How fast would you need to be going in order to orbit the Earth at an altitude where you just skimmed the surface?
Assume a circular orbit around the equator. Assume air resistance isn’t a factor. Also assume that the Earth is homogenous enough that local variations in the gravitational field don’t introduce perturbations.
Using an online calculator and a height above the surface of 1 meter, I got a velocity of 7.90973 km/s.
This is the calculator I used:
ETA:
This site gives the same result and has a bit more explanation about the formulas involved.
http://hyperphysics.phy-astr.gsu.edu/hbase/orbv3.html
Interestingly enough, Arthur C. Clarke’s short story “Maelstrom II” concerns:
An astronaut is accidentally thrown into a highly eccentric orbit around the moon. It’s going to take some time to get a ship into an orbit capable of intersecting and rescuing him, but his orbit is so eccentric that it actually intersects a mountaintop.
They blow the mountain up, letting him orbit through the place where the mountain was, giving him more time.
You might think there’d be a big difference, but it’s quite minor.
The speed of an orbit is proportional to the semi-major axis. The earth’s radius is about 4000 miles, and the differences in length of axes is without rounding error. So if you look at a satellite at 4100 from the center, or 100 miles above the surface, the difference is minimal. The calculator gives 7.8116 km/sec. Pretty close.
And here I was expecting a link to “The Holes Around Mars”.
I remembered the story, and the name of the moon, but not the name of the story. 
Thanks for the link.
I saw this thread, and did a double-take. I asked pretty much the same question, seven years ago. Neat! 
Cool post username combo.