Built on Facts

Orbits and Escape

How about a quick little circular motion exercise, since that’s what I’m teaching in my recitation at the moment? We know the force equals mass time acceleration, so how about we put the force of gravity on the right side and the force required for uniform circular motion on the left:


And we’ll solve that for the velocity:


That’s the velocity you need to reach in order to attain circular orbit at a distance r above the earth’s center. M is the mass of the Earth (or whatever you happen to be orbiting, and m is the mass of the thing in orbit. Of course m cancels anyway. But what if we find the escape velocity by setting the initial kinetic energy equal to the potential energy at the surface of the planet? We get this:


Solve for v again:


Looks a lot like the expression for orbital velocity! We can compare them directly by dividing. Let me label orbital velocity with “o” and escape velocity with “e”:


So no matter your mass, or the mass of the planet, or the escape velocity, or any other detail the escape velocity is always 1.41 times greater than the circular orbit velocity. Tremendously important? Nah. But a cool problem nonetheless.


  1. #1 Winter Toad
    September 23, 2008

    It’s kind of important, it’s really just the virial theorem expressed in another form.

  2. #2 Paul Johnson
    September 23, 2008

    that is pretty cool

  3. #3 Jolly Bloger
    September 23, 2008

    So what does the path look like if you “orbit” (v perpendicular to g) with a velocity between 1 and 1.41 times stable orbiting velocity? Would you boomerang out and come straight back for a splat, or would you manage to circle a few times before spiraling in?

  4. #4 Winter Toad
    September 23, 2008

    In response to #3, if you gave the orbiting object a tangential velocity between 1 and 1.41 times the circular orbital velocity, it would end up in an elliptical orbit, returning to its starting point after one complete orbit. It would not hit the surface at all, but would repeat this elliptical orbit forever or until some perturbation changed the conditions. The point where you gave it the push would be the periapsis (point of closest approach to the central gravitational body), and a point 180 degrees around the orbit would be the apoapsis, at some greater distance from the central body.

  5. #5 Anthony
    September 23, 2008

    It’s not actually necessary to specify ‘circular'; it’s true for any orbit where your current distance from the primary is equal to the semimajor axis of the orbit.

    Or, you can just note that v^2 – 2GM/r = -GM/s, where s is the semimajor axis of the orbit. Escape velocity is equivalent to setting s to infinity, and thus v^2 – 2GM/r = 0.

    Of course, this all assumes relativistic effects are negligible.

  6. #6 Uncle Al
    September 23, 2008

    Are test bodies’ inertial and gravitation masses indistinguishable? Sure! – Galileo and inclined planes to PSR J1903+0327 (arxiv/0805.2396) 11% lightspeed equatorial spin magnetized neutronium versus hydrogen plasma. Also, perhaps not.

    The Equivalence Principle posulates isotropic vacuum. Like Euclid’s Fifth Postulate, it don’t gotta be. A massed sector chiral vacuum background (left foot) is inert in all prior observations. It is active for chemically identical opposite parity mass distributions (left foot fitted with left and right shoes). Thermodynamics constrains EP parity violation to less than 10^(-12) difference/average.

    That would be 20X S/N in a parity Eotvos experiment contrasting space groups P3(1)21 and P3(2)21 quartz, multiplied by 400 for 99.97% of test mass being active mass (versus EP composition experiments, all killed by the pulsar binary). The best EP test can run with 8000X higher sensitivity contrasting teleparallel vs. metric gravitation. Somebody should look.

    (No problem with quartz-locked watches: only space group P3(2)21 is commercially grown)

  7. #7 andy
    September 23, 2008

    One of the interesting implications of this is that it is harder to get to the Sun than it is to reach interstellar space. To get to the Sun, you have to cancel all of the Earth’s orbital velocity. To get to interstellar space, you need to reach escape velocity, which is a velocity change of “only” 41% of the orbital velocity…

  8. #8 Winter Toad
    September 23, 2008

    To extend the point of andy in #7, it requires, in the absence of gravitational slingshot maneuvers, less delta-v to reach almost any other place in the ecliptic plane than to reach our sun. It takes less impulse to go to any of many thousands of stars than it takes to go to our own sun. Of course, it takes a very long time to travel to another star on a barely-greater-than-zero-energy trajectory, but it takes less fuel to get to, say, one of the stars of Regulus than to get to the local star.

  9. #9 Carl Brannen
    September 24, 2008

    Another interesting ratio: The time required to make a circular orbit, with the time required to fall through zero and make it back (i.e. the most non circular orbit with the same maximum height). And the same problem, but with the assumption that the point falling to zero and back goes through a hole drilled through the core of the planet.

  10. #10 Bing McGhandi
    September 24, 2008

    Hey! I followed that! Sorry at such an enthusiasm over such a simple equation, but you should see a bunch of English Ph.D.s trying to figure out a tip. Sigh.

    From my POV, however, your commentators are bibbling.


  11. #11 andy.s
    September 24, 2008

    A long time ago, I read (in a book intended for high school students) that you could compute the orbital velocity by figuring out how fast it would take to reach your horizon while falling vertically.

    If you fall for a time t, you fall a distance of s = 1/2 at^2.

    The distance to the horizon at this height is sqrt(2Rs) sqrt(Ra)t, so the velocity needed is sqrt(Ra).

    Since a = GM/R^2, v_c = sqrt(GM/R), as expected.

    You drop 16 ft in one second, and the earth surface falls away 16 ft in 5 miles. So travel that 5 miles in 1 second and you’ll have the same altitude as you did before: i.e., you’re in orbit.

  12. #12 CCPhysicist
    September 25, 2008

    No one has mentioned that this means (in terms of energy) that orbit is halfway to anywhere? A rather famous Asimov (?) observation that explains the value of orbital assembly of interplanetary spacecraft.

  13. #13 joe nahhas
    February 20, 2009

    Astrophysicists Puzzle # 1: As Camelopardis Binary Stars Apsidal motion solution
    By Joe Nahhas
    The 30 years most studied Binary Stars motion puzzle that Einstein Harvard MIT Cal-Tech Stanford Princeton Oxford Cambridge University of Moscow University of Okinawa NASA and all other 100,000 space-time Physicists and Astrophysicists could not solve by any said or published Physics including 109 years of Nobel Prize winner Physics and physicists and 400 years of astronomy and is dedicated to the two DRS KH. F. Khailullin and V.S. Kozyreva of Moscow University who posted this motion puzzle in 1983 as not solvable by any Published Physics and still posted as motion puzzle on Smithsonian-NASA website SAO/NASA.

    Universal Mechanics Solution: For 350 years Physicists Astronomers and Mathematicians missed Kepler’s time dependent equation introduced here and transformed Newton’s equation into a time dependent Newton’ equation and together these two equations explain Quantum – Relativistic effects; it combines classical mechanics and quantum mechanics into one mechanics and explains “relativistic” effects as the difference between time dependent measurements and time independent measurements of moving objects and in practice it amounts to “Visual” effects.

    All there is in the Universe is objects of mass m moving in space (x, y, z) at a location
    r = r (x, y, z). The state of any object in the Universe can be expressed as the product

    S = m r; State = mass x location:

    P = d S/d t = m (d r/dt) + (dm/dt) r = Total moment
    = change of location + change of mass
    = m v + m’ r; v = velocity = d r/d t; m’ = mass change rate

    F = d P/d t = d²S/dt² = Total force
    = m(d²r/dt²) +2(dm/dt)(d r/d t) + (d²m/dt²)r
    = mγ + 2m’v +m”r; γ = acceleration; m” = mass acceleration rate

    In polar coordinates system

    r = r r(1) ;v = r’ r(1) + r θ’ θ(1) ; γ = (r” – rθ’²)r(1) + (2r’θ’ + rθ”)θ(1)

    r = r [cosθ î + sinθĴ] = r r (1); r (1) = cosθ î + sinθ Ĵ
    v = d r/d t = r’ r (1) + r d[r (1)]/d t = r’ r (1) + r θ'[- sinθ î + cos θĴ]

    v = r’ r (1) + r θ’ θ (1)

    Then θ (1) = -sine θ î +cosine θ Ĵ; r(1) = cosine θ î + sine θ Ĵ

    And, d [θ (1)]/d t= θ’ [- cosine θ î – sine θ Ĵ= – θ’ r (1)
    And, d[r (1)]/d t = θ’ [-sine θ’ î + cosine θ Ĵ] = θ’ θ(1)

    γ = d [r’ r (1) + r θ’ θ (1)] /d t
    = r” r (1) + r’d [r (1)]/d t + r’ θ’ r (1) + r θ” r (1) +r θ’ d [θ (1)]/d t

    γ = (r” – rθ’²) r (1) + (2r’θ’ + r θ”) θ (1)

    F = m[(r”-rθ’²)r(1) + (2r’θ’ + rθ”)θ(1)] + 2m'[r’r(1) + rθ’θ(1)] + (m”r) r(1)

    = [d²(mr)/dt² – (mr)θ’²]r(1) + (1/mr)[d(m²r²θ’)/dt]θ(1) = [-GmM/r²]r(1)

    d²(mr)/dt² – (mr)θ’² = -GmM/r² Newton’s Gravitational Equation (1)
    d(m²r²θ’)/dt = 0 Central force law (2)

    (2) : d(m²r²θ’)/d t = 0 < ==> m²r²θ’ = [m²(θ,0)φ²(0,t)][ r²(θ,0)ψ²(0,t)][θ'(θ, t)]
    = [m²(θ,t)][r²(θ,t)][θ'(θ,t)]
    = [m²(θ,0)][r²(θ,0)][θ'(θ,0)]
    = [m²(θ,0)]h(θ,0);h(θ,0)=[r²(θ,0)][θ'(θ,0)]
    = H (0, 0) = m² (0, 0) h (0, 0)
    = m² (0, 0) r² (0, 0) θ'(0, 0)
    m = m (θ, 0) φ (0, t) = m (θ, 0) Exp [λ (m) + ì ω (m)] t; Exp = Exponential
    φ (0, t) = Exp [ λ (m) + ỉ ω (m)]t

    r = r(θ,0) ψ(0, t) = r(θ,0) Exp [λ(r) + ì ω(r)]t
    ψ(0, t) = Exp [λ(r) + ỉ ω (r)]t

    θ'(θ, t) = {H(0, 0)/[m²(θ,0) r(θ,0)]}Exp{-2{[λ(m) + λ(r)]t + ì [ω(m) + ω(r)]t}} ——I
    Kepler’s time dependent equation that Physicists Astrophysicists and Mathematicians missed for 350 years that is going to demolish Einstein’s space-jail of time

    θ'(0,t) = θ'(0,0) Exp{-2{[λ(m) + λ(r)]t + ỉ[ω(m) + ω(r)]t}}

    (1): d² (m r)/dt² – (m r) θ’² = -GmM/r² = -Gm³M/m²r²

    d² (m r)/dt² – (m r) θ’² = -Gm³ (θ, 0) φ³ (0, t) M/ (m²r²)

    Let m r =1/u

    d (m r)/d t = -u’/u² = -(1/u²)(θ’)d u/d θ = (- θ’/u²)d u/d θ = -H d u/d θ
    d²(m r)/dt² = -Hθ’d²u/dθ² = – Hu²[d²u/dθ²]

    -Hu² [d²u/dθ²] -(1/u)(Hu²)² = -Gm³(θ,0)φ³(0,t)Mu²
    [d²u/ dθ²] + u = Gm³(θ,0)φ³(0,t)M/H²

    t = 0; φ³ (0, 0) = 1
    u = Gm³(θ,0)M/H² + Acosθ =Gm(θ,0)M(θ,0)/h²(θ,0)

    mr = 1/u = 1/[Gm(θ,0)M(θ,0)/h(θ,0) + Acosθ]
    = [h²/Gm(θ,0)M(θ,0)]/{1 + [Ah²/Gm(θ,0)M(θ,0)][cosθ]}

    = [h²/Gm(θ,0)M(θ,0)]/(1 + εcosθ)
    mr = [a(1-ε²)/(1+εcosθ)]m(θ,0)

    r(θ,0) = [a(1-ε²)/(1+εcosθ)] m r = m(θ, t) r(θ, t)
    = m(θ,0)φ(0,t)r(θ,0)ψ(0,t)

    r(θ,t) = [a(1-ε²)/(1+εcosθ)]{Exp[λ(r)+ω(r)]t} Newton’s time dependent Equation ——–II

    If λ (m) ≈ 0 fixed mass and λ(r) ≈ 0 fixed orbit; then

    θ'(0,t) = θ'(0,0) Exp{-2ì[ω(m) + ω(r)]t}

    r(θ, t) = r(θ,0) r(0,t) = [a(1-ε²)/(1+εcosθ)] Exp[i ω (r)t]

    m = m(θ,0) Exp[i ω(m)t] = m(0,0) Exp [ỉ ω(m) t] ; m(0,0)

    θ'(0,t) = θ'(0, 0) Exp {-2ì[ω(m) + ω(r)]t}


    = 2πa² [√ (1-ε²)]/T a² (1-ε) ²; θ'(0, 0) = 2π [√ (1-ε²)]/T (1-ε) ²

    θ'(0,t) = {2π[√(1-ε²)]/T(1-ε)²}Exp{-2[ω(m) + ω(r)]t

    θ'(0,t) = {2π[√(1-ε²)]/(1-ε)²}{cos 2[ω(m) + ω(r)]t – ỉ sin 2[ω(m) + ω(r)]t}

    θ'(0,t) = θ'(0,0) {1- 2sin² [ω(m) + ω(r)]t – ỉ 2isin [ω(m) + ω(r)]t cos [ω(m) + ω(r)]t}

    θ'(0,t) = θ'(0,0){1 – 2[sin ω(m)t cos ω(r)t + cos ω(m) sin ω(r) t]²}

    – 2ỉ θ'(0, 0) sin [ω (m) + ω(r)] t cos [ω (m) + ω(r)] t
    Δ θ (0, t) = Real Δ θ (0, t) + Imaginary Δ θ (0.t)

    Real Δ θ (0, t) = θ'(0, 0) {1 – 2[sin ω (m) t cos ω(r) t + cos ω (m)t sin ω(r)t]²}
    W(ob) = Real Δ θ (0, t) – θ'(0, 0) = – 2 θ'(0, 0){(v°/c)√ [1-(v*/c) ²] + (v*/c)√ [1- (v°/c) ²]}²

    v ° = spin velocity; v* = orbital velocity; v°/c = sin ω (m)t; v*/c = cos ω (r) t
    If v°/c < < 1; (v°/c) ² ≈ 0; And If v*/c << 1; (v*/c) ² ≈ 0 W (ob) = - 2[2π √ (1-ε²)/T (1-ε) ²] [(v° + v*)/c] ² W (ob) = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² radians W (ob) = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² degrees; Multiplication by 180/π W° (ob) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² degrees/100 years The circumference of an ellipse: 2πa (1 - ε²/4 + 3/16(ε²)²- --.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4) v (m) = √ [GM²/ (m + M) a (1-ε²/4)] v (M) = √ [Gm² / (m + M)a(1-ε²/4)] As Camelopardis Apsidal motion solution: Data T=3.431; r(m) =0.1499 m=3.3M(0) R(m) =2.57R(0) [v°(m),v°(M)]=[40,30] ε = 0.1695; 1-ε = 00.8305; r(M) =0.1111 M=2.5M(0) R(M) = 2.5R(0) ;m + M=5.8M(0) G=6.673x10^-11; M (0) = 1.98892x10^30kg; R (0) = 0.696x10^9m 1- ε²/4 = 0.9928; [√ (1-ε²)]/ (1-ε) ² = 1.43 a = [R (m)/r (m)] = (2.57/0.1499) (0.696x10^9) m With a (1-ε²/4) = (2.57/0.1499) (0.696x10^9) (0.9988) = 11.8470x10^9m And v (m) = √ [GM²/ (m + M) a (1-ε²/4)] = 110.1786325km/sec And v (M) = √ [Gm²/ (m + M) a (1-ε²/4)] = 145.435795km/sec And v (cm) = ∑m v/∑m = 125.3756853km/sec σ =√ {∑ [v-v (cm)] ²/2} =√ {[(110.1786325-125.3756853)² + (145.435795-125.7356853)²]/2} σ = 25.1659669 Spin: v° = v° (m) + v° (M) = 40km/s + 30km/s = 70km/sec And v (m) = 110.1786325km/sec Also, v (M) = 145.435795km/sec; v (cm)= 125.3756853km/sec σ = 25.1659669km/sec 1- With v* = v (m) + v (M) +σ=350,7803944km/sec; [√ (1-ε²)]/ (1-ε) ² = 1.43 T = 3.431days W° (observed) = (-720x36526/T) x {√ [(1-ε²)] (1-ε) ²} {[v* + v°]/c} ²= 15.0°/century Dr Guinan: W°= 15°/century 1989 2- With v * = 2v (cm) + σ = And v* = 2v (cm) + σ = 2[m v(m) + M v(M)]/(m + M) + √{{[v(m)-v(cm)]² + [v(M)-v(cm)]²}/2} = 275.9176729km/sec Then v* + v° = 275.9176729 + 70 = 345.9176729km/sec W° = (-720x36526/T) x {[√ (1-ε²)]/ (1-ε) ²} {[v* + v°]/c} ²= 14.6°/100 years 3- Khailullin: 1983 v (p) =110.4; v(s) = 145.8; σ=25.2685 2∑ m v/∑m + σ + 70=346.0185 W°= 14.6 °/century same as reported [same as published] References: Apsidal motion of As Cameloparids by Khailullin: 1983 Apsidal motion of As Camelopardis Edward Guinan and Frank Maloney: 1986 Joenahhas1958@yahoo.com all rights reserved

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