What would your weight be in a hole 1 000 miles deep

[Alan McDougall]

If we could dig a hole a 1 000 miles deep, what effect would it have on our weight,if we compared our weights from that measured on bathroom type scale. (bottom and surface)?

 

I was a mining Engineer so this led me to propose this question

 

(please overlook the ramifications of heat lava etc)

 

Alan

[timo]

The weight decreases linearly until it reaches zero at the center; so it would be W*(R-1000 mi)/R, with W being the original weight and R being the radius of earth.

[swansont]

The weight decreases linearly until it reaches zero at the center; so it would be W*(R-1000 mi)/R, with W being the original weight and R being the radius of earth.

 

Assuming constant density, so this is approximately correct but there will be some deviations from the linear decrease.

[Daecon]

So what is the radius of the Earth? About 4,000 miles?

 

So that would be Weight*3000/4000, or 3/4 of it's current weight at the surface (approximately)?

[D H]

The weight decreases linearly until it reaches zero at the center; so it would be W*(R-1000 mi)/R, with W being the original weight and R being the radius of earth.

That assumes the Earth has a uniform density: A nice, simple freshman physics problem. Unfortunately, that is all this assumption is good for. It does not reflect reality. Inside the real world, Earth's gravitational force initially increases with depth. This increase continues down to the mantle/outer core boundary; only then does it start dropping.

 

Think of it this way: The Earth's core represents about 9.4% of the total volume of the Earth but represents about 32% of the total mass of the Earth.

[Alan McDougall]

That assumes the Earth has a uniform density: A nice, simple freshman physics problem. Unfortunately, that is all this assumption is good for. It does not reflect reality. Inside the real world, Earth's gravitational force initially increases with depth. This increase continues down to the mantle/outer core boundary; only then does it start dropping.

 

Think of it this way: The Earth's core represents about 9.4% of the total volume of the Earth but represents about 32% of the total mass of the Earth.

 

Einsteins Theories did not reflect reality until they were proved correct by the eclipse of the sun in 1919? where the astronomers observes planet Mercury in front of the sun instead of where it should have been nicely hidden behind the sun. Gosh space which was considered an empty void could bend , what a silly thought is that

 

The assumption was a solid earth, but what about the huge mass at the two sides of the hole and the increasing mass above ones head??

 

Would it cancel out nicely as some say in response?

[insane_alien]

Alan, only if it was a uniform sphere, the earth, i'm sure you will agree is not a uniform sphere.

 

the gravity initially increases because the effect of being closer to most of the mass over powers the effect of the material above you. this is simply because the mass is concentrated towards the center.

 

eventually however, the mass above you will counteract this and your weight will start to decrease as you approach the center.

[timo]

That assumes the Earth has a uniform density: A nice, simple freshman physics problem. Unfortunately, that is all this assumption is good for. It does not reflect reality.

But you have to admit that it is the best quantitative answer to the question so far; even though that is because it is the only one.

 

Inside the real world, Earth's gravitational force initially increases with depth. This increase continues down to the mantle/outer core boundary; only then does it start dropping.

Nice. I wasn't aware of that. So how about the rest? It is a reasonable approximation to split earth in two pieces, core and mantle, and assume sphericallity and uniformity for both of them? That would still lead to a relatively simple answer to the question (sum of two forces).