Question

How does Earth’s curvature relate to the speed needed for a projectile to orbit the Earth?

Answer 1

In short ,

V = √gR {where V = orbital speed at radius distance R = radius from Earth's center of orbiting object}

The idea of relating orbital speed to Earth's curvature comes right from Newton's Principia.

Newton realized, having formulated his law of universal gravitation, that you could draw this connection. His derivation of it went like this.

If a cannon on top of a tall mountain, fires a cannonball horizontally, it falls as it goes out, tracing a path that curves downward. At the usual speeds of such a firing, it will travel maybe a few miles before that curve brings it to the ground.

But if you keep increasing the charge of gunpowder to keep increasing the launch speed, it will travel farther and farther each time. Eventually, the cannonball goes so fast horizontally, that its vertical fall rate makes its curved path match the Earth's curvature.

And because as it goes around, the force of gravity keeps changing direction, always pointing toward the center of the Earth, the curve of its path is actually a circle — the cannonball just keeps going around in a circular orbit.

So the (linear) curvature of Earth's surface, is
κ⊕ = 1/R; R = Earth's radius.
The curvature of the cannonball's trajectory, as it emerges from the cannon, is
κ = g/v₀² ; v₀ = launch speed
g = acceleration of gravity = GM/R² ; M = Earth's mass, G = universal gravitational constant

So the necessary speed for close Earth orbit, is where the curvature of the trajectory matches that of the Earth:

g/v₀² = 1/R
v₀² = gR = (9.807 m/s²)(6.371·10⁶ m) = 6.248·10⁷ m²/s²
v₀ = 7904 m/s

And that is the close-Earth orbital speed.