Circular motion

. It is a role played by an actual interaction, such as tension or an electric force, rather than an additional interaction to add to a free-body diagram. A purely radial centripetal force is perpendicular to the velocity, so it does no work and leaves the speed unchanged.

Circular orbit

 0  0

A circular orbit maintains constant distance from its centre of attraction. Its velocity changes direction even when its speed stays constant. The inward centripetal acceleration has magnitude

v^{2} / r

, so the radial force must satisfy

F_{r} = - m v^{2} / r

. For a Newtonian gravitational field with standard gravitational parameter

μ

, this gives

v = μ / r

Circular orbital speed from surface gravity ( $v = g_{s} R^{2} / r$ )

 0  0

For a spherical central body of radius

R

with surface gravity

g_{s} = GM / R^{2}

, a circular test-particle orbit at radius

r

has

v = g_{s} R^{2} / r

. surface gravity is used to infer

GM

, not incorrectly kept constant at orbital height.

Circular orbit condition for a radial kinetic Lagrangian ( $p_{θ}^{2} = - 4 b (R)^{2} c^{'} (R) / b^{'} (R)$ )

 0  0

Let a planar Lagrangian have the form

L = a (r) \overset{r}{˙}^{2} + b (r) \dot{θ}^{2} + c (r)

, with smooth coefficients and

b (r) > 0

. Its cyclic coordinate

θ

has constant conjugate momentum

p_{θ} = 2 b (r) \dot{θ}

. For a circular orbit

r = R

, the radial Euler-Lagrange equation reduces to

b^{'} (R) \dot{θ}^{2} + c^{'} (R) = 0

. If

b^{'} (R) \neq = 0

, eliminating

\dot{θ}

gives the displayed condition. A real orbit requires

- c^{'} (R) / b^{'} (R) \geq 0

; the radial kinetic coefficient

a

does not enter this constant-radius condition, although it affects radial perturbations. If

b^{'} (R) = 0

, the original condition must be used rather than divided by zero.

Circular-orbit energy ratio in a power-law potential ( $2 K = s U (U = A r^{s})$ )

 0  0

A circular orbit in a central potential

U (r) = A r^{s}

requires inward force

U^{'} (r) = m v^{2} / r > 0

. Thus

2 K = r U^{'} (r) = s U

, giving

K / U = s /2

for the specified zero of potential. The ratio depends on that choice of zero; adding a constant to

U

leaves the orbit unchanged but changes

K / U

. This is the circular special case of the virial theorem.

Circular orbit as continuous free fall ( $r (t) = a (cos ω t, sin ω t), ω^{2} = μ / a^{3}$ )

 0  0

For a fixed spherical gravitating source, place a test particle at

(a, 0)

with initial velocity

(0, μ / a)

, where

a

is outside the source. Its exact circular orbit is

r (t) = a (cos ω t, sin ω t), ω = \frac{μ}{a ^{3}} .

(1)

Differentiating twice gives

\ddot{r} = - ω^{2} r = - μ r / a^{3}

, precisely the Newtonian gravitational field at each point of that circle. There is no outward physical force to cancel gravity in this inertial frame. Gravity supplies the inward centripetal acceleration that continually changes the velocity's direction.

Figure 1.
A circular free-fall orbit, inward acceleration, and the straight path the same initial velocity would produce without gravity
. **Details**.

Over a short interval, the displacement from the straight tangent path begins inward with magnitude

\frac{1}{2} (μ / a^{2}) t^{2}

. The curvature of the orbit matters at the same order: the particle moves sideways while continually falling away from its instantaneous tangent. The full circular solution remains at radius

a

rather than striking the source.

Its speed is

μ / a

, its oscillation period is

2 π a^{3} / μ

, and its specific orbital energy is

- μ / (2 a)

. Removing gravity while keeping the initial velocity would give the straight path

(a, aω t)

, rather than a stationary particle. This contrast explains why orbital weightlessness is compatible with a substantial gravitational force: a freely coasting spacecraft and its contents fall together. ^[ref]

Circular-orbit energy gradient ( $d ε = Ω \cdot d h$ )

 0  0

In a spherically symmetric gravitational potential

Φ (r)

, a circular orbit has

r Ω^{2} = Φ^{'} (r)

, specific orbital energy

ε = Φ + r^{2} Ω^{2} /2

and specific angular momentum

h = r^{2} Ω n

, with

∣ n ∣ = 1

. Differentiating gives

ε^{'} = 2 r Ω^{2} + r^{2} Ω Ω^{'} = Ω (r^{2} Ω)^{'}

. Changes in orbital orientation have

n \cdot d n = 0

, so

Ω \cdot d h = Ω d (r^{2} Ω) = d ε

. For a particle of variable mass

M

and total angular momentum

H = M h

, this implies

d (Mε) = (ε - Ω h) d M + Ω \cdot d H

. Moreover

d (ε - Ω h) / d r = - h Ω^{'}

, which is positive when the orbital angular velocity decreases outward.

Circular orbits in an exponential central potential ( $l^{2} = r^{3} e^{- r}$ )

 0  0

For unit mass with specific angular momentum

l

V (r) = - e^{- r}

, the effective potential is

- e^{- r} + l^{2} / (2 r^{2})

. A positive-radius circular orbit satisfies the displayed equation. Since

r^{3} e^{- r}

has its unique maximum

27/ e^{3}

r = 3

, exactly two radii occur for

0 < l^{2} < 27/ e^{3}

. Their radial stability follows from

V_{eff}^{''} = e^{- r} (3/ r - 1)

: the inner orbit is stable at fixed angular momentum, the outer unstable. The case

l = 0

has no finite-radius circular orbit despite satisfying the upper bound.

Circular-orbit stability for a power-law central force ( $Ω_{r}^{2} = (n + 3) k a^{n - 1}$ )

 0  0

For the radial acceleration

- k r^{n}

with

k > 0

, a circular orbit of radius

a

has squared specific angular momentum

h^{2} = k a^{n + 3}

. At fixed

h

, its radial small oscillations have squared frequency

(n + 3) k a^{n - 1}

, so the orbit is stable for

n > - 3

. This is a special case of the effective potential stability criterion.

Small radial oscillations in a power-law potential ( $ω_{r} /Ω = k + 2 (V = β r^{k})$ )

 0  0

For a central potential

V = β r^{k}

and fixed nonzero angular momentum

L

, the effective potential

L^{2} / (2 m r^{2}) + β r^{k}

has a nondegenerate stable circular orbit when

β k > 0

and

k > - 2

. Its radius obeys

r_{0}^{k + 2} = L^{2} / (m β k)

. The radial small oscillation frequency satisfies

ω_{r}^{2} = (k + 2) Ω^{2}

, where

Ω = ∣ L ∣/ (m r_{0}^{2})

. Rational frequency ratios close the linearized near-circular sketch, but generally not the exact finite-amplitude orbit; Bertrand's theorem distinguishes the exceptional harmonic and Kepler potentials.

Circular orbit in a quadratic central potential ( $r_{0}^{4} = h^{2} / (2 k m^{2})$ )

 0  0

For

V (r) = km r^{2}

, fixed nonzero angular momentum

h

gives the effective potential

V_{eff} = h^{2} / (2 m r^{2}) + km r^{2}

. A positive-radius circular orbit exists when

k > 0

, at the displayed radius, and

V_{eff}^{''} (r_{0}) = 8 km > 0

proves radial stability by the effective potential stability criterion. Its angular speed has magnitude

2 k

and its small radial oscillation frequency is

8 k

. If

k \leq 0

, no such orbit with nonzero angular momentum exists; for

h = 0, k > 0

, the sole minimum is at the origin, not a positive-radius circular orbit.

 Ancestors (4)

 Incoming links (29)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (1)

Circular motion by

Wikipedia Bot  1

 View more

Circular motion refers to the movement of an object along the circumference of a circular path. This type of motion can be classified into two main categories: 1. **Uniform Circular Motion (UCM)**: In uniform circular motion, an object moves around a circle at a constant speed. Although the speed remains constant, the velocity is constantly changing because the direction of the object's motion is continually changing. This is due to centripetal acceleration, which acts towards the center of the circular path.

 Read the full article

  See all articles in the same topic Create my own version

 Discussion (0)  Subscribe (1) New discussion

 Discussion (0) New discussion