I'm at work so not really giving my full attention. I may have used km as m...

ETA: This is why I gave it algebraically at first: going back to that, the force you'd feel relative to the force at the surface will just be r/R, as defined originally. It's just linear, with 0 force at the centre and at the surface, the usual 700N or whatever you weigh.

Edited by RobM77 on Tuesday 29th November 15:31

RobM77

35,349 posts

233 months

Tuesday 29th November 2016
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Abbott said:
Hmm interesting. If you were a free body at the centre of the earth then the gravitaional effecrt would be zero but would you not be subject to the gravitational pull of the moon/sun which would drag you away from the centre. If so by how much.
(Disclaimer: I could be talking B*****ks)
Yes, the earth's forces would sum to zero, but you'd still be subject to gravity from other bodies, the most notable being the gravity from the moon and the sun. These lunar and solar forces would pull you away from the centre of the earth slightly until the force was balanced by the earth's gravity trying to pull you back again. You'd also have an effect from the earth's rotation of course. This would of course change constantly as the moon orbits the earth and the earth turns. As for how much all this sums to, I'm at work so don't have time to calculate it, sorry!

CrutyRammers

13,735 posts

197 months

Tuesday 29th November 2016
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MiggyA said:
CrutyRammers said:
The closest end would have more force acting on it than the furthest. But unless the difference in force was enough to pull it apart, the pin would be accelerated as a single object and hence all effectively weigh the same at both ends....I thunk.
Your first sentence is correct but the second one isn't really. Maybe put it this way: think about a pile of bricks rather than a single object. Pile them up high and hopefully it's clear that the ones at the top weigh less than the ones at the bottom, even though they're all pushing down through each other? Would adding mortar to bind them together change that? Of course not. Same situation with the atoms making up this theoretical pin. It doesn't have to be pulled apart for the thing to weigh less when stood on end.
I'm not convinced by that, because the top one doesn't "push down", it's pulled. The bottom one is pulled too, with greater force. But if they are all solidly connected, then the top one must have the same force acting on it as the bottom one. If they are not connected then the column stretches apart as per Eric's black hole example.
The argument about centre of gravity makes more sense. In which case, it depends if you stand the pin up, or simply rotate it around the CoG.

V8LM

5,166 posts

208 months

Tuesday 29th November 2016
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Derek Smith said:
will an elongated object, such as a 1,000 mile pin, weigh less if stood on end rather than placed horizontally on the ground.
Yes, as explained above, because the integral of all the forces from every point in the pin to every point on the earth is the same as the force as if all the mass of the earth and pin were at their respective CoG (Newton's Universal law) (approximate centre) and so the pin on its end has its CoG further from the Earth's CoG when stood on its end than when laid flat, and so the gravitational attraction less, so weigh less.

Edited by V8LM on Tuesday 29th November 23:40

Monty Python

4,812 posts

196 months

Wednesday 30th November 2016
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By weight we mean the force of attraction between the earth and the object. Therefore, they would not weigh the same because the vertical bar has most of its mass farther from the center of the earth than the horizontal rod does. The weight of a point mass on the surface of the earth is W1=mMG/R2 where M is the mass of the earth, R is the radius of the earth, G is the universal gravitational constant, and m is the mass of the point mass. This is usually written as W1=mg. But, if the point mass is a distance H above the surface, its weight is smaller, W2=mMG/(R+H)2=mg/(1+(H/R))2. If you know integral calculus it is not hard to show that the weight of a vertical uniform bar of length L and mass m is W3=mg/(1+(L/R)). In your example L=1000 mi is not small compared to R≈4000 mi, so W3=mg/(1.25)=0.8W1. But, because L is so large, your horizontal bar does not have a weight of mg either unless it bends to conform with the curved surface of the earth. But, you can see from the figure that the distances from the center of the earth are much smaller than for the vertical bar, so it will surely have a larger weight, just not quite as big as mg.

andy_s

19,397 posts

258 months

Wednesday 30th November 2016
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Hugo a Gogo said:
What is the force of gravity one metre from the centre of the earth then?
About the same as from a 2m diameter asteroid.