[Caveat for this answer: it (both parts) is almost literally a transcript of a back-of-the-envelope calculation: there may be mistakes.]
The calculation for a distant camera not co-rotating with the Earth
A 50mm lens on 35mm film has about a 40 degree angle of view. Let's assume we're pointing that lens at the Earth so the Earth fills this angle of view, we are looking down at the equator, and the camera is not co-rotating with the Earth. Rather than do any complicated sums we'll assume that the end points on a line drawn through the centre of the planet and ending at the surface subtend 40 degrees to the camera. If we assume the radius of the earth is $R$ and the angle of view of the lens is $2\theta$, this gives us
$$B = \frac{R}{\tan\theta}$$
where $B$ is the distance from the camera to the centre of the earth. From this we get
$$b = R\left(\frac{1}{\tan\theta} - 1\right)$$
where $b$ is the distance from the point on the surface of the Earth directly under the camera to the camera.
Now we want to calculate the angular velocity of this point with respect to the camera, $\omega_C$, in terms of $\omega_E$, the angular velocity of the Earth. Well, we can do this by equating the distance it moves in terms of $\omega_C$ to that it moves in terms of $\omega_E$ in some short time $\delta t$:
$$\omega_C \delta t b = \omega_E \delta t R$$
or
$$\omega_C = \frac{\omega_E R}{b}$$
or
$$\omega_C = \frac{\omega_E}{\frac{1}{\tan\theta} - 1}$$
So, we know $\omega_E$ and $\theta$, and so we know $\omega_C$.
The next thing we want to know is the angular size of a pixel for the camera. If there are $N$ pixels across the field of view, then at the centre of the field of view a pixel subtends an angle of about $(2\tan\theta)/N$ (I might have this wrong).
So, now, finally, the time for a point on the surface of the Earth directly under the camera to move across one pixel is
$$\frac{\left(\frac{2\tan\theta}{N}\right)}{\left(\frac{\omega_E}{\frac{1}{\tan\theta} - 1}\right)} = \frac{2-2\tan\theta}{N\omega_E}$$
So, OK, plug in $\theta=\pi/9$, $N=5000$ and $\omega_E=2\pi/(3600\times 24)$, and we get about 3.5 seconds (note I previously had both the expression here wrong (I had $\omega_E = 2\pi/3600$) and also the result was hopelessly wrong for some reason).
So, in other words, it takes a point on the equator of the Earth about 3.5 seconds to move a single pixel across the image for a camera with a 25M pixel sensor and with with a normal lens, taking a picture such that the Earth fills the entire picture, if the camera is not co-rotating with the Earth. A typical exposure might be a couple of milliseconds.
This is why the Earth does not seem to be blurred when viewed like this.
It's worth noting, as pointed out by Jibb Smart in a comment, that the radius of the earth vanishes above: the parameters which control the motion blur are $\omega_E$, the angular velocity of the Earth, $\theta$, half the angle of view of the camera and $N$, the number of pixels, or equivalently, the resolution of the image if that is dominated by some other factor such as the lens. So this result applies to a photograph of any spherical, rotating object (it would need to be corrected for very wide angles of view as my assumption that you can see the ends of a line through the planet becomes seriously wrong in that case: fixing this is just a matter of doing slightly more correct trigonometry though, I was just lazy).
The calculation for low Earth orbit
Errol Hunt pointed out in a comment that a more plausible case is to consider a camera on a satellite in LEO, so let's do that.
We know that satellites in LEO orbit the Earth in about 90 minutes. This means that we can just ignore the Earth's rotation to a good first approximation, so we'll do that.
For a light object in a circular orbit about a point mass at a distance $r$, the speed of the object is given by
$$v = \sqrt{\frac{G M}{r}}$$
The Earth is well-approximated by a point mass because of Newton's shell theorem, so for a satellite a distance $h$ above the Earth we have
$$v = \sqrt{\frac{G M}{R + h}}$$
Where $G$ is Newton's gravitational constant, $M$ is the mass of the Earth, & $R$ is its radius.
If the satellite is looking down at the Earth directly below it, then in time $\delta t$ it sees the surface move by $v\delta t$. Assuming that $\delta t$ is sufficiently small, then the image will move by an angle
$$\delta\theta \approx \frac{v\delta t}{h}$$
So again, we want to know how long $\delta t$ can be for this to be the same as a pixel at the centre of the image. From above this means that
$$\frac{2 \tan\theta}{N} = \frac{v\delta t}{h}$$
(where now $\theta$ is half the angle of view again, sorry), and so
$$\delta t = \frac{2 h\tan\theta}{Nv}$$
Or, plugging in $v$ in terms of $h$:
$$\delta t = \frac{2h\sqrt{h + R}\tan\theta}{N\sqrt{GM}}$$
And, once more, we can plug in $\theta = \pi/9$, $N=5000$ and, say $h=200\,\mathrm{km}$ (this is a very low orbit: things only get better as we go up) as well as standard values for $G$, $M$ & $R$ and we get $\delta t \approx 4\times 10^{-3}\,\mathrm{s}$: about $1/250\,\mathrm{s}$ in other words. This is a completely reasonable exposure time for any reasonably modern sensor (or film!) looking down at the Earth.
Again, this is why the Earth is not blurred when we take pictures of it from space.
