Visual walkthrough — Orbit types — LEO, MEO, GEO, HEO, SSO, Molniya

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Step 1 — What "in orbit" even means: falling forever

WHAT. A satellite is not "up where there is no gravity". Gravity is almost as strong up there as on the ground. A satellite is an object thrown sideways so fast that as it falls, the ground curves away underneath it just as fast. It keeps missing the Earth. That is an orbit.

WHY start here. Every equation below is really just "the falling exactly matches the curving away". If you believe this picture, the algebra is only bookkeeping.

PICTURE. The red ball is dropped straight down — it lands. Throw it sideways harder and harder (black arcs) and it lands farther away. Throw it hard enough and the fall curve never meets the ground: it circles.

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Step 2 — Naming the picture: $r$, $v$, and $m$

WHAT. Before any formula, we label the drawing. Draw the Earth's centre as a dot. The satellite is a point going around a circle.

• $r$ = the distance from Earth's centre to the satellite (not from the ground!). Radius of the circle.
• $v$ = the satellite's speed along the circle (how many metres of arc per second).
• $m$ = the satellite's mass (how much stuff it is made of).
• $M$ = the Earth's mass. $G$ = the gravitational constant, the fixed number that sets how strong gravity is anywhere in the universe.

WHY these four. Gravity depends on both masses and the distance. Motion in a circle depends on the speed and the radius. These are exactly the knobs.

PICTURE. The red satellite, the radius arrow $r$ from the centre, and the velocity arrow $v$ drawn tangent (touching the circle, pointing along the direction of travel).

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Step 3 — Two forces, one balance

WHAT. Two physical facts meet here.

Fact A — gravity pulls inward. Newton's law of gravitation says the pull between Earth and satellite is $F_{\text{grav}}=\frac{GMm}{r^2}$

• $GMm$ on top: bigger Earth ($M$), bigger satellite ($m$), or stronger gravity ($G$) → stronger pull.
• $r^2$ on the bottom: double the distance and the pull drops to a quarter. This is the famous inverse-square law — the pull spreads out over a sphere whose area grows as $r^2$.

Fact B — going in a circle requires an inward force. Anything moving in a circle is constantly changing direction, and changing direction is acceleration pointed toward the centre. The force needed to supply it is $F_{\text{centripetal}}=\frac{m{v}^2}{r}$

• $v^2$ on top: faster means the direction whips around harder, needing more force.
• $r$ on the bottom: a tighter (smaller) circle turns you harder, needing more force.

WHY set them equal. In Step 1 we said the fall exactly matches the curving. In force language that is: the inward force gravity supplies is precisely the inward force the circle demands. No leftover, no shortfall. So $\underset{\text{what gravity gives}}{\underbrace{\frac{GMm}{r^2}}}=\underset{\text{what the circle needs}}{\underbrace{\frac{m{v}^2}{r}}}$

PICTURE. One red arrow (gravity, inward) sitting exactly on top of the dashed black arrow (the centripetal requirement, inward). Same length = balance.

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Step 4 — Solving for speed: $v=\sqrt{\mu /r}$

WHAT. Now pure algebra on the balance equation. First cancel $m$ (it sits on both sides — the satellite's own mass does not matter, a bowling ball and a pea orbit identically): $\frac{GM}{r^2}=\frac{v^2}{r}$ Multiply both sides by $r$ to clear one power of $r$: $\frac{GM}{r}=v^2⟹v=\sqrt{\frac{GM}{r}}$

We bundle $GM$ into a single symbol $\mu$ (the "standard gravitational parameter") because $G$ and $M$ always travel together and $\mu$ is measured far more precisely than either alone: ${\mu }_{\oplus }\approx 3.986\times 10^{14}{\text{m}}^3/{\text{s}}^2$.

$\boxed{v=\sqrt{\frac{\mu }{r}}}$

WHY the square root, and WHY does bigger $r$ give a smaller $v$? $r$ is on the bottom inside the root. Grow $r$ → the fraction shrinks → its root shrinks → $v$ shrinks. Physically: far out, gravity is weak (that $r^2$ denominator in Fact A), so only a lazy, slow circle is needed to stay balanced. This kills the highway intuition "farther must be faster".

PICTURE. The curve of $v$ against $r$: a steep drop that flattens. Red dot marks LEO (fast), black dot marks GEO (slow).

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Step 5 — From speed to period: how long is one lap?

WHAT. The period $T$ is the time for one full trip around. One trip covers the circumference $2\pi r$ (the perimeter of the circle) at speed $v$. Time = distance ÷ speed: $T=\frac{2\pi r}{v}$

• $2\pi r$: the total length of the circular path.
• dividing by $v$: metres of path ÷ metres per second = seconds.

Now substitute the $v$ we just found, $v=\sqrt{\mu /r}$: $T=\frac{2\pi r}{\sqrt{\mu /r}}=2\pi r\sqrt{\frac{r}{\mu }}=2\pi \sqrt{\frac{r^3}{\mu }}$

The middle step: dividing by $\sqrt{\mu /r}$ is multiplying by $\sqrt{r/\mu }$; then $r\cdot \sqrt{r}=\sqrt{r^2}\cdot \sqrt{r}=\sqrt{r^3}$.

$\boxed{T=2\pi \sqrt{\frac{r^3}{\mu }}}$

WHY we care about $r^3$. Square both sides: $T^2=\frac{4{\pi }^2}{\mu }{r}^3$, i.e. $T^2\propto r^3$ — this is exactly Kepler's 3rd law, which we have now built from a rock on a string. (For stretched ellipses replace $r$ with the semi-major axis; see Two-Body Problem and the Vis-Viva Equation.)

PICTURE. Period climbing steeply with radius; horizontal line marks one sidereal day; where it meets the curve is the GEO radius we are hunting.

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Step 6 — The trick: demand $T$ = one sidereal day

WHAT. "Geostationary" means the satellite appears to hang still over one spot on the equator. That happens only if it goes around exactly as fast as the Earth spins. So we set $T$ equal to Earth's rotation period.

But which rotation period? Not the 24-hour solar day. That is the time for the Sun to return overhead, and it is a little long because Earth also creeps along its orbit around the Sun, so it must spin a bit extra to face the Sun again. The true spin relative to the fixed stars is the sidereal day: $T_{\text{sid}}=86164\text{s}(\approx 23\text{ h }56\text{ m }4\text{ s})$

WHY sidereal. The satellite doesn't care where the Sun is; it must match the Earth's actual spin in space. Using $86400$ s instead overshoots the altitude by ~50 km.

PICTURE. Two clocks: a solar day (Sun returns overhead) is slightly longer than a sidereal day (a fixed star returns overhead), because of Earth's motion around the Sun over that day.

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Step 7 — Invert Kepler and get the number

WHAT. We have $T$ and want $r$, so flip the period formula. Start from $T=2\pi \sqrt{\frac{r^3}{\mu }}.$ Square both sides: $T^2=4{\pi }^2\frac{r^3}{\mu }$. Solve for $r^3$: $r^3=\frac{\mu T^2}{4{\pi }^2}$. Take the cube root: $\boxed{r={\left(\frac{\mu T^2}{4{\pi }^2}\right)}^{1/3}}$

Plug in $\mu =3.986\times 10^{14}$ and $T=86164$ s: $r={\left(\frac{3.986\times 10^{14}\cdot (86164{)}^2}{4{\pi }^2}\right)}^{1/3}\approx 4.216\times 10^7\text{m}=42164\text{km}.$

Finally subtract Earth's radius to convert to altitude above the ground (Step 2's warning!): $h=r-R_{\oplus }=42164-6371\approx 35793\mathbf{km}(\approx 35786\text{to standard precision}).$

WHY the cube root. $T$ depends on $r^3$; to undo a cube you take a cube root. That is the whole reason the exponent is $1/3$.

PICTURE. The number line of orbits: LEO low, MEO in the middle (GPS), and the single red tick at GEO, $35786$ km, where the period hits one sidereal day.

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Step 8 — The edge and degenerate cases (never skipped)

WHAT & WHY — walk every scenario the reader could hit:

• $e=0,i=0$ (the true GEO). Perfect circle in the equatorial plane. The ground track collapses to a single point — the satellite hangs motionless. This is the case we solved.
• $e=0$ but $i\ne 0$ (tilted geosynchronous). Same period, same $r$, but the tilt makes it drift north–south each day, tracing a figure-8 analemma across the sky. Same altitude, not stationary. See Satellite Ground Tracks.
• $i=0$ but $e\ne 0$ (eccentric geosynchronous). Same average $a$, but it speeds up and slows down (Kepler's 2nd law), sliding east–west daily — again a figure-8, not a fixed point.
• $r\to R_{\oplus }$ (degenerate low limit). As $r$ shrinks to Earth's surface, $T\to 2\pi \sqrt{R_{\oplus }^3/\mu }\approx 84$ min — the fastest any circular orbit can be, and there $v$ is largest. Below that you're inside the planet: no orbit.
• $r\to \infty$ (degenerate high limit). $v\to 0$ and $T\to \infty$: infinitely far, infinitely slow, infinitely long. GEO is not a special place in the physics — it is just the one $r$ where $T$ happens to equal a day.
• Below ~2000 km: atmospheric drag steals energy each pass; a "circular" orbit slowly spirals down. Our frictionless balance (Step 3) is an idealisation that only holds above the sensible atmosphere.

PICTURE. Three sky-tracks side by side: a still red dot (true GEO), a figure-8 (tilted), and an east–west smear (eccentric) — all at the same altitude.

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The one-picture summary

Everything above, compressed: balance the two forces → solve for $v$ → convert to $T$ → demand one sidereal day → invert to get $r$ → subtract $R_{\oplus }$.

Recall Feynman retelling — say it back in plain words

A satellite is just a thing thrown so fast sideways it keeps missing the ground. Going in a circle needs a constant pull toward the centre, and gravity is exactly that pull — so I set "the pull gravity gives" equal to "the pull the circle needs". The satellite's own weight cancels out (a pea and a boulder orbit the same), leaving speed = square-root of (Earth's gravity number ÷ distance). Because distance is on the bottom, higher orbits are slower, not faster. One lap takes the circle's perimeter divided by that speed, which tidies up to $T=2\pi \sqrt{r^3/\mu }$ — Kepler's law, born from a rock on a string. To hang still over one spot, I make one lap last exactly one Earth-spin — the sidereal day of 86 164 s, not the 24-hour solar day. I flip the period formula (cube root, because time depended on distance cubed), get a radius of 42 164 km, subtract Earth's own radius, and out drops the magic altitude: about 35 786 km. And if the orbit is tilted or stretched, same altitude but it traces a figure-8 instead of standing still.

Recall Quick self-check

Why is $v$ smaller for larger $r$? ::: $r$ is inside the denominator of $\sqrt{\mu /r}$, so growing $r$ shrinks $v$; far out, gravity is weak and only a slow circle is needed to balance it. Why sidereal, not solar, day for GEO? ::: The satellite must match Earth's true spin relative to the stars ($86164$ s); the solar day is longer because Earth also orbits the Sun. What does subtracting $R_{\oplus }$ do? ::: Converts centre-distance $r$ into altitude-above-ground $h$. What happens at the same altitude if $i\ne 0$? ::: Still geosynchronous, but it traces a figure-8 analemma — not geostationary.