Circular orbit — velocity, period, energy

WHY does a circular orbit need a specific speed?

WHAT must be true? For circular motion at constant speed, the net force must be a centripetal force — pointing to the center, with magnitude $\dfrac{mv^2}{r}$ . The only force available is gravity. So gravity must supply exactly the centripetal requirement. Not more, not less. That single balance fixes the speed.

HOW to derive the orbital velocity (from scratch)

Step 1 — Write the two forces that must be equal.

Gravity on $m$ : $F_g = \frac{GMm}{r^2}$

Why this step? Newton's law of gravitation — the inward pull at distance $r$ .

Centripetal requirement for uniform circular motion: $F_c = \frac{mv^2}{r}$

Why this step? Any object moving in a circle of radius $r$ at speed $v$ needs this inward force; it's kinematics, not a new force.

Step 2 — Set them equal (gravity is the centripetal force): $\frac{GMm}{r^2} = \frac{mv^2}{r}$

Step 3 — Cancel and solve. Cancel $m$ (so the answer is independent of the satellite's mass!) and one power of $r$ : $\frac{GM}{r} = v^2$

HOW to get the period $T$

Step 1 — Distance / speed. In one orbit the satellite travels the circumference $2\pi r$ at speed $v$ : $T = \frac{2\pi r}{v}$

Why this step? Period = time for one lap = (distance per lap) / (speed). Pure definition.

Step 2 — Substitute $v=\sqrt{GM/r}$ : $T = \frac{2\pi r}{\sqrt{GM/r}} = 2\pi r \cdot \sqrt{\frac{r}{GM}} = 2\pi \sqrt{\frac{r^3}{GM}}$

HOW to get the energies

Kinetic energy. Just plug $v^2 = GM/r$ into $K=\tfrac12 mv^2$ : $K = \frac{1}{2}m\frac{GM}{r} = \frac{GMm}{2r}$

Potential energy. Gravitational PE (zero at infinity, negative when bound): $U = -\frac{GMm}{r}$ Why negative & why $1/r$ ? $U$ is the work to bring $m$ in from infinity; pulling-in releases energy, so bound states sit below zero. (Derivable as $U=-\int_\infty^r F_g\,dr$ .)

Total mechanical energy. Add them: $E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$

Figure — Circular orbit — velocity, period, energy

Worked Examples

Worked example 1. Low Earth Orbit (ISS-like)

$M_\oplus = 5.97\times10^{24}\,$ kg, $G=6.674\times10^{-11}$ , altitude 400 km so $r = 6.37\times10^6 + 4.0\times10^5 = 6.77\times10^6\,$ m.

Speed: $v=\sqrt{\frac{(6.674\times10^{-11})(5.97\times10^{24})}{6.77\times10^6}}=\sqrt{5.88\times10^7}\approx 7.67\times10^3\,$ m/s ≈ 7.67 km/s. Why this step? Direct plug into $v=\sqrt{GM/r}$ — no mass of station needed.

Period: $T = \frac{2\pi r}{v} = \frac{2\pi(6.77\times10^6)}{7670} \approx 5.55\times10^3\,$ s ≈ 92 min. ✓ matches the real ISS.

Worked example 3. Energy to raise an orbit

Move a 1000 kg satellite from $r_1=7\times10^6$ to $r_2=8\times10^6$ m around Earth. $\Delta E = -\frac{GMm}{2r_2}+\frac{GMm}{2r_1}=\frac{GMm}{2}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$ $=\frac{(6.674\times10^{-11})(5.97\times10^{24})(1000)}{2}\left(\frac{1}{7\times10^6}-\frac{1}{8\times10^6}\right)\approx 3.56\times10^9\,$ J. Why this step? Energy needed = difference of total orbital energies. Higher orbit has higher (less negative) $E$ , so it costs energy. (Counter-intuitively the higher orbit is slower!)

Common mistake Steel-manned misconceptions

"Higher orbit → faster, because it has more energy." Why it feels right: higher orbit has greater total energy $E$ (less negative) and larger PE, so "more energy = more speed" sounds natural. The fix: The extra energy went into potential energy. Kinetic energy actually drops: $K=\frac{GMm}{2r}$ decreases as $r$ grows, and $v=\sqrt{GM/r}$ decreases too. Higher = slower but more total energy.

"The satellite's mass matters for the speed." Why it feels right: heavier things feel more gravity. The fix: They also need more centripetal force, and $m$ cancels in $GM/r^2 = v^2/r$ . Speed and period are independent of $m$ . ( $K$ , $U$ , $E$ do scale with $m$ .)

" $U = mgh$ works for orbits." Why it feels right: it's the formula from school physics. The fix: $mgh$ is the flat-Earth, constant- $g$ approximation. Over orbital distances $g$ varies, so you must use $U=-GMm/r$ .

Recall Feynman: explain it to a 12-year-old

Imagine throwing a ball really, really hard sideways. Normally it falls and hits the ground. But the Earth is round, so the ground curves away. If you throw the ball just the right speed (about 8 km every second!), the ground curves away exactly as fast as the ball falls — so the ball keeps "missing" the Earth and falls around it forever. That's an orbit: falling without ever landing. Throw it from higher up and it can go slower, because gravity up there is weaker. The faster-falling part and the gravity-pulling part have to balance perfectly — that's the only speed that works at each height.

Flashcards

What balance defines a circular orbit?

Gravity supplies exactly the centripetal force:

GMm/r^2 = mv^2/r

Formula for circular orbital speed?

v=\sqrt{GM/r}

(independent of orbiting mass

m

Why does satellite mass not affect the speed or period?

m

cancels in

GMm/r^2 = mv^2/r

Formula for the period of a circular orbit?

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

, i.e.

T^2\propto r^3

(Kepler III).

Kinetic energy of a circular orbit?

K=+GMm/(2r)

Potential energy of a circular orbit?

U=-GMm/r

Total mechanical energy of a circular orbit?

E=-GMm/(2r)

(negative ⇒ bound).

Relations between K, U, E?

E=-K

and

U=2E=-2K

(virial theorem for

1/r

As orbit radius increases, what happens to v, T, E?

v decreases, T increases, E increases (less negative).

Does a higher orbit move faster or slower?

Slower (

v\sim r^{-1/2}

), but it has more total energy.

Energy to move between circular orbits

r_1\to r_2

\Delta E=\frac{GMm}{2}\left(\frac1{r_1}-\frac1{r_2}\right)

Why is

E<0

physically meaningful?

Bound orbit; you'd need to add

+GMm/2r

to reach escape (

E=0

Connections

Newton's Law of Universal Gravitation — the force we set equal to centripetal.
Centripetal Force and Uniform Circular Motion — supplies $mv^2/r$ .
Kepler's Three Laws — $T^2\propto r^3$ is a special case here.
Gravitational Potential Energy — origin of $U=-GMm/r$ .
Escape Velocity — the $E=0$ boundary; $v_{esc}=\sqrt{2}\,v_{circ}$ .
Elliptical Orbits and the Vis-viva Equation — generalizes $v$ , $E$ to non-circular orbits.
Virial Theorem — explains $U=2E=-2K$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, circular orbit ka pura raaz ek hi balance mein chhupa hai: gravity ka force exactly utna hona chahiye jitna centripetal force chahiye. Yaani $GMm/r^2 = mv^2/r$ . Is equation mein satellite ki mass $m$ dono taraf cut ho jaati hai — isiliye orbit ki speed satellite ke weight par depend hi nahi karti! Feather ho ya bus, same height pe same speed se ghoomenge. Solve karne par milta hai $v=\sqrt{GM/r}$ .

Intuition yeh hai: satellite asal mein hamesha "gir" raha hai, par itni speed se side mein ja raha hai ki Earth uske niche se curve hoke nikal jaati hai — toh woh kabhi zameen pe girta hi nahi, bas girte-girte Earth ke around ghoomta rehta hai. Agar zyada upar jaoge (bada $r$ ), toh gravity weak hoti hai, isliye kam speed chahiye — matlab upar wala orbit dheema hota hai, fast nahi. Yeh point exam mein bahut log galat karte hain.

Period nikalna easy hai: ek chakkar ki doori $2\pi r$ ko speed se divide karo, $T = 2\pi\sqrt{r^3/GM}$ — yahi Kepler ka teesra law hai ( $T^2 \propto r^3$ ). Aur energy: $K = +GMm/2r$ , $U = -GMm/r$ , total $E = -GMm/2r$ . Total energy negative hai matlab orbit "bound" hai — bhaagne ke liye energy add karni padegi. Yaad rakho relation: $E = -K$ aur $U = 2E$ . Bas yeh teen formula aur ek balance equation — pura topic clear.

Circular orbit — velocity, period, energy

WHY does a circular orbit need a specific speed?

HOW to derive the orbital velocity (from scratch)

HOW to get the period $T$

HOW to get the energies

Worked Examples

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Orbital Mechanics & Astrodynamics

Connections