Orbital velocity for circular orbit — derivation

WHAT are we finding?

"Circular" matters: speed is constant in magnitude, only direction changes.
$r$ is measured from the centre of the planet, not the surface: $r = R + h$ where $R$ is the planet's radius and $h$ the altitude.

WHY does it work? (the physical logic)

Two facts collide:

Newton's 2nd law for circular motion. Anything moving in a circle of radius $r$ at speed $v$ must be accelerating inward (toward the centre) with magnitude $a_c = \dfrac{v^2}{r}$ . Something must supply the force $F = m a_c$ .
Newton's law of gravitation. The Earth pulls the satellite inward with $F_g = \dfrac{GMm}{r^2}$ .

If the only force is gravity, then gravity must equal the required centripetal force. Set them equal — that single equation pins down the speed.

HOW to derive it (from first principles)

Formula Step-by-step derivation

Step 1 — Required centripetal force. $F_{c} = \frac{m v_o^2}{r}$ Why this step? Uniform circular motion demands an inward force of exactly this size; otherwise the path isn't a circle.

Step 2 — Available gravitational force. $F_{g} = \frac{G M m}{r^2}$ Why this step? This is the only real force acting on the orbiting body (no thrust, ignore drag).

Step 3 — Force balance (the key physical claim). $F_g = F_c \;\Rightarrow\; \frac{GMm}{r^2} = \frac{m v_o^2}{r}$ Why this step? Gravity is the centripetal force; we are not adding two forces, we are saying one plays the role of the other.

Step 4 — Cancel and solve. Cancel $m$ (orbit doesn't depend on satellite mass!) and one power of $r$ : $\frac{GM}{r^2} = \frac{v_o^2}{r} \;\Rightarrow\; v_o^2 = \frac{GM}{r}$ $\boxed{\,v_o = \sqrt{\dfrac{GM}{r}}\,}$

Surface-gravity form

At the surface $g = \dfrac{GM}{R^2} \Rightarrow GM = gR^2$ . Substituting: $v_o = \sqrt{\frac{gR^2}{r}}\quad\xrightarrow{\;h=0,\;r=R\;}\quad v_o=\sqrt{gR}$ Why useful? You may not know $G$ and $M$ separately, but you always know $g$ and $R$ .

Figure — Orbital velocity for circular orbit — derivation

Common mistakes (Steel-manned)

Active recall

Recall Cover the answers and test yourself

What force provides the centripetal force in orbit? ⇒ Gravity alone.
Write $v_o$ in terms of $G,M,r$ . ⇒ $v_o=\sqrt{GM/r}$ .
Why doesn't satellite mass appear? ⇒ $m$ cancels — both $F_g$ and $F_c \propto m$ .
How does $v_o$ change as $r$ increases? ⇒ Decreases, as $1/\sqrt r$ .
What's $r$ for altitude $h$ ? ⇒ $r=R+h$ (from Earth's centre).
Relation to escape velocity? ⇒ $v_{esc}=\sqrt2\,v_o$ .

Recall Feynman: explain to a 12-year-old

Imagine throwing a ball really, really fast sideways from a tall mountain. Gravity always pulls it down toward the ground. If you throw gently it lands nearby. Throw harder, it lands further. Throw it insanely hard — about 8 km every second — and as it falls, the round Earth curves away below it by exactly the same amount. So it keeps "missing the ground" forever and goes all the way around! It isn't being held up by anything; it's just falling sideways so fast it never lands. Bigger or smaller ball, doesn't matter — the magic speed is the same, because gravity tugs heavy and light things the same way.

Flashcards

Orbital velocity formula (circular)

v_o=\sqrt{\dfrac{GM}{r}}

What provides the centripetal force for a satellite?

Gravity, and gravity alone.

The equation set up at the heart of the derivation

\dfrac{GMm}{r^2}=\dfrac{mv_o^2}{r}

Why does satellite mass not appear in

v_o

It cancels — both gravitational and required centripetal forces are proportional to

m

How does orbital speed vary with orbital radius?

v_o\propto 1/\sqrt r

— bigger orbit ⇒ slower.

Orbital velocity using surface gravity (near surface)

v_o=\sqrt{gR}\approx 7.9

km/s.

What is

r

for a satellite at altitude

h

r=R+h

, measured from Earth's centre.

Relation between orbital and escape velocity

v_{esc}=\sqrt2\,v_o

Does gravity do work on a circular-orbiting satellite?

No — it's perpendicular to velocity, so speed stays constant.

LEO orbital speed approximate value

~7.9 km/s (≈ 28,000 km/h).

Connections

Newton's Law of Universal Gravitation
Centripetal Force and Uniform Circular Motion
Escape Velocity
Kepler's Third Law — combine $v=2\pi r/T$ with $v_o$ to get $T^2\propto r^3$ .
Acceleration due to gravity g and GM = gR²
Geostationary Orbit

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, orbital velocity ka funda simple hai: satellite Earth ke around isliye ghoomta hai kyunki gravity use andar ki taraf kheech rahi hai — wahi gravity centripetal force ka kaam kar rahi hai. Koi alag "aage dhakelne wala" force nahi hota. Satellite continuously neeche gir raha hai, lekin uski side-ki speed itni perfect hai ki Earth ki curve neeche-neeche hat jaati hai, aur woh kabhi zameen pe nahi girta. Bas yahi orbit hai.

Derivation me hum sirf do cheezein barabar karte hain. Centripetal force chahiye $\frac{mv_o^2}{r}$ , aur gravity deti hai $\frac{GMm}{r^2}$ . Dono equal kar do, $m$ cancel ho jaata hai, ek $r$ cancel ho jaata hai, aur mil jaata hai $v_o=\sqrt{GM/r}$ . Sabse important baat — satellite ka apna mass matter hi nahi karta! Bhaari ho ya halka, same radius pe same speed chahiye.

Do cheezein hamesha yaad rakho. Pehla: $r$ hamesha Earth ke centre se naapo, yaani $r=R+h$ , sirf height $h$ mat use karna — bahut bachche yahin galti karte hain. Doosra: $v_o \propto 1/\sqrt{r}$ , matlab neeche wale (low) orbit tez hote hain aur door wale (Moon jaise) dheere. Surface ke paas shortcut: $v_o=\sqrt{gR}\approx 7.9$ km/s.

Aur ek bonus: escape velocity exactly $\sqrt2$ guna hoti hai orbital velocity ki — orbit karne ke liye 7.9 km/s, lekin Earth chhodne ke liye ~11.2 km/s. Exam me ye relation seedha aata hai, toh ratt lo: $v_{esc}=\sqrt2\,v_o$ .

Orbital velocity for circular orbit — derivation

WHAT are we finding?

WHY does it work? (the physical logic)

HOW to derive it (from first principles)

Surface-gravity form

Common mistakes (Steel-manned)

Active recall

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Newton's Laws & Dynamics

Connections