Satellites: Speed, Period, Energy & Weightlessness

In October 1945, a young RAF radar officer named Arthur C. Clarke published a four-page article in Wireless World proposing something outrageous: put a relay station in an orbit exactly $35{,}800\ \mathrm{km}$ up, over the equator, and it will circle the Earth in precisely one day — hanging apparently motionless in the sky, able to beam radio to half the planet. Three such stations, he noted, would cover the world. Clarke computed that altitude with nothing beyond the physics in this article: gravity supplies exactly the centripetal force a circular orbit demands, and that one condition welds a satellite's speed, period and altitude into a single package. Today several hundred spacecraft occupy "Clarke's orbit," and your weather forecast arrives from it daily. A satellite is the purest exam laboratory in mechanics: one force (gravity), one geometry (a circle, to start), and everything else — how fast, how long, how much energy, why astronauts float — falls out of two lines of algebra. Every board asks it: IGCSE wants $v = 2\pi r/T$ and the geostationary story, CBSE and NEET want orbital speed and energy, JEE wants the strange negative-energy ledger where losing energy makes a satellite speed up, and AP and IB want weightlessness explained properly, without the phrase "no gravity in space." This article builds the whole package: speed, period, the $T^2 \propto r^3$ law in embryo, the energy ledger $E = -K$ , the unique geostationary radius, and the honest meaning of weightlessness.

Key takeaways

A circular orbit has one condition: gravity is the centripetal force, $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ . Everything follows from it.
Orbital speed $v = \sqrt{GM/r}$ — fixed by the radius, not chosen; lower orbits are faster. Skimming Earth's surface: $7.9\ \mathrm{km/s}$ , period $\approx 84$ min.
Period $T = 2\pi\sqrt{r^3/GM}$ — the seed of Kepler's third law, $T^2 \propto r^3$ .
The energy ledger of an orbit: $K = \dfrac{GMm}{2r}$ , $U = -\dfrac{GMm}{r}$ , total $E = -\dfrac{GMm}{2r} = -K$ . Negative total = bound.
Geostationary orbit: demand $T =$ one sidereal day and the algebra returns exactly one radius, $r \approx 42{,}200\ \mathrm{km}$ ( $h \approx 35{,}800\ \mathrm{km}$ ), over the equator only.
Weightlessness in orbit is free fall, not absence of gravity — at ISS altitude, $g$ is still about $89\%$ of its surface value.
Key idea: a satellite is a projectile falling around the Earth forever — the orbit condition fixes speed, period and energy the moment the radius is chosen.

The core insight

For a circular orbit of radius $r$ about mass $M$ , the single condition "gravity supplies the centripetal force" fixes everything: $v = \sqrt{GM/r}$ , $T = 2\pi\sqrt{r^3/GM}$ , and total energy $E = -GMm/2r$ — negative because the satellite is bound, and equal in magnitude to its kinetic energy.

Where this lives today

Orbits are now zoned like a city. Low Earth orbit ( $300$ – $1200\ \mathrm{km}$ , periods $\sim90$ – $110$ min) hosts the ISS and the imaging and broadband constellations — this article's $v \approx 7.7\ \mathrm{km/s}$ world, where the drag paradox is an operations budget: the ISS loses about $2\ \mathrm{km}$ of altitude a month and must regularly re-boost, gaining energy to slow down. Medium orbits near $20{,}200\ \mathrm{km}$ ( $T = 12$ h, two GPS laps a day) carry navigation. And the Clarke belt at $42{,}200\ \mathrm{km}$ is such prime real estate that slots are allocated by international treaty, degree by degree of longitude. Equation (2) has also become a scale: timing an orbit weighs the central body, which is how the masses of exoplanet-hosting stars and the black hole at the Milky Way's centre (from the star S2's 16-year orbit) are measured. When a mission ends, the same energy ledger runs in reverse — operators either lower $E$ until the atmosphere claims the craft, or pay the energy bill to lift dead geostationary satellites into a graveyard orbit $300\ \mathrm{km}$ higher.

The physical picture

Newton put the essential picture in the Principia itself. Stand on a very tall mountain and throw a stone horizontally: it falls and lands. Throw harder: it lands farther, because while gravity pulls it down the same distance per second, it covers more ground. Now throw at exactly the speed where the Earth's surface curves away as fast as the stone falls — about $8\ \mathrm{km}$ of ground makes the surface drop $5\ \mathrm{m}$ , and a stone at $7.9\ \mathrm{km/s}$ falls $5\ \mathrm{m}$ in its first second. The stone now falls continuously and misses the ground forever. That is an orbit: not an escape from gravity but a permanent surrender to it, arranged sideways.

Figure

Newton's cannonball — fall fast enough sideways and you miss the ground forever

Three consequences of this picture do heavy exam duty. First, the speed is not a choice. At radius $r$ there is exactly one speed that makes the fall match the curvature; a satellite moving slower drops into a lower path, one moving faster climbs. Radius picks speed — equation (1) below — which is why "at what speed must a satellite at height $h$ travel?" has a unique answer. Second, everything aboard falls together. The astronaut, the station, and a dropped spanner all accelerate at the same $g(r)$ toward Earth's centre; nothing presses on anything, scales read zero, and that shared free fall is the entire content of "weightlessness" — gravity at the International Space Station's altitude is still $8.7\ \mathrm{m/s^2}$ , about $89\%$ of the surface value. Third, the geometry is self-balancing only if the pull points at the circle's centre — so the orbital plane must contain the Earth's centre. That innocent remark is why a geostationary satellite can hover only over the equator: a circle above Delhi's latitude would need a centre north of Earth's centre, and gravity refuses to point there.

The math: one condition, four results

The whole subject is the sentence "gravity supplies the centripetal force," made quantitative. For mass $m$ circling mass $M$ at radius $r$ (measured from Earth's centre — the most-lost mark in the chapter):

\frac{GMm}{r^2} = \frac{mv^2}{r} \quad\Longrightarrow\quad v = \sqrt{\frac{GM}{r}} \qquad (1)

where: $v$ — the unique orbital speed at radius $r$ . The satellite's mass has cancelled: bolts and space stations share an orbit at the same speed. Read the scaling: $v \propto 1/\sqrt r$ — lower is faster. A skimming orbit at $r \approx R$ gives $v = \sqrt{gR} = \sqrt{9.8 \times 6.37\times10^6} \approx 7.9\ \mathrm{km/s}$ ; the Moon, 60 Earth radii out, ambles at about $1\ \mathrm{km/s}$ . Divide the circumference by the speed and the period appears:

T = \frac{2\pi r}{v} = 2\pi\sqrt{\frac{r^3}{GM}} \qquad (2)

where: $T$ — the time for one revolution. Square it and the law of the next article is already visible: $T^2 = \dfrac{4\pi^2}{GM}\,r^3$ — periods squared grow as radii cubed, with a constant that depends only on the central body. Skimming Earth: $T = 2\pi\sqrt{R/g} \approx 84$ minutes; the ISS at $400\ \mathrm{km}$ takes $92.5$ ; the Moon takes $27.3$ days. And demanding $T =$ one day returns the one radius that hovers — worked example 3. The energy ledger closes the package. Kinetic energy from (1), potential energy from the gravity well:

K = \frac{GMm}{2r}, \qquad U = -\frac{GMm}{r}, \qquad E = K + U = -\frac{GMm}{2r} \qquad (3)

where: $E$ — total mechanical energy, negative because the satellite is bound (a positive- $E$ object escapes, as the escape-velocity article showed). Notice the fixed ratios: $U = 2E$ , $K = -E = |E|$ . The magnitude $|E|$ is the binding energy — the top-up that would set the satellite free. These locked ratios generate the chapter's best paradox: drag makes $E$ more negative, so $r$ shrinks — and by (1) the speed increases. Friction speeds the satellite up; the archetype module on orbit energetics drills the consequences.

Figure

The energy ledger of a circular orbit — K, E and U locked in ratio 1 : −1 : −2

you want	use	watch for
speed at radius $r$	$v = \sqrt{GM/r} = \sqrt{gR^2/r}$	$r = R + h$ , from the centre
period	$T = 2\pi\sqrt{r^3/GM}$	$T^2 \propto r^3$ for ratio questions
energy / binding energy	$E = -GMm/2r$ ; binding $= \|E\| = K$	the 2 in the denominator
geostationary radius	set $T = 86{,}164\ \mathrm{s}$ in (2)	sidereal day; equator only
speed from period (IGCSE)	$v = 2\pi r/T$	pure geometry, no $G$ needed

Derivation AOrbital speed and period from the one condition

We derive equations (1) and (2) — the speed and period of a circular orbit — from the single dynamical fact that gravity provides the centripetal force. The strategy is Newton's second law along the radius, plus circle geometry; nothing else is available and nothing else is needed. Watch two things: the satellite's mass cancelling (orbits are mass-democratic), and the radius $r$ being centre-to-satellite — surface altitude enters only through $r = R + h$ .

Step 1. Newton's second law toward the centre. The only force on the satellite is gravity, magnitude $GMm/r^2$ , pointing at Earth's centre; circular motion at speed $v$ demands centripetal acceleration $v^2/r$ :

\frac{GMm}{r^2} = \frac{mv^2}{r}

Step 2. Cancel $m$ and one power of $r$ :

v^2 = \frac{GM}{r} \quad\Longrightarrow\quad v = \sqrt{\frac{GM}{r}}

Why $m$ had to cancel: gravity's pull is proportional to the mass it accelerates — the same proportionality that makes all objects fall alike makes all objects orbit alike.

Step 3. Geometry gives the period — distance over speed:

T = \frac{2\pi r}{v} = 2\pi r\sqrt{\frac{r}{GM}} = 2\pi\sqrt{\frac{r^3}{GM}}

Step 4. Package for exams. With $GM = gR^2$ (surface gravity in disguise), $v = R\sqrt{g/r}$ and $T = \dfrac{2\pi}{R}\sqrt{\dfrac{r^3}{g}}$ — forms that need no $G$ and no $M$ . Why the $gR^2$ trick matters: boards rarely hand you $M_E$ ; they hand you $g$ and $R$ . Converting fluently is worth a minute per problem. *Verified: $r = R$ : $v = \sqrt{gR} = \sqrt{9.8 \times 6.37\times10^6} = 7.9\ \mathrm{km/s}$ , $T = 2\pi\sqrt{R/g} = 5{,}060\ \mathrm{s} \approx 84\ \mathrm{min}$ ✓ — the ground-skimming pair that also appeared as the critical-spin day in the variation-of-g article. Squaring Step 3 gives $T^2 = (4\pi^2/GM)r^3$ : Kepler's third law, one article early ✓.*

Go deeper · Derivation B · 1 minThe energy ledger — why a bound orbit has E = −K

Derivation BThe energy ledger — why a bound orbit has E = −K

We derive equation (3), the energy bookkeeping of a circular orbit, and extract the drag paradox. The strategy is assembly, not new physics: kinetic energy comes from the orbit condition (Derivation A), potential energy from the gravity well $U = -GMm/r$ established two modules ago. The thing to watch is the factor 2: kinetic energy is exactly half the well depth at that radius, and every famous ratio in the topic descends from that.

Step 1. Kinetic energy from the orbit condition. Since $v^2 = GM/r$ :

K = \frac12 mv^2 = \frac{GMm}{2r}

Step 2. Potential energy from the well (owned by the potential-energy module, applied here):

U = -\frac{GMm}{r} = -2K

Step 3. Add:

E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}

Why negative: bound means "cannot reach infinity," and infinity costs $E \ge 0$ . The magnitude $|E| = GMm/2r$ is the binding energy — the energy gap between this orbit and freedom.

Step 4. The drag paradox. Air drag does negative work, lowering $E$ — making it more negative, which by Step 3 means a smaller $r$ . But Step 1 says smaller $r$ carries larger $K$ : the satellite descends and speeds up. The bookkeeping balances because falling releases potential energy at twice the rate drag drains the total: one share pays the drag bill, one share becomes speed. Why examiners adore this: "a satellite experiences drag; what happens to its speed?" defeats intuition trained on cars. The energy ledger, not the force picture, answers cleanly. *Verified: ratios $K : E : U = 1 : -1 : -2$ hold at every $r$ ✓; at $r = 7000\ \mathrm{km}$ with $m = 400\ \mathrm{kg}$ : $K = 1.14\times10^{10}\ \mathrm{J}$ , $E = -1.14\times10^{10}\ \mathrm{J}$ — recomputed in worked example 2 ✓.*

Limiting cases

case	result	meaning
$r \to R$ (skimming orbit)	$v = \sqrt{gR} = 7.9\ \mathrm{km/s}$ , $T \approx 84\ \mathrm{min}$	the fastest, shortest Earth orbit
$r$ large	$v \propto 1/\sqrt r \to$ slow, $T \propto r^{3/2} \to$ long	the Moon: $1\ \mathrm{km/s}$ , $27.3$ days
$T = 86{,}164\ \mathrm{s}$	$r = 42{,}200\ \mathrm{km}$ , unique	geostationary: hovers, equator only
$v \to \sqrt2\,v_{\rm orb}$ at any $r$	$E \to 0$ : escapes	the $\sqrt2$ ladder from the previous article
$E$ decreased (drag, retro-burn)	$r$ shrinks, $v$ grows	the drag paradox
two orbits, $r_2 = 4r_1$	$v_2 = v_1/2$ , $T_2 = 8T_1$	the ratio grammar of MCQs

Common trap“Astronauts float because there is no gravity in space”

At the ISS's $400\ \mathrm{km}$ altitude, gravity is $g = GM/r^2 = 8.7\ \mathrm{m/s^2}$ — a mere 11% below the surface value. If gravity were absent, the station would not orbit at all; it would sail off in a straight line. Astronauts float because station, astronaut and every loose object are in identical free fall, endlessly falling around the Earth together — so nothing pushes on anything, and every scale reads zero. Weightlessness is the absence of the normal force, not of gravity. The same physics, one article back, made a lift's passenger weightless when the cable snapped; an orbit is a cable-snap that never ends. Say "free fall," never "zero gravity," and the conceptual mark is yours on every board.

Worked examples

1 · Board level — speed and period at altitude (CBSE/ISC pattern, 3 marks)

Find the orbital speed and period of a satellite in a circular orbit at a height equal to the Earth's radius. Take $R = 6.37\times10^6\ \mathrm{m}$ , $g = 9.8\ \mathrm{m/s^2}$ .

Step 1. Radius from the centre — the mark most candidates drop: $r = R + h = 2R = 1.274\times10^7\ \mathrm{m}$ .

Step 2. Speed via the $gR^2$ form of (1): $v = \sqrt{\dfrac{gR^2}{2R}} = \sqrt{\dfrac{gR}{2}} = \sqrt{\dfrac{9.8 \times 6.37\times10^6}{2}} = \sqrt{3.12\times10^7}$ .

Step 3. Evaluate: $v = \boxed{5.6\ \mathrm{km/s}}$ — slower than the $7.9\ \mathrm{km/s}$ skimming value by exactly $\sqrt2$ , since $r$ doubled.

Step 4. Period: $T = \dfrac{2\pi r}{v} = \dfrac{2\pi \times 1.274\times10^7}{5.59\times10^3} = 1.43\times10^4\ \mathrm{s} \approx \boxed{4.0\ \mathrm{hours}}$ .

2 · Board level — the energy ledger (NEET/ISC pattern)

A $400\ \mathrm{kg}$ satellite orbits at $r = 7000\ \mathrm{km}$ . Find its kinetic energy, total energy, and binding energy. ( $GM = 3.99\times10^{14}\ \mathrm{m^3/s^2}$ .)

Step 1. Kinetic energy: $K = \dfrac{GMm}{2r} = \dfrac{3.99\times10^{14} \times 400}{2 \times 7\times10^6} = \boxed{1.14\times10^{10}\ \mathrm{J}}$ .

Step 2. Total energy is its negative — no new arithmetic: $E = -K = \boxed{-1.14\times10^{10}\ \mathrm{J}}$ .

Step 3. Binding energy is the magnitude: $|E| = \boxed{1.14\times10^{10}\ \mathrm{J}}$ — supply this and the satellite just escapes.

Step 4. Sanity: the speed is $v = \sqrt{GM/r} = \sqrt{5.70\times10^7} = 7.55\ \mathrm{km/s}$ , and $\tfrac12(400)(7546)^2 = 1.14\times10^{10}\ \mathrm{J}$ ✓ — the ledger's ratios $K : E : U = 1 : -1 : -2$ cost nothing to verify.

3 · Competitive level — Clarke's orbit from scratch (JEE-Main pattern, ~3 minutes)

Find the radius, altitude and speed of a geostationary orbit. Sidereal day $T = 86{,}164\ \mathrm{s}$ , $GM = 3.986\times10^{14}\ \mathrm{m^3/s^2}$ , $R = 6371\ \mathrm{km}$ .

Step 1. Invert equation (2) for the radius: $r^3 = \dfrac{GMT^2}{4\pi^2}$ .

Step 2. Numbers: $T^2 = 7.42\times10^9\ \mathrm{s^2}$ , so $r^3 = \dfrac{3.986\times10^{14} \times 7.42\times10^9}{39.48} = 7.50\times10^{22}\ \mathrm{m^3}$ .

Step 3. Cube root: $r = 4.22\times10^7\ \mathrm{m} = \boxed{42{,}200\ \mathrm{km}}$ ; altitude $h = r - R = \boxed{35{,}800\ \mathrm{km}}$ — Clarke's number.

Step 4. Speed: $v = \dfrac{2\pi r}{T} = \dfrac{2.65\times10^8}{86164} = \boxed{3.07\ \mathrm{km/s}}$ . Timing note: the only slow step is the cube root — remember $42{,}000\ \mathrm{km}$ -ish and use options to your advantage. Two flags examiners plant: the day must be sidereal ( $86{,}164\ \mathrm{s}$ , not $86{,}400$ — the difference matters at the fourth figure), and there is exactly ONE such radius, over the equator only. "Geostationary above Mumbai" is always a wrong option.

4 · Competitive level — ratio grammar and the drag paradox (JEE/NEET stretch, ~3 minutes)

Satellites P and Q circle Earth at radii $r$ and $4r$ . (a) Find the ratios $v_P/v_Q$ and $T_P/T_Q$ . (b) P's orbit then decays slowly by air drag to a slightly lower altitude. State, with reason, what happens to its speed, period, kinetic energy and total energy.

Step 1. Part (a), speeds from $v \propto 1/\sqrt r$ : $\dfrac{v_P}{v_Q} = \sqrt{\dfrac{4r}{r}} = \boxed{2}$ — the inner satellite is twice as fast.

Step 2. Periods from $T \propto r^{3/2}$ : $\dfrac{T_P}{T_Q} = \left(\dfrac{1}{4}\right)^{3/2} = \boxed{\dfrac{1}{8}}$ .

Step 3. Part (b), run the ledger, not intuition. Drag removes mechanical energy: $E$ decreases (more negative). Since $E = -GMm/2r$ , $r$ decreases — consistent with "orbit decays."

Step 4. At smaller $r$ : $v = \sqrt{GM/r}$ increases, $K = -E$ increases, and $T \propto r^{3/2}$ decreases. The satellite ends up faster while continuously losing energy — the potential energy released by falling exceeds the drag loss, and the surplus becomes speed. Timing note: part (a) is pure exponent reading — $\tfrac12$ for speed, $\tfrac32$ for period — and should take 40 seconds. Part (b) is a memorised outcome for rankers: drag ⇒ lower, faster, shorter period, larger K, smaller E. Reconstructing it mid-exam from the ledger costs a minute; knowing the pattern costs five seconds.

Exam traps

Height versus radius. Every formula in this article uses $r$ measured from Earth's centre. A satellite "at a height $R$ " orbits at $r = 2R$ . Under time pressure, candidates substitute $h$ straight into $\sqrt{GM/r}$ — the examiner has an option waiting for exactly that.

The missing 2. Total energy is $-GMm/2r$ , potential energy is $-GMm/r$ . Dropping the 2 (or applying it to $U$ ) scrambles every energy answer. Anchor: $K$ is half the well depth; $E = -K$ .

Solar day versus sidereal day. Geostationary means one revolution per sidereal day, $86{,}164\ \mathrm{s}$ — the Earth's true rotation period relative to the stars. Using $86{,}400\ \mathrm{s}$ shifts the radius by about $80\ \mathrm{km}$ ; options are sometimes spaced to catch it.

Geostationary anywhere. No circular orbit can hover over a non-equatorial city: the orbital plane must pass through Earth's centre. A "geostationary satellite above 28° N" is a contradiction — standard assertion-reason bait.

Drag intuition. "Friction slows things down" fails in orbit; drag makes a satellite faster (worked example 4). Conversely, to raise an orbit you fire engines and yet end up slower — energy up, speed down. Trust $E = -GMm/2r$ , never the car analogy.

Orbital speed is not escape speed. $\sqrt{GM/r}$ orbits; $\sqrt{2GM/r}$ escapes. Options offering $11.2\ \mathrm{km/s}$ for a low-orbit speed (or $7.9$ for escape) appear every year — the $\sqrt2$ ladder sorts them instantly.

Mass of the satellite in $v$ or $T$ . Neither contains $m$ — a heavier satellite needs a bigger rocket, not a different speed. If your ratio answer depends on satellite masses, you have mixed the energy formulas (which do carry $m$ ) into the kinematics (which do not).

IGCSE scope note. CIE 0625 asks $v = 2\pi r/T$ , the qualitative orbit story, and geostationary versus polar-orbit uses — not $\sqrt{GM/r}$ or energies. Know your paper's ceiling; don't import machinery the question didn't ask for.

Origins · 1 minArthur C. Clarke and how this was discovered

Arthur C. Clarke

1917–2008 · Somerset, London & Colombo

Arthur C. Clarke was a radar instructor and physics-trained futurist before he became one of science fiction's giants. His October 1945 Wireless World paper, "Extra-Terrestrial Relays," computed that a satellite at $35{,}800\ \mathrm{km}$ over the equator would orbit in exactly one day — hovering, from the ground's viewpoint — and proposed three such stations to relay radio worldwide. The idea was dismissed as fantasy economics; nineteen years later Syncom 3 reached that very orbit, and today several hundred spacecraft occupy what the International Astronomical Union formally calls the Clarke Orbit. Clarke never patented it ("a patent is really a licence to be sued," he joked, estimating he'd have earned nothing anyway before it expired). He moved to Sri Lanka in 1956, wrote 2001: A Space Odyssey with Kubrick, and lived to see his relay ring carry the world's weather pictures and television — the rare case of a physics homework problem becoming a trillion-dollar piece of infrastructure.

History

1687

— Newton's Principia presents the cannonball thought experiment: orbital motion is perpetual falling.

1945

— Clarke's Wireless World article computes the geostationary relay orbit at 35,800 km.

1957

— Sputnik 1: the first artificial satellite, a 96-minute orbit announced by beeps.

1963

— Syncom 2 (geosynchronous) and, in 1964, Syncom 3 (truly geostationary) realise Clarke's orbit.

1978

— the first GPS satellites launch into 12-hour orbits — timing as infrastructure.

1998

— the International Space Station's first module; continuous human free fall since 2000.

2019

— Starlink begins filling low orbits by the thousand; megaconstellations make orbital traffic management a discipline.

Going further

Texts. NCERT Physics Class 11, Ch. 8 (Gravitation) · HC Verma, Concepts of Physics Vol. 1, Ch. 11 · Clarke's original 1945 paper, "Extra-Terrestrial Relays" (four pages, freely available, and readable with exactly this article's toolkit).

Provenance. All results are standard canon. A verifier should recompute: $v = \sqrt{gR} = 7.9\ \mathrm{km/s}$ and $T = 84$ min at $r = R$ ; example 1's $5.6\ \mathrm{km/s}$ and $4.0$ h at $r = 2R$ ; example 2's $K = 1.14\times10^{10}\ \mathrm{J}$ ledger; example 3's $r^3 = 7.50\times10^{22}\ \mathrm{m^3}$ , $r = 42{,}200$ km, $v = 3.07\ \mathrm{km/s}$ ; example 4's ratios 2 and 1/8.

Prerequisites. Newton's Law of Universal Gravitation · Circular Motion: Centripetal & Tangential Acceleration · Gravitational Potential Energy & Potential. Leads to. Kepler's Laws of Planetary Motion · Raising & Switching Orbits (archetype) · Orbit Classification & Energy (UG deep dive).

Why can't a satellite orbit at any speed I choose at a given height?

Because the orbit condition is an equality, not a suggestion. At radius $r$ , gravity provides a fixed centripetal budget $GM/r^2$ ; a circle of that radius at speed $v$ demands $v^2/r$ . Only one $v$ makes demand equal supply. Go faster and gravity can't bend the path enough — the satellite rises into an ellipse; slower, and it curves down toward a lower point. The circular orbit is the knife-edge where the two match, which is exactly why radius alone answers "how fast?"

Is the 84-minute period the same 84 minutes as the tunnel through the Earth and the critical spin day? That can't be coincidence.

It isn't. All three are the number $2\pi\sqrt{R/g} \approx 84$ min wearing different costumes: the skimming orbit (this article), the day-length at which equatorial gravity just fails to hold objects down (variation-of-g), and the oscillation period of a body dropped through a tunnel in a uniform Earth (the tunnel archetype ahead). Deep reason: a circular orbit viewed edge-on is an oscillation with the same $\omega^2 = g/R = GM/R^3$ . One frequency, three famous problems.

When a satellite fires its engines to reach a higher orbit, it ends up moving slower. Where did the work go?

Into the gravitational account, twice over. Raising the orbit increases total energy $E$ (engines do positive work), but the higher circle needs less kinetic energy — $K = GMm/2r$ falls as $r$ grows. The ledger balances because potential energy rose by twice what kinetic energy lost: $U = -2K$ keeps the ratio locked. So the burn's energy all lands in $U$ , with some of the old speed converted too. "Energy up, speed down" feels wrong precisely because everyday driving has no $U$ ledger of this kind.

Why must the geostationary orbit be over the equator? Couldn't we tilt it to hover over India?

Gravity always points at Earth's centre, so it can only supply centripetal force toward that centre — meaning every orbital plane must contain the centre of the Earth. A circle hovering over 20° N would need its centre at 20° N inside the Earth, and no force points there. The best you can do is a geosynchronous tilted orbit: one-day period, but tracing a daily figure-eight in the sky north and south of the equator. Truly motionless in the sky = equatorial, one radius, full stop. (India's regional navigation system, NavIC, uses exactly those inclined geosynchronous figure-eights.)

Do heavier satellites need to move faster to stay up?

No — $v = \sqrt{GM/r}$ contains the Earth's mass, not the satellite's. Both the gravitational pull and the centripetal demand scale with satellite mass, so it cancels: a paint fleck and a 400-tonne station coexist happily at the same radius and speed (which is also why orbital debris is so dangerous — it shares your speed neighbourhood). What mass does change is the energy and fuel to get there: $E = -GMm/2r$ scales with $m$ , so heavier payloads cost proportionally more to lift and to move.

What actually happens if a satellite is suddenly slowed to below its orbital speed? Does it fall straight down?

No — it falls forward. Reduce the speed a little and the path becomes an ellipse dipping below the original circle; the satellite is still moving several kilometres per second sideways. This is precisely how de-orbiting works: a retro-burn of a few hundred m/s lowers one side of the orbit into the atmosphere, and drag finishes the job over the following revolutions. Falling "straight down" would require cancelling the entire 7-plus km/s of tangential speed — vastly more fuel than any deorbit burn carries.

The Moon is a satellite too — do these formulas really cover it?

Completely, with one honesty note. Equation (2) with the Moon's $r = 3.84\times10^8\ \mathrm{m}$ gives $T = 2\pi\sqrt{r^3/GM} \approx 27.4$ days — the sidereal month, correct to within the approximations (the Moon's orbit is slightly elliptical, and strictly both bodies orbit their common centre of mass, which sits about 4700 km from Earth's centre — still inside the Earth). Newton's own "Moon test" ran this logic in reverse: the same law that drops an apple predicts the Moon's monthly circuit. Satellites just repeat the Moon's trick closer in.