<p>$\pi$ is just a number. It doesn't vary with <em>anything</em>.</p>

<p>The ratio of the circumference to the diameter of a circle varies with geometry, so in the general case, this definition doesn't hold. But there are other definitions, such as
$$
\pi = \int_{-1}^{1}\frac{dx}{\sqrt{1-x^2}},
$$
or
$$
\mathrm{The\,smallest\,positive\,number} \, x \, \mathrm{for\,which}\, \arccos(2x) = 0.
$$
The latter definition can also be expressed a power series, which is independent of geometry.</p>

CC BY-SA 3.0
2016-03-18T14:29:20.857Z
2016-03-18T14:29:20.857Z
pela
<p>Will Pi have the same value on the surface of say Jupiter, compared to the value of PI on Earth. Or does Pi vary with gravity?</p>

CC BY-SA 3.0
2016-03-18T17:44:25.073Z
I'm voting to close this question as off-topic because it concerns a mathematical constant.
James K
<gravity>
Does Pi change when gravity changes
Jakob
# Does Pi change when gravity changes

(Asked by: username_0 on 2016-03-18)

Will Pi have the same value on the surface of say Jupiter, compared to the value of PI on Earth. Or does Pi vary with gravity?

**Tags:** gravity

> **Comments:**
> **username_1**: I'm voting to close this question as off-topic because it concerns a mathematical constant.  

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## 1 answers

### Answer by username_2 on 2016-03-18. Score: 3 | Accepted answer

$\pi$ is just a number. It doesn't vary with *anything*.

The ratio of the circumference to the diameter of a circle varies with geometry, so in the general case, this definition doesn't hold. But there are other definitions, such as
$$
\pi = \int\_{-1}^{1}\frac{dx}{\sqrt{1-x^2}},
$$
or
$$
\mathrm{The\,smallest\,positive\,number} \, x \, \mathrm{for\,which}\, \arccos(2x) = 0.
$$
The latter definition can also be expressed a power series, which is independent of geometry.


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