How far do we need to go from earth to view it as a complete sphere?

Question

How far do we need to go from earth so that our view makes us believe that the earth is just a sphere without any terrains and depressions? It must look like a completely uniform spherical land mass.

Answer 1

@Bernhard has a point: because the Earth is an oblate spheroid with a relative error of ~0.5%, rather than a true sphere, you will never see it as a "uniform sphere" if your tolerance for relative error is smaller than that. But I see you as talking about terrain, rather than the overall shape of the Earth, so for my comment I will assume that the Earth is a sphere.

Suppose you are at a height $h$ from the surface of a spherical planet of radius $R$ that also has features (geological or buildings) of size $\delta R$. The interesting question as I see it is, how does the relative angle subtended by those features (relative to the angle of the planet overall) vary as a function of $h$?

Draw the Earth as a circle, and your vantage as a point $h$ above the surface, and draw lines which are tangent to the circle that intersect your point: because they're tangent to the circle, they'll form a right triangle with hypotenuse $R + h$ and "opposite" side $R$, hence half the angle subtended by the Earth is given by the arcsine$$\theta = \sin^{-1}\frac R{R+h}.$$Measured in radians, the angle $\delta\theta$ subtended by the features $\delta R$ is given by an arctangent:$$\delta\theta = \tan^{-1} \frac{\delta R}{\sqrt{(R + h)^2 - R^2}},$$where we get the denominator, the "adjacent" side of the triangle, from applying the Pythagorean theorem to our previous triangle.

Then it becomes useful to write $\eta = h / R$ and $\epsilon = \delta R/R$, to write these as$$\theta = \sin^{-1} \frac 1{1 + \eta};~~~~\delta\theta = \tan^{-1} \frac{\epsilon}{\sqrt{\eta^2 + 2 \eta}}.$$Now taking $\delta R$ as the 8.85km height of Everest above sea level and Earth's average radius of 6,370 km we can say $\epsilon \approx 0.0014$.

This is actually simple enough that you can program it in mostly any programming language you want; I threw something together in JS:

function bisectDecreasing(bounds, target, fn) { 
    var lo = bounds[0], hi = bounds[1], mid = (lo + hi) / 2; 
    return fn(mid) < target ? [lo, mid] : [mid, hi]; }
}
function find(target) {
    var b = [0, 1e6]; 
    function f(h) {
        return Math.atan(0.0014 / Math.sqrt(h*h + 2*h))/Math.asin(1/(1 + h));
    }
    while ((b[1] - b[0]) > 1e-10) { 
        b = bisectDecreasing(b, target, f);
    }
    return (b[0] + b[1]) / 2;
}

Running find(0.01) on this tells me $\eta \approx 0.004486\dots$; multiplying this through by $R$ suggests that after only 29km height, even Everest-scale features become less than 1% of the relative size in the Earth as you see it. Then find(0.005) will give you the accuracy at which the Earth is seen to be a spheroid, rather than a perfect sphere; my program gives 0.021 for this, or 132 km.

So from low-earth orbit, terrain will already be beneath 0.5% of the angular variation in what you see. You of course can't get this less than $\epsilon = 0.14%$ (which diverges $h \rightarrow \infty$); but to get it down to 0.2% still only requires $h \approx 0.26 ~R$, which is not a very serious orbit.

Answer 2

Never.

The radius of the earth varies from 6353 to 6384 kilometers. Neglecting terrain will not lead to the earth looking like a sphere.

This is obviously unrelated to human factors, such a sight, but that would make it more of a biology than a physics question.

Answer 3

Practically speaking, on the order of several hundred kilometers, probably in the range of 1000.

Astronauts in low earth orbit have not reported seeing the horizon as anything other than smooth. This is because, at low angles of incidence, the atmosphere blurs the outlines of objects such as mountains. However, this image (sorry for the link rather than the image itself, but it's a very large one) https://losdragonesdeleden.files.wordpress.com/2013/11/himalaya-iss035-e-21917.jpg shows the Himalayas from low orbit. If you examine the horizon at the upper left corner, you can actually see slight variations.