<blockquote>
<p>Would my weight change if the earth was a different (radially-symmetric) shape?</p>
</blockquote>
<blockquote>
<p>If you weigh 120 pounds standing on a sphere, and you distort the sphere into an ellipsoid but you're always standing on the tip of it, will your weight change, even when it's stretched millions of miles and you're on the tip, and even if it's flattened out into an inch-thick disk?</p>
</blockquote>
<p>Yes, it would. For example, earth currently isn't a perfect sphere. In addition to all these useless things like humans, buildings, etc, the Earth also bulges more at the equator and less at the poles. This actually does result in a (slightly) different weight depending on where on Earth you are. This effect also exists between sea level and the peak of a mountain (although on a lesser scale).</p>
<p>You're scenario takes this to the extreme. I don't know exactly how to calculate how much you would weigh (and you've also presented multiple measurements for this earth slab, and you haven't told me how long it is). However, <a href="https://www.wtamu.edu/%7Ecbaird/sq/2014/01/07/do-i-weigh-less-on-the-equator-than-at-the-north-pole/#:%7E:text=If%20we%20use%20a%20more,198%20pounds%20at%20the%20equator." rel="nofollow noreferrer">a difference of 21 kilometers decreases you're weight by 1%</a>.</p>
<p>Based on my rough estimates (in the small font below), to weigh 10% of your weight, you'll need the earth slab to have a radius (on the long end) of 13563 kilometers. That's slightly over twice Earth's current radius.</p>
<p>If you want any other value, then just change the equations I did, and replace the 230. Doubling it will give results for 1% of your weight, quadrupaling 0.1%, and so on. Halfing will give results for the square root of 10 times 10% (about 31.5%).</p>
<p>I'm curious what wins the title for most work used to lose weight, this idea or <a href="https://nautil.us/lose-weight-the-slow-and-incredibly-difficult-way-239703/" rel="nofollow noreferrer">removing mass from the Earth</a>.</p>
<p><sub>0.99 to the power of 230 is about 10%. This means that it would take 230 increases like the one caused by Earth's bulge to decrease your weight by 90%. Considering that Earth's bulge is about 1/3 of a percent an increase in radius, we need to perform 1.0033 to the power of 230 to the get how much bigger the radius of the new earth slab is. The answer comes out to about 2.1335. 2.1335x6357 (earth's normal radius) is about 13563 kilometers. </sub></p>

CC BY-SA 4.0
2024-11-13T03:17:16.380Z
2024-11-13T03:20:48.287Z
2024-11-13T03:20:48.287Z
Starship
<p>I'm on a huge sphere, so most of the mass of the Earth pulls from the left and right rather than straight down.</p>
<p>The horizontal vectors cancel due to symmetry. But the mass one mile down and 100 miles away doesn't pull me down very much.</p>
<p>I <em>might</em> ask: if the earth was a cylinder 3 feet wide and  I was standing on one end of it, what would my 120 pound weight be?</p>
<p>My guess is, 120 pounds, because physics answers usually end that way. Like, Newton discovered that if you're inside a hollow metal sphere floating in space, the pull of gravity from the wall of the sphere is always zero, no matter how close to the wall you are, and  no matter how thick and massive the metal is.</p>
<p>I suspect that my cylinder Earth situation is similar, because all of the cylinder's mass is directly under you, but most  of it is much farther away than the 8,000-mile diameter of the Earth.</p>
<p>HOWEVER, surely this depends on the shape, right? If the Earth was an inch-thin disk a zillion miles wide, I'm sure I'd weigh almost nothing. Right?</p>
<p><strong>SO MY QUESTION IS:</strong></p>
<p>If you weigh 120 pounds standing on a sphere, and you distort the sphere into an ellipsoid but you're always standing on the tip of it, will your weight change, even when it's stretched millions of miles, and even if it's flattened out into an inch-thick disk?</p>

CC BY-SA 4.0
2024-11-13T01:01:01.937Z
your weight at the earth's pole is more than at the equator, so yes, your weight will change.
James K
CC BY-SA 4.0
2024-11-13T01:02:20.260Z
According to https://physics.stackexchange.com/a/264311/123208 an infinite slab with the same mean density and surface gravity as the Earth is ~4300 km thick.
PM 2Ring
<gravity>
Would my weight change if the earth was a different (radially-symmetric) shape?
Miss Understands
# Would my weight change if the earth was a different (radially-symmetric) shape?

(Asked by: username_0 on 2024-11-12)

I'm on a huge sphere, so most of the mass of the Earth pulls from the left and right rather than straight down.

The horizontal vectors cancel due to symmetry. But the mass one mile down and 100 miles away doesn't pull me down very much.

I *might* ask: if the earth was a cylinder 3 feet wide and I was standing on one end of it, what would my 120 pound weight be?

My guess is, 120 pounds, because physics answers usually end that way. Like, Newton discovered that if you're inside a hollow metal sphere floating in space, the pull of gravity from the wall of the sphere is always zero, no matter how close to the wall you are, and no matter how thick and massive the metal is.

I suspect that my cylinder Earth situation is similar, because all of the cylinder's mass is directly under you, but most of it is much farther away than the 8,000-mile diameter of the Earth.

HOWEVER, surely this depends on the shape, right? If the Earth was an inch-thin disk a zillion miles wide, I'm sure I'd weigh almost nothing. Right?

**SO MY QUESTION IS:**

If you weigh 120 pounds standing on a sphere, and you distort the sphere into an ellipsoid but you're always standing on the tip of it, will your weight change, even when it's stretched millions of miles, and even if it's flattened out into an inch-thick disk?

**Tags:** gravity
> **Comments:**
> **username_1**: your weight at the earth's pole is more than at the equator, so yes, your weight will change.  
> **username_2**: According to https://physics.stackexchange.com/a/264311/123208 an infinite slab with the same mean density and surface gravity as the Earth is ~4300 km thick.  

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## 1 answers

### Answer by username_3 on 2024-11-13. Score: 2 | Accepted answer

> Would my weight change if the earth was a different (radially-symmetric) shape?

> If you weigh 120 pounds standing on a sphere, and you distort the sphere into an ellipsoid but you're always standing on the tip of it, will your weight change, even when it's stretched millions of miles and you're on the tip, and even if it's flattened out into an inch-thick disk?

Yes, it would. For example, earth currently isn't a perfect sphere. In addition to all these useless things like humans, buildings, etc, the Earth also bulges more at the equator and less at the poles. This actually does result in a (slightly) different weight depending on where on Earth you are. This effect also exists between sea level and the peak of a mountain (although on a lesser scale).

You're scenario takes this to the extreme. I don't know exactly how to calculate how much you would weigh (and you've also presented multiple measurements for this earth slab, and you haven't told me how long it is). However, [a difference of 21 kilometers decreases you're weight by 1%](https://www.wtamu.edu/%7Ecbaird/sq/2014/01/07/do-i-weigh-less-on-the-equator-than-at-the-north-pole/#:%7E:text=If%20we%20use%20a%20more,198%20pounds%20at%20the%20equator.).

Based on my rough estimates (in the small font below), to weigh 10% of your weight, you'll need the earth slab to have a radius (on the long end) of 13563 kilometers. That's slightly over twice Earth's current radius.

If you want any other value, then just change the equations I did, and replace the 230. Doubling it will give results for 1% of your weight, quadrupaling 0.1%, and so on. Halfing will give results for the square root of 10 times 10% (about 31.5%).

I'm curious what wins the title for most work used to lose weight, this idea or [removing mass from the Earth](https://nautil.us/lose-weight-the-slow-and-incredibly-difficult-way-239703/).

0.99 to the power of 230 is about 10%. This means that it would take 230 increases like the one caused by Earth's bulge to decrease your weight by 90%. Considering that Earth's bulge is about 1/3 of a percent an increase in radius, we need to perform 1.0033 to the power of 230 to the get how much bigger the radius of the new earth slab is. The answer comes out to about 2.1335. 2.1335x6357 (earth's normal radius) is about 13563 kilometers.


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