Circular orbit — velocity, period, energy: Question bank

3.2.13 · D5 · Physics › Orbital Mechanics & Astrodynamics › Circular orbit — velocity, period, energy

Neeche ke har cheez ke liye jo char core facts hain, taaki scroll na karna pade: $v = \frac{GM}{r}, T = 2 π \frac{r ^{3}}{GM}, E = - \frac{GM m}{2 r}, K = + \frac{GM m}{2 r}, U = - \frac{GM m}{r}$ Yahan $G$ Newton's gravitational constant hai, $M$ badi central mass hai, $m$ orbiting mass hai, $r$ orbit radius hai (center se measure hoti hai, surface se nahi), $v$ speed hai, $T$ period hai, $K$ / $U$ / $E$ kinetic/potential/total energy hain.

True ya false — justify karo

Same altitude par ek heavier satellite ek light wale se slower orbit karta hai.

False.

\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}

mein satellite mass

m

dono sides par cancel ho jaata hai, isliye

v = GM / r

mein koi

m

nahi hai — ek pankh aur ek bus same

r

par same speed aur period share karte hain.

Ek higher circular orbit mein zyada total energy hoti hai par kam kinetic energy.

True.

E = - \frac{GM m}{2 r}

badhe

r

ke saath zero ki taraf badhta hai (zyada total energy), phir bhi

K = \frac{GM m}{2 r}

ghatta hai — extra energy saari (negative) potential energy

U

ko raise karne mein chali gayi.

Agar total energy

E

exactly zero hai, toh orbit ek bahut bada circle hai.

False.

E = 0

escape boundary hai: path ek parabola hai, closed circle nahi. Ek circle ko

E = - \frac{GM m}{2 r} < 0

chahiye; kisi bhi circle ki energy strictly negative hoti hai.

Orbit radius double karne se period double ho jaati hai.

False.

T \propto r^{3/2}

, isliye

r

double karne par

T

2^{3/2} \approx 2.83

se multiply hoti hai, 2 se nahi. Sirf woh quantities jo

r

mein linear hain woh double hongi.

Ek circular orbit ke liye, kinetic energy minus total energy ke barabar hoti hai.

True.

K = + \frac{GM m}{2 r}

aur

E = - \frac{GM m}{2 r}

, isliye

K = - E

. Yeh

K = - E

U = 2 E

pattern

1/ r

force ke liye virial theorem hai — dekho Virial Theorem.

Moon aur ek low satellite har kilogram par same gravity feel karte hain, isliye woh same speed par orbit karte hain.

False. Gravity per kilogram

GM / r^{2}

hai, jo

r

par depend karta hai. Moon bahut door hai (bada

r

), isliye uska

g

bahut kamzor hai aur uski orbital speed

v = GM / r

ek low satellite ke muqable mein bahut kam hai.

E = 0

par object "just barely escaped" hai, matlab woh infinity par rest mein pahunchta hai.

True.

E = K + U

; jab

r \to \infty

U \to 0

, isliye

E = 0

force karta hai

K \to 0

, yaani woh vanishing speed ke saath infinity tak coast karta hai. Dekho Escape Velocity (

E = 0

boundary,

v_{esc} = 2 v_{c i r c}

Ek given radius par escape speed circular speed ki exactly do guna hoti hai.

False. Yeh

2

guna hai,

2

guna nahi.

v_{esc} = 2 GM / r = 2 GM / r = 2 v_{c i r c}

— factor square root ke andar hai.

Error dhundho

"400 km par satellite ki potential energy nikalne ke liye

U = m g h

use karo."

Error flat-Earth constant-

g

formula use karna hai. Orbital distances par

g = GM / r^{2}

bahut change hota hai, isliye tumhe

U = - \frac{GM m}{r}

use karna hoga;

m g h

sirf surface ke paas tangent-line approximation hai — dekho Gravitational Potential Energy.

"Kyunki higher orbit mein zyada energy hoti hai, rocket ko satellite ko wahan pahuanchane ke liye speed up karna hoga."

Deceptive hai. Final orbit slower hoti hai (

v = GM / r

r

ke saath ghatta hai). Badi energy

U

raise karti hai,

K

nahi. Forward fire karna transfer shuru karta hai, par destination circle ki kinetic energy start se kam hoti hai.

"Gravity centripetal force plus ek extra outward centrifugal force hai jo ise balance karti hai."

Inertial frame mein koi real outward force nahi hai. Sirf gravity centripetal force hai; kuch bhi "balance" nahi karta — ek unbalanced inward force exactly wahi hai jo straight path ko circle mein mod deti hai. Dekho Centripetal Force and Uniform Circular Motion.

"Kyunki

T^{2} \propto r^{3}

GM

par depend karta hai, toh yeh satellite ki mass

m

par bhi depend karta hai."

Constant

\frac{4 π ^{2}}{GM}

hai — sirf central mass

M

appear karta hai, orbiting

m

nahi. Newton's law deta hai

\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}

aur

m

's cancel ho jaate hain

T

compute hone se pehle.

"Satellite isliye upar rehta hai kyunki gravity wahan tak pahunchne ke liye itni kamzor hai."

Gravity bilkul wahan tak pahunchti hai — wahi inward pull exactly orbit ko curve karta hai. Satellite isliye upar rehta hai kyunki uski sideways speed itni hai ki ground utni hi tezi se curve ho jaata hai jitna woh girta hai, is liye nahi ki gravity absent hai.

K = \frac{1}{2} m v^{2}

aur

U = - \frac{GM m}{r}

, isliye

E = K + U = + \frac{GM m}{2 r}

Sign slip hai.

K = + \frac{GM m}{2 r}

aur

U = - \frac{GM m}{r}

, isliye

E = \frac{GM m}{2 r} - \frac{GM m}{r} = - \frac{GM m}{2 r}

— negative, kyunki

∣ U ∣ = 2 K

Why questions

Satellite ki mass speed se kyun drop out ho jaati hai par energy se nahi?

Speed ek force balance se aati hai jahan

m

dono sides par appear hota hai aur cancel ho jaata hai. Energies

K, U, E

actual joules hain jo store hote hain, aur mass double karne par genuinely har ek double ho jaata hai — cancel karne ke liye koi doosra

m

nahi hota.

Har bound orbit ki total energy negative kyun hoti hai?

Hum

U = 0

infinity par set karte hain; ek bound object us reference ke neeche baithta hai kyunki use infinity se andar kheenchne par energy release hui. Negative

E

literally matlab hai "use free karne ke liye tumhe energy add karni padegi."

T^{2} \propto r^{3}

Newton se seedha kyun aa jaata hai, jabki Kepler ko yeh data se discover karna pada?

v = GM / r

T = 2 π r / v

mein substitute karne par automatically

T = 2 π r^{3} / GM

milta hai. Kepler ne planetary data mein pattern dekha; Newton ne dikhaya ki yeh zaroor ek inverse-square force se follow karta hai — dekho Kepler's Three Laws aur Newton's Law of Universal Gravitation.

Circle ke liye gravity ko exactly

\frac{m v ^{2}}{r}

supply kyun karni chahiye — na zyada, na kam?

Zyada inward force path ko circle se tighter curve karta hai (woh spiral in ho jaata hai); kam hone par body bahar drift karti hai (ek ellipse ya escape). Sirf exact match

r

ko constant rakhta hai, jo ki circle ki definition hai.

Higher orbit slower kyun hoti hai jabki usmein zyada energy hoti hai?

v

aur

E

dono

r

ke saath change hote hain par opposite sense mein:

v \sim r^{- 1/2}

ghatta hai, jabki

E \sim - r^{- 1}

zero ki taraf badhta hai. Badhti energy potential energy ke roop mein store hoti hai, isliye speed ghat sakti hai jabki total energy badhti hai.

Hum gravity ko poori orbit ke liye exactly center ki taraf point karte hue kyun treat kar sakte hain?

Ek circle mein har point center se same distance

r

par hota hai, aur ek spherical mass se gravity hamesha radially inward point karti hai. Isliye force constant magnitude

GM m / r^{2}

rakhti hai aur hamesha center ki taraf aim karti hai — perfectly constant centripetal pull supply karne ke liye.

Edge cases

Kya hota hai

v

ko jab

r \to \infty

(infinitely badi orbit)?

v = GM / r \to 0

. Infinitely door circular orbit ki speed vanishing hoti hai aur,

T \sim r^{3/2}

se, infinite period hoti hai — yeh barely move karta hai.

Kya hota hai

v

ko jab

r \to 0

(center ko skim karte hue)?

v = GM / r \to \infty

. Mathematically required speed blow up karti hai; physically pehle surface se takra jaoge, isliye real orbits ki minimum radius hoti hai (body ki radius plus atmosphere).

Agar tum orbital speed set karo par

r

actual surface radius lene do, kya yeh valid orbit hai?

Sirf idealized airless, mountain-free limit mein. Ek real planet par surface, terrain, aur atmosphere grazing orbits ko impossible banate hain; formula valid rehta hai par geometry nahi.

E > 0

(positive total energy) kya correspond karta hai?

Ek unbound trajectory — ek hyperbola. Body ke paas infinity tak pahunchne ke liye zyada than enough energy hai speed ke saath, isliye woh kabhi return nahi karta; yeh circular regime se aage hai aur Elliptical Orbits and the Vis-viva Equation se link karta hai.

Us orbit mein kya special hai jahan

v

local escape speed ke barabar hoti hai?

Koi aisa circular orbit exist nahi karta: us radius par

v_{esc} = 2 v_{c i r c}

tak pahunchne par

E = 0

milta hai, jo ek parabola hai, circle nahi. Us

r

par circular se upar koi bhi speed path ko ellipse, parabola, ya hyperbola mein khol deti hai.

Agar do satellites ka period same ho, toh unke radii ke baare mein kya true hona chahiye?

Unki radius same honi chahiye.

T = 2 π r^{3} / GM

fixed

M

ke liye

r

mein one-to-one hai, isliye equal periods force karte hain equal

r

(assuming woh same central body orbit karte hain).

Recall Is page ko close karne se pehle ek-line self-test

Zor se bolo: "Higher orbit slower kyun hai phir bhi higher-energy?" Answer ::: Kyunki $v \sim r^{- 1/2}$ ghatta hai jabki $E \sim - r^{- 1}$ zero ki taraf badhta hai — extra energy potential energy ke roop mein store hoti hai, speed ke roop mein nahi.

Connections

Parent: Circular orbit — velocity, period, energy
Newton's Law of Universal Gravitation — woh inward force jise hum balance karte hain.
Centripetal Force and Uniform Circular Motion — $m v^{2} / r$ requirement supply karta hai.
Kepler's Three Laws — kyun $T^{2} \propto r^{3}$ coincidence nahi hai.
Gravitational Potential Energy — $U = - GM m / r$ ki origin aur kyun $m g h$ fail karta hai.
Escape Velocity — $E = 0$ boundary, $v_{esc} = 2 v_{c i r c}$ .
Elliptical Orbits and the Vis-viva Equation — jahan $E > 0$ aur $E = 0$ paths rehte hain.
Virial Theorem — explain karta hai $K = - E$ , $U = 2 E$ .