This is probably a question more suited to the maths SE, but here is an answer anyway.

Think about a sphere, radius $R$, with a circle drawn on it. Here is a very scrappy picture:

So, what we want to do is work out the relation between the circumference of the circle and its 'radius', where by 'radius' we mean the arc drawn on the surface of the sphere.

Well, OK, the circle is traced out by a line from the centre of the sphere to the surface making an angle, $\theta$, with the line to the centre of the circle (see the diagram). We can do some elementary trigonometry to find two things:

• the real radius of the circle is $R\sin\theta$ and the circumference of it is therefore $c = 2\pi R \sin\theta$;
• the length of the arc drawn on the sphere is $R\theta$: call this $r$.

So $r = R\theta$, or in other words $\theta = r/R$. So the circumference of the circle is given in terms of $r$ by

$$c = 2\pi R \sin\left(\frac{r}{R}\right)$$

And this is valid for $\theta\in [0, \pi]$ or $r \in [0, \pi R]$.

Well, $\sin x \le x$, so

$$\begin{align} c &= 2 \pi R \sin\left(\frac{r}{R}\right)\\ &\le 2 \pi r \end{align}$$

In other words the circumference of the circle is always less than it would be in flat space.

It's useful to look at a couple of particular cases.

First of all, if $r$ is small, then $\sin\left(r/R\right) \approx r/R$ and so

$$\begin{align} c &\approx 2 \pi R \times r/R\\ &= 2 \pi r \end{align}$$

In other words when the circle is small its circumference is approximately what it would be in flat space.

If $\theta = \pi/2$, on the equator of the sphere, then $r = R\pi/2$ and

$$\begin{align} c &= 2 \pi R \sin\left(\frac{R \pi}{2 R}\right)\\ &= 2 \pi R \end{align}$$

While what we would get in flat space is $2 \pi \times R\pi/2$. So the circumference is 'too small' by a factor of $\pi/2$.

Finally if $\theta = \pi$, then $r = \pi R$ and

$$\begin{align} c &= 2 \pi R \sin\left(\frac{\pi R}{R}\right)\\ &= 2 \pi R \sin \pi\\ &= 0 \end{align}$$

This is because we are now at the south pole.

This is probably a question more suited to the maths SE, but here is an answer anyway.

Think about a sphere, radius $R$, with a circle drawn on it. Here is a very scrappy picture:

So, what we want to do is work out the relation between the circumference of the circle and its 'radius', where by 'radius' we mean the arc drawn on the surface of the sphere.

Well, OK, the circle is traced out by a line from the centre of the sphere to the surface making an angle, $\theta$, with the line to the centre of the circle (see the diagram). We can do some elementary trigonometry to find two things:

• the real radius of the circle is $R\sin\theta$ and the circumference of it is therefore $c = 2\pi R \sin\theta$;
• the length of the arc drawn on the sphere is $R\theta$: call this $r$.

So $r = R\theta$, or in other words $\theta = r/R$. So the circumference of the circle is given in terms of $r$ by

$$c = 2\pi R \sin\left(\frac{r}{R}\right)$$

And this is valid for $\theta\in [0, \pi]$ or $r \in [0, \pi R]$.

Well, $\sin x \le x$, so

$$\begin{align} c &= 2 \pi R \sin\left(\frac{r}{R}\right)\\ &\le 2 \pi r \end{align}$$

In other words the circumference of the circle is always less than it would be in flat space.

It's useful to look at a couple of particular cases.

First of all, if $r$ is small, then $\sin\left(r/R\right) \approx r/R$ and so

$$\begin{align} c &\approx 2 \pi R \times r/R\\ &= 2 \pi r \end{align}$$

In other words when the circle is small its circumference is approximately what it would be in flat space.

If $\theta = \pi/2$, on the equator of the sphere, then $r = R\pi/2$ and

$$\begin{align} c &= 2 \pi R \sin\left(\frac{R \pi}{2 R}\right)\\ &= 2 \pi R \end{align}$$

While what we would get in flat space is $2 \pi \times R\pi/2$. So the circumference is 'too small' by a factor of $\pi/2$.

Finally if $\theta = \pi$, then $r = \pi R$ and

$$\begin{align} c &= 2 \pi R \sin\left(\frac{\pi R}{R}\right)\\ &= 2 \pi R \sin \pi\\ &= 0 \end{align}$$

This is because we are now at the south pole.