Satellites are falling all the time, but they move sideways so fast that Earth's curved surface drops away beneath them as quickly as they fall.
Here is the surprise: a satellite is falling. It never stops falling. It just never reaches the ground.
NASA explains that a satellite stays in orbit when its speed is balanced by the pull of Earth's gravity. Too slow, and it falls back to Earth. Too fast, and it flies off into space.
Throw a ball sideways from a tall tower. It curves down and lands nearby. Throw it harder and it lands farther away.
Now imagine throwing it so hard that, while it falls, the curved Earth bends away underneath it by the same amount. The ball keeps falling but the ground keeps dropping away. NASA puts it this way: an orbiting spacecraft moves at just the right speed so the curve of its fall matches the curve of Earth. That endless fall is an orbit.
The International Space Station flies about 400 km up and goes once around Earth in about 92 minutes (ESA). From Earth's mass and its distance, its speed works out to about 7.7 km per second.
In one second the station moves 7.7 km sideways. In that same second, gravity (about 8.7 m/s² at that height) pulls it down by roughly 4.3 metres. And over a 7.7 km stretch, Earth's surface curves away below a straight line by — again — about 4.3 metres.
The fall and the curve cancel. Second after second, the station falls 4.3 m and stays the same height above the ground.
At low heights there is still a trace of air. It drags on satellites and slowly lowers them. NASA notes that the exact height of the ISS depends on how often it is reboosted to a higher orbit. A satellite that is never boosted will keep sinking until it re-enters the atmosphere.
Watch the Velocity card stay near 7.66 km/s — that sideways speed is what keeps the station from falling to the ground. Open the ISS tracker
For a circular orbit of radius r, gravity must supply exactly the centripetal acceleration:
v² / r = GM / r² → v = √(GM / r)
With GM = 398,600 km³/s² and the ISS at 400 km above the equatorial radius of 6,378 km, r = 6,778 km. v = √(398,600 ÷ 6,778) = √58.81 ≈ 7.67 km/s, about 27,600 km/h.
The one-second picture can be checked with geometry. A straight line of length d tangent to a circle of radius r leaves the circle behind by a sag s ≈ d² / (2r). With d = 7.67 km: s = 58.81 ÷ 13,556 ≈ 0.00434 km = 4.3 m.
In the same second, a body falling from rest with acceleration g covers ½gt². Local gravity at r = 6,778 km is GM/r² ≈ 0.00868 km/s², so ½ × 8.68 × 1² ≈ 4.3 m.
The two numbers are equal by construction: substituting v = √(GM/r) into d²/(2r) gives (GM/r)/(2r) = ½(GM/r²). That identity is the orbit.
What if the speed is wrong? Below v the path becomes an ellipse whose low point is inside the circle; if that low point is inside the atmosphere or Earth, the satellite comes down. Above v the path becomes a larger ellipse. At √2 × v (about 10.85 km/s at this height) the path opens into a parabola and the object escapes — the subject of the escape-velocity page.
It is falling, but its sideways speed is so high that Earth's curved surface drops away beneath it as fast as it falls, so it never gets closer to the ground.
Its path would become an ellipse dipping closer to Earth. If the low point were deep enough in the atmosphere, drag would bring it down.
r = 6,778 km, so v = √(398,600 ÷ 6,778) ≈ 7.67 km/s.