# What is the percentage of Earth's hemisphere seen from different orbital heights or distances?

We often see photos of Earth from space, but it is rarely clear what percentage of Earth's hemisphere is actually visible from any particular orbit.

• Is there a graphic or diagram that would show the percentage of Earth's hemisphere that is visible from low Earth orbit? • high Earth orbit? • geosynchronous orbit? • lunar orbit? • 1 million miles? • the Sun?

• Alternatively, is there a graphic or diagram that would show the percentage of a mathematical hemisphere "seen" from various points at multiples of a sphere's radii?

• Is there a website with interactive graphics that can calculate the above?

• I am a fan of space exploration (and accurate graphics). My working knowledge of math and geometry is limited.

• Assuming the surface of the earth is a sphere, or more realistically? – kimchi lover May 22 at 22:52
• $\frac{h/2}{r+h}$ of the surface of a sphere with radius $r$ can be seen when viewed from an altitude of $h$. I'll write this up. – robjohn May 22 at 23:03
• $\frac{h}{r+h}$ of the surface of a hemisphere of a sphere with radius $r$ can be seen when viewed from an altitude of $h$. I've written this up. – robjohn May 23 at 3:58

Suppose we are at an altitude of $$h$$ above a sphere with radius $$r$$. $$\triangle ABD\cong\triangle DCB$$; therefore, $$\frac{BC}{BD}=\frac{BD}{AD}$$ Since $$AD=h+r$$ and $$BD=r$$, we get $$BC=\frac{r^2}{h+r}$$. Subtracting from $$r$$ gives the width of the cap to be $$\frac{hr}{h+r}$$. The surface area of any cap or band on a sphere is the width of the cap or band times the circumference of the sphere: $$2\pi r\frac{hr}{h+r}$$. Since the surface area of the hemisphere is $$2\pi r^2$$, the portion of the hemisphere area in the cap is $$\bbox[5px,border:2px solid #C0A000]{\frac{h}{h+r}}$$ 