Newton's Cannonball demonstrates that an orbit requires immense horizontal speed to match the Earth's rigid curvature. How do we scientifically calculate exactly what that velocity must be?
Any object moving in a perfect circle must have a fundamental force pulling it toward the center, otherwise, its velocity vector would carry it off in a straight line out into deep space. This inward pull is called Centripetal Force ($F_c$).
For a satellite in orbit, what physical force is providing this exact required inward pull? Gravity. Because we established in Module 1.1 that gravity provides an absolute force of $F = G \frac{m_1 m_2}{r^2}$, we can perfectly equate the two formulas:
Once again, the mass of the rocket ($m_r$) cancels out entirely from both sides. We multiply both sides by $r$ to isolate $v^2$, and then take the square root to reveal the ultimate secret to spaceflight:
The International Space Station resides at $400\text{ km}$ above the surface.
What exact sideways velocity ($v$) must it maintain to avoid falling out of the sky?
$G = 6.674 \times 10^{-11}$
$M_{earth} = 5.972 \times 10^{24}\text{ kg}$
$R_e = 6,371\text{ km} = 6,371,000\text{ m}$
$h = 400\text{ km} = 400,000\text{ m}$
Conclusion: The ISS must fly sideways at nearly 7.6 kilometers *per second* (~17,100 mph) to sustain orbit. Note that the mass of the ISS (420 tons) was entirely irrelevant.
Examine the final velocity equation: $v = \sqrt{GM/r}$.
Because the radius ($r$) sits strictly in the denominator, the higher your altitude climbs, the slower you must travel to maintain orbit.
A satellite ripping through the lowest possible edges of Earth's atmosphere must travel roughly $7.8\text{ km/s}$ to avoid crashing. However, the Moon is orbiting roughly $384,000\text{ km}$ away. At that extreme altitude, Earth's inward gravitational tug is profoundly weak, meaning the Moon only needs heavily reduced centripetal force to sustain its circle. As a result, the Moon gracefully strolls through its orbit at a leisurely $1.0\text{ km/s}$.