Capiert Posted January 20, 2023 Posted January 20, 2023 (edited) (Please) take a look at the math. If I reduce (=decrease) the eccentricity to almost zero (then) Kepler's_orbit_Period (3rd law) is (still) TKepler=ra1.5; but (however) at zero eccentricity e=0 (then) both the semi_major_axis ra & the semi_minor_axis rb are equal to the same (=identical), radius r=ra=rb if e=0 (but) having the circular orbit_Period Tcircle=2*Pi*((r/g)^0.5). Surely, for an Earth's satellite, both Orbit_Periods TKepler_(e=0)#Tcircle r1.5#2*Pi*((r/g)0.5), ^2 are NOT the same &/or can NOT be the same. Squaring both sides r3#4*Pi2*r/g, /r r2#4*Pi2/g, ^0.5 r#2*Pi/g, *g g*r#2*Pi, let the centrifugal_acceleration ac=g equal the gravitational free_fall (for an orbit) ac*r#2*Pi, because the centrifugal_acceleration ac=vc2/r multiplied by the common radius r as ac*r=vc2 is the (circular) orbit_speed squared vc2; & (that vc2) is (definitely) NOT 2*Pi. vc2= ac*r=2*Pi*(2*Pi*r2/T2) is a factor of *(2*Pi*r2/T2) larger than 2*Pi. Thus, (as far as I can see) Kepler's math is NONSENSE! (=it makes NO SENSE to me in its existing form); it has NO consistency with regular (Newtonian) Physics; & must have a different meaning. I hope that explains my problem (with Kepler('s laws)) well enough. 2 different (circular Orbit_period) formulas, which 1 is correct? (Kepler's or Newton's?) Please DON'T tell me Newton('s centrifugal_acceleration) is WRONG! ..because that is from where I derived my circular Orbit_Period from. Edited January 20, 2023 by Capiert Typo.
swansont Posted January 21, 2023 Posted January 21, 2023 1 hour ago, Capiert said: If I reduce (=decrease) the eccentricity to almost zero (then) Kepler's_orbit_Period (3rd law) is (still) TKepler=ra1.5; Kepler’s 3rd law is a proportionality, not an equality If your equation for period doesn’t have the form T^2/R^3 = constant, you’ve done it wrong. (g should not be in your equation. Put it in terms of M, R and G)
Janus Posted January 21, 2023 Posted January 21, 2023 Newtonian Physics says the period of an orbiting object is T = 2pi R^(3/2)/(GM)^(1/2) Thus 1/(2pi) = R^(3/2) / T(GM)^(1/2) Square both sides: 1/(2pi)^2 = R^3 / GM T^2 Move GM to the left side of the equation: 1/GM(2pi)^2 = R^3 / T^2 Invert both sides GM(2pi)^2 = T^2/R^3 So what Kepler's law states is that for any central body, there is a specific relationship between R and T Newton keeps the relationship. It just includes the mass of the central body, so if you know any two of T, R, or M. you can find the third. 1
Capiert Posted December 7, 2023 Author Posted December 7, 2023 (edited) On 1/21/2023 at 1:40 AM, swansont said: Kepler’s 3rd law is a proportionality, not an equality Thanks, that'( i)s exactly what I needed. Tkepler~ra1.5 Tkepler=k*ra1.5 On 1/21/2023 at 1:40 AM, swansont said: If your equation for period doesn’t have the form T^2/R^3 = constant, you’ve done it wrong. Ok. Rooted gives k=T/(r^1.5). On 1/21/2023 at 1:40 AM, swansont said: (g should not be in your equation. But, I do NOT see why (NOT). Surely you mean ONLY the end result; NOT the starting basis. ? On 1/21/2023 at 1:40 AM, swansont said: Put it in terms of M, R and G) From Janus's post On 1/21/2023 at 2:40 AM, Janus said: Newtonian Physics says the period of an orbiting object is T = 2pi R^(3/2)/(GM)^(1/2) Independent of this thread's presentation syntax: (as side_track) I quickly (also) see** (in her 1st line) (that) Newton's gravitational_force Fg=G*(M/R)*(m/r) is 2 mass_per_distance (linear_)ratios (M/R)*(m/r) multiplied together, & then multiplied by a proportionality constant G, where their (common) center_to_center distances R=r are (intended as) identical(ly the same). (I.e. Caution (poor) syntax, (clash): Distance, NOT just a radius. It's a ruff approximation. NOT to be confused with each (masses') radius separately (with their own (different) radius size); but instead the sum of both radii (distances + 1 height), (all) together. E.g. r=rm+rM+h=R, h=separation_height, surface to surface. Disclaimer: That'( i)s how I intuitively interpreted Fg, in a flash. That example of (symbol) r has NOTHING to do with my orbit radius r, later.) ** The Key (move=mano[e]uver): is shift (=move) the rooted_G*M (denominator) 1/((G*M)^0.5) (to) under the 2*π (numerator) (to obtain the proportionality constant k). T=(2*Pi/((G*M)^(1/2)))*(r^(3/2)). On 1/21/2023 at 2:40 AM, Janus said: Thus 1/(2pi) = R^(3/2) / T(GM)^(1/2) Square both sides: 1/(2pi)^2 = R^3 / GM T^2 Move GM to the left side of the equation: GM/(2pi)^2 = R^3 / T^2 (is (surely) enough info to follow thru correctly, & ignore the typo) On 1/21/2023 at 2:40 AM, Janus said: 1/GM(2pi)^2 = R^3 / T^2 Invert both sides ((2pi)^2)/GM = T^2 / R^3 On 1/21/2023 at 2:40 AM, Janus said: GM(2pi)^2 = T^2/R^3 On 1/21/2023 at 2:40 AM, Janus said: So what Kepler's law states is that for any central body, there is a specific relationship between R and T. Newton keeps the relationship. It just includes the mass of the central body, so if you know any two of T, R, or M, you can find the third. Great inspiration. Thanks. --- So giving it another go, again. A circular orbit_Period, is Tcircle=2*Pi*((R/g)^0.5). Equating, the Weight (Force) Wt=m*g of a mass m, with Newton's gravity_Force Fg=G*M*m/(R^2) for the Earth's_mass M, separated by their total (radial) distance R (center to center), (& due to Newton's 3rd law of opposite & equal reaction) we have (equal & opposite forces, balancing (out)) Wt=Fg m*g=G*M*m/(R^2), /m & dividing both sides by the (small(er)) mass m we get the free_fall (gravitational) acceleration g=G*M/(R^2). (Please Not(ic)e: that g is typically measured near the Earth's surface; but (g) gets smaller as the separation ((e.g. orbit_)radius R) gets large(r), e.g. to a GSO (geo_stationary orbit's) radius RGSO~g*(T^2)/(2*(Pi^2)) where the weightless(ness) gGSO=0 is zero.) Inserting that (g, as inverse factor 1/g=(R^2)/(G*M)) into the circular orbit_Period Tcircle=2*Pi*((R*(1/g))^0.5), gives Tcircle=2*Pi*((R*(R^2)/(G*M))^0.5) & we get R^3 under the root(_sign) Tcircle=2*Pi*((R^3)/(G*M))^0.5). The Key (move=mano[e]uver): is shift (=move) the rooted_G*M (denominator) 1/((G*M)^0.5) (to) under the 2*π (numerator) (to obtain the proportionality constant k). Tcircle=((2*Pi/(G*M))^0.5)*((R^3)^0.5), rooting the (R^3) to (R^3)^0.5=R^(3/2)=R^1.5 we get Tcircle=(2*Pi/((G*M)^0.5)*(R^3/2), 3/2=1.5 Tcircle=(2*Pi/((G*M)^0.5)*(R^1.5) or ingesting(=incorporating) the 2*π into under the root_sign, as rooted 4*(Pi^2); we have Tcircle=((4*(Pi^2)/(G*M))^0.5)*(R^1.5). Janus's (wonderful method) tells me: Newtonian Physics says the period of an orbiting object is T=2*Pi*(R^(3/2))/((G*M)^0.5), *1/(T*2*Pi) Thus 1/(2*Pi)=(R^(3/2))/(T*((G*M)^0.5)), ^2=Square both sides 1/((2*Pi)^2)=(R^3)/((T^2)*G*M), *G*M Move G*M to the left side of the equation (by multiplying both sides by G*M, gives): (G*M)/((2*Pi)^2)=(R^3)/(T^2), invert both sides ((2*Pi)^2)/(G*M)=(T^2)/(R^3), swap sides (T^2)/(R^3)=((2*Pi)^2)/(G*M), multiply by R^3 T^2=(((2*Pi)^2)/(G*M))*(R^3), ^0.5=root both sides T=(2*Pi/((G*M)^0.5)*(R^1.5) So Swansont's (searched (for (proportionality))) constant, is k=2*Pi/((G*M)^0.5) for the circular orbit_period T=k*(R^1.5) with orbit_radius R. Edited December 7, 2023 by Capiert -1
swansont Posted December 7, 2023 Posted December 7, 2023 14 hours ago, Capiert said: But, I do NOT see why (NOT). g is the acceleration due to gravity on the surface of the earth. It depends on the mass and radius of the earth. Those are not orbital parameters, and are specific to the earth, not other celestial bodies. There’s no reason for it to show up in an orbital equation; objects generally do not orbit at the surface of the earth. 14 hours ago, Capiert said: Please Not(ic)e: that g is typically measured near the Earth's surface; but (g) gets smaller as the separation ((e.g. orbit_)radius R) gets large(r), e.g. to a GSO (geo_stationary orbit's) radius RGSO~g*(T^2)/(2*(Pi^2)) where the weightless(ness) gGSO=0 is zero.) g is small at GSO, but it’s not zero. Objects in orbit are in freefall. That’s why they are weightless. It’s not because the local g is zero. (if there were no acceleration, it could not be in orbit. By Newton’s first law, if there is no acceleration, its motion would be in a straight line)
Capiert Posted December 8, 2023 Author Posted December 8, 2023 (edited) On 12/7/2023 at 5:20 PM, swansont said: g is the acceleration due to gravity wrt On 12/7/2023 at 5:20 PM, swansont said: on the surface of the earth. Motion is always relative to some reference. In that (acceleration motion) case (which is the (experimental) observable) it sure looks to me like wrt the Earth's surface. The rest (=explanation) is theory. On 12/7/2023 at 5:20 PM, swansont said: It depends on the mass and radius of the earth. As the formula says. On 12/7/2023 at 5:20 PM, swansont said: Those are not orbital parameters, Why NOT? They are variables that can be mannipulated for other examples. On 12/7/2023 at 5:20 PM, swansont said: and are specific to the earth, Their values are specific e.g. to Earth; but (as) the variables they do NOT need to be specific to (ONLY) the Earth. Nature does NOT prefer for her (natural) laws. Her laws are universal. Thus other (different) examples (must) exist. I'm NOT telling you anything new. You know that already. On 12/7/2023 at 5:20 PM, swansont said: not other celestial bodies. Other celestial bodies have their own values for the parameters. On 12/7/2023 at 5:20 PM, swansont said: There’s no reason for it to show up in an orbital equation; The same variables can be used more or less universally. You just have NOT seen the connection yet. Or am I wrong? You will naturally say yes if I am NOT mistaken. On 12/7/2023 at 5:20 PM, swansont said: objects generally do not orbit at the surface of the earth. If you are on the Earth, then you are (also) moving with it ((as) circular motion), without a (visible) change in height. An orbit can be equated (at least by me) to circular motion (which is) without height change. At least I can attempt to (try &) do that (if you can NOT). On 12/7/2023 at 5:20 PM, swansont said: g is small at GSO, but it’s not zero. If there is NO_fall vertically then I see NO acceleration (vertically, either). On 12/7/2023 at 5:20 PM, swansont said: Objects in orbit are in freefall. But a GSO has NO vertical_fall. It (vertical_"fall") is NOT observable. I think you are confusing that objects are "free" (NOT bound) to fall if they could; but they do NOT fall ("down", vertically) (perhaps because they are moving?). Each orbit radius (value) r=(vc^2)/ac has its own circumferential_speed vc=cir/T=2*Pi*r/T; but that centrifugal_acceleration ac=(vc^2)/r is only a math_construct, anyway. It stems from squaring the circumferential_speed vc^2=ac*r & then splitting that into an acceleration ac & (r radius_)distance product. It'( i)s otherwise total NONSENSE to express an orbit in "area" (units) per time_squared; when (circumferential_)speed vc=cir/T will do (already). The ac*r product was only created for convenience. On 12/7/2023 at 5:20 PM, swansont said: That’s why they are weightless. Weightless per word definition is "less" weight; NOT NO weight. But here I use it as otherwise intended, meaning: Weightless (as like floating) indicates zero (vertical_)acceleration. E.g. Einstein's Equivalence. I DON'T care what you "believe" (to explain), I am interested in the (experimental) observables (in order to formulate). If the weight Wt=m*g but the mass m is NOT falling (e.g. NOT changing its vertical_position) then its weight is (also) zero. Its (=The mass's) g=0 wrt the Earth's surface. It (=The mass) does NOT change vertical_height h=constant. Its (=The mass's) vertical motion is zero wrt the Earth's surface. That means (both): NO speed, & NO acceleration wrt the Earth's surface. (I can NOT understand why you think so rigidly. I suspect you forget that you are using only (math_)constructs.) On 12/7/2023 at 5:20 PM, swansont said: It’s not because the local g is zero. What do you mean by local (there)? Up in the sky at the mass? (y?); or down on the Earth's surface (n?). The reference is the Earth's surface; but it'( i)s observing the mass (from there=Earth's_surface). On 12/7/2023 at 5:20 PM, swansont said: (if there were no acceleration, it could not be in orbit. Why NOT? Orbit is ZERO vertical_acceleration (observed). On 12/7/2023 at 5:20 PM, swansont said: By Newton’s first law, if there is no acceleration, its motion would be in a straight line) You can still have (1D) "linear" acceleration (or deceleration, completely) without (circular) orbits. (Thus) That is NOT an exclusive decisive example to rely on. It (=Your(=That (particular)) argument) does NOT decide ANYTHING. (It'( i)s NOT a double blind proof.) I would need a (much) better example than that to convince me otherwise. Sorry. Edited December 8, 2023 by Capiert
swansont Posted December 8, 2023 Posted December 8, 2023 27 minutes ago, Capiert said: wrt Motion is always relative to some reference. This is about acceleration. Acceleration is not motion. You can have an acceleration when v = 0 Acceleration is not relative. If an object is accelerating, you can tell. 27 minutes ago, Capiert said: In that (acceleration motion) case (which is the (experimental) observable) it sure looks to me like wrt the Earth's surface. At the earth's surface. g = GM/r^2 where r is the radius of the earth, i.e. it is determined at the earth's surface 27 minutes ago, Capiert said: The rest (=explanation) is theory. As the formula says. Why NOT? They are variables that can be mannipulated for other examples. g does not depend on anything being in orbit. You can use g in an equation, but then you have to do things like correct for the fact that the actual acceleration is not g if you are not on the surface of the earth, which is an unnecessary complication of the formula. Follow the KISS principle (Keep It Simple, Stupid) 27 minutes ago, Capiert said: Their values are specific e.g. to Earth; but (as) the variables they do NOT need to be specific to (ONLY) the Earth. But causes unnecessary complication. Compare how many lines is took Janus to derive the equation and how many lines it took you to do it. 27 minutes ago, Capiert said: Other celestial bodies have their own values for the parameters. Indeed they do. 27 minutes ago, Capiert said: If you are on the Earth, then you are (also) moving with it ((as) circular motion), without a (visible) change in height. If you are standing on earth you are not in orbit. You are not in freefall. 27 minutes ago, Capiert said: An orbit can be equated (at least by me) to circular motion (which is) without height change. At least I can attempt to (try &) do that (if you can NOT). An orbit has more conditions than a circular motion. 27 minutes ago, Capiert said: If there is NO_fall vertically then I see NO acceleration (vertically, either). Any object moving in circular motion has an acceleration toward the center of the circle 27 minutes ago, Capiert said: But a GSO has NO vertical_fall. It (vertical_"fall") is NOT observable. The vertical fall is observable. But there is an equal amount of sideways motion as well, which is why the path is circular, and why vertical (y) and horizontal (x) aren't the most useful descriptions. In a circular coordinate system you use radial and tangential. 27 minutes ago, Capiert said: I think you are confusing that objects are "free" (NOT bound) to fall if they could; but they do NOT fall ("down", vertically) (perhaps because they are moving?). An object in orbit is not free; it must be in a bound state. You have to add energy to get it to an arbitrary distance Freefall just means you are acceleration at the local gravitational acceleration. 27 minutes ago, Capiert said: Each orbit radius (value) r=(vc^2)/ac has its own circumferential_speed vc=cir/T=2*Pi*r/T; but that centrifugal_acceleration ac=(vc^2)/r is only a math_construct, anyway. The acceleration is centripetal (center-seeking) 27 minutes ago, Capiert said: It stems from squaring the circumferential_speed vc^2=ac*r & then splitting that into an acceleration ac & (r radius_)distance product. It'( i)s otherwise total NONSENSE to express an orbit in "area" (units) per time_squared; when (circumferential_)speed vc=cir/T will do (already). It's not expressed as an area per time squared. 27 minutes ago, Capiert said: The ac*r product was only created for convenience. Weightless per word definition is "less" weight; NOT NO weight. Weightless (especially in this context) means no weight. 27 minutes ago, Capiert said: But here I use it as otherwise intended, meaning: Weightless (as like floating) indicates zero (vertical_)acceleration. Again, we should speak of radial, and there is a radial acceleration. I refer you again to Newton's first law. If there was no acceleration, the object would travel in a straight line. Do you deny the correctness of Newton's laws? 27 minutes ago, Capiert said: It (=The mass) does NOT change vertical_height h=constant. Its (=The mass's) vertical motion is zero wrt the Earth's surface. That means (both): NO speed, & NO acceleration wrt the Earth's surface. (I can NOT understand why you think so rigidly. I suspect you forget that you are using only (math_)constructs.) No, I'm using physics definitions. If you're going to use physics terminology, you have to use the same definitions. If you make up your own definitions you can't communicate ideas. 27 minutes ago, Capiert said: What do you mean by local (there)? At the location under discussion 27 minutes ago, Capiert said: Why NOT? Orbit is ZERO vertical_acceleration (observed). Patently untrue 27 minutes ago, Capiert said: You can still have (1D) "linear" acceleration (or deceleration, completely) without (circular) orbits. Acceleration is a change in velocity. It works in more than one dimension. Velocity is a vector; it has a magnitude (the speed) and a direction. If you change direction, there is an acceleration, even if speed is constant. If the discussion is about circular orbits, you can't be looking at this in one dimension. A circle has two dimensions
Capiert Posted December 9, 2023 Author Posted December 9, 2023 (edited) On 12/8/2023 at 8:56 PM, swansont said: This is about acceleration. Acceleration is not motion. That'( i)s new to me. What then is your definition of motion? Mine is, a change of position with time. On 12/8/2023 at 8:56 PM, swansont said: You can have an acceleration when v = 0. Please explain. I can NOT imagine acceleration without a change of position (wrt time). On 12/8/2023 at 8:56 PM, swansont said: Acceleration is not relative. Please fill me in, there. (=That (statement) does NOT make sense to me. E.g. I can NOT have a chicken without the (prerequisite=)egg, 1st. I can NOT build(=continue) upon NO foundation.) On 12/8/2023 at 8:56 PM, swansont said: If an object is accelerating, you can tell. As I said (=implied), I need help there. On 12/8/2023 at 8:56 PM, swansont said: At the earth's surface. g = GM/r^2 where r is the radius of the earth, i.e. it is determined at the earth's surface So it (=g) is wrt radius r. (y/n?) On 12/8/2023 at 8:56 PM, swansont said: g does not depend on anything being in orbit. g is only an acceleration (it's called free_fall acceleration), & it is vertical. (That means:) It has magnitude, & direction. (So it must (also) be a vector, too. ?) On 12/8/2023 at 8:56 PM, swansont said: You can use g in an equation, but then you have to do things like correct for the fact that the actual acceleration is not g That sentence is a head_twister for me. Would you mind explaining a bit better. I assume your 1st "g" is a symbol, but your 2nd "g" makes NO sense to me. My examples stated how (my) g varied, e.g. wrt height, because they were wrt to a reference. At least I knew (exactly) what & where I was dealing with. But with your definitions I go off into nirvana (because they seem not_founded e.g. NOT specified enough, or arbitrarily ambiguous). I DON'T mind an extra complication if it helps me understand clearly, instead of get lost. On 12/8/2023 at 8:56 PM, swansont said: if you are not on the surface of the earth, which is an unnecessary complication of the formula. Unnecessary? complicated? maybe for you; but NOT for me. On 12/8/2023 at 8:56 PM, swansont said: Follow the KISS principle (Keep It Simple, Stupid) I DON'T want to make it stupid; I want to make it thorough. It'( i)s too easy to get lost if a definition is lost (=missing). On 12/8/2023 at 8:56 PM, swansont said: But causes unnecessary complication. I DON'T like the complexity either; but your (physics) definitions (e.g. calculus) which dominate the scene, force me into a more (extensive) complex syntax just to make the (algebraic) distinction. Considering it (=my syntax) is only algebra I should NOT (even) need to state "average_" every time; NOR delta, etc. for simple differences. But I do (have to) (just to make the (algebraic) distinction.) It'( i)s that simple! but has become that complicated. On 12/8/2023 at 8:56 PM, swansont said: Compare how many lines is took Janus to derive the equation and how many lines it took you to do it. Yes! 2..3 lines. Tkepler=k*ra1.5 On 1/21/2023 at 1:40 AM, swansont said: If your equation for period doesn’t have the form T^2/R^3 = constant, you’ve done it wrong. Rooted gives k=T/(r^1.5). T=(2*Pi/((G*M)^(1/2)))*(r^(3/2)). Wt=Fg m*g=G*M*m/(R^2), /m g=G*M/(R^2). NO big deal. Right? On 12/8/2023 at 8:56 PM, swansont said: Indeed they do. If you are standing on earth you are not in orbit. You are not in freefall. Naturally, NOT falling. Standing, was (ONLY) an analogy (to an orbit equivalent). Take it or leave it. I find it useful. On 12/8/2023 at 8:56 PM, swansont said: An orbit has more conditions than a circular motion. Yes. Ellipse, etc. That'( i)s why I kept it simple KIS to start with ONLY circular_motion. Once the basics have been established correctly, the (complexity) details may be (included later &) improved upon. On 12/8/2023 at 8:56 PM, swansont said: Any object moving in circular motion has an acceleration toward the center of the circle That'( i)s an interesting point. On 12/8/2023 at 8:56 PM, swansont said: The vertical fall is observable. But there is an equal amount of sideways motion as well, I DON'T deny it. (I mean I will have to consider & deal with it, later, of course.) On 12/8/2023 at 8:56 PM, swansont said: which is why the path is circular, and why vertical (y) and horizontal (x) aren't the most useful descriptions. There you go preferring a (particular, alternative) coordinate system; when I simply convert (if needed) as option. On 12/8/2023 at 8:56 PM, swansont said: In a circular coordinate system you use radial and tangential. Why is that (suppose to be) better? I assume you mean simpler.? On 12/8/2023 at 8:56 PM, swansont said: An object in orbit is not free; Thus it can NOT fall.? On 12/8/2023 at 8:56 PM, swansont said: it must be in a bound state. What does that mean (more exactly, please)? On 12/8/2023 at 8:56 PM, swansont said: You have to add energy to get it to an arbitrary distance Yes. Faster (orbit_speed) is a larger radius. So if things like a sattelite go faster, then they automatically (ascend) & go up to a higher radius. By the same token if they slow (down) their orbit_speed vc=cir/T then they will (automatically) decrease their (orbit_)radius. On 12/8/2023 at 8:56 PM, swansont said: Freefall just means you are acceleration accelerating(?)(y/n?) On 12/8/2023 at 8:56 PM, swansont said: at the local gravitational acceleration. ? I assume you mean to say: Freefall just means you are accelerating at the local gravitational acceleration. But I am still NOT clear on what you mean by local & how you measure that acceleration, e.g. what reference do "you" use (to measure, with). How is such a measurement done=performed? On 12/8/2023 at 8:56 PM, swansont said: The acceleration is centripetal (center-seeking) Well done! On 12/8/2023 at 8:56 PM, swansont said: It's not expressed as an area per time squared. But its units are so. Energy too. (Kilogram) Meters_squared per second_squared. Should that mean otherwise. Orthogonality is NOT stated mathematically for a rectangle. On 12/8/2023 at 8:56 PM, swansont said: Weightless (especially in this context) means no weight. Yes, that is (also) my intended meaning for this thread. (But I also tolerate a NON_zero decrease as well, NOT that its usage is wrong, because it is NOT, but because I am (lazy &) accustomed to using inappropriate usage. E.g. I have bad habits. On 12/8/2023 at 8:56 PM, swansont said: Again, we should Please fill me in as to the necessity!, if any? On 12/8/2023 at 8:56 PM, swansont said: speak of radial, and there is a radial acceleration. I'm still trying to picture that. E.g. From what perspective. You talk about the (circle's) center.(?) But that tends to erase (some) things (perhaps definitions?) in my head. On 12/8/2023 at 8:56 PM, swansont said: I refer you again to Newton's first law. If there was no (center seeking) On 12/8/2023 at 8:56 PM, swansont said: acceleration, the object would travel in a straight line. Yes, most likely (I think, perhaps?). On 12/8/2023 at 8:56 PM, swansont said: Do you deny the correctness of Newton's laws? NO. I DON'T think so. I suspect they are accurate (observations=laws). (Sidetrack: My only beef (=complaint), is he (=Newton) did NOT always use them (3 laws), although they were the best ((universal) observations). E.g. for the (~12 h) tides.) Newton's 1st law of inertia is, objects in motion (tend to) stay in motion (thus (they) stay moving in a straight line); & objects at rest (tend to) stay at rest (in other words they do NOT change (what they were doing or NOT doing); til (Newton's 2nd law) they are accelerated or decelerated; by a (repelling) collision recoil-Newton's 3rd law. There is a particular detail concerning the in_line (NO_angle) affect which I have forgotten. Newton's 1st law says (indirectly) that there must be a reason for any (speed_)change that happens; a speed_change does NOT happen automatically on its own for NO reason. The 1st law is basically conservation of (average_)momentum. Or the law of constant(=consistent)_speed. The 2nd law is the law of acceleration. Particularly linear_acceleration. I interpret it (=The 2nd law) as the average_momentum squared. On 12/8/2023 at 8:56 PM, swansont said: No, I'm using physics definitions. If you're going to use physics terminology, you have to use the same definitions. If you make up your own definitions you can't communicate ideas. I CAN'T communicate to Physicists because they will be the last to understand, otherwise. (They seem to me, to be on their own (isolated) island, NOT always good.) A normal (NON_physicist) person usually gets my drift (faster, easier). On 12/8/2023 at 8:56 PM, swansont said: At the location under discussion Yes, but a ("local") location is usually typically wrt to some reference that is meaning 2 (different) points, NOT just 1; otherwise why a distinction. On 12/8/2023 at 8:56 PM, swansont said: Why NOT? Orbit is ZERO vertical_acceleration (observed). On 12/8/2023 at 8:56 PM, swansont said: Patently untrue Then I must conclude (=interpret, centripetal_acceleration) zero_seeking is (instead) deceleration. Disclaimer: I can NOT explain it (=that (radial), vertical orbit position, constant height) any other way. How otherwise can you get more from less? How (else) can you get a (linear, vertical) "acceleration" from a constant (circular_)"speed" vc=cir/T, cir=2*Pi*r. Normally (=Typically) it is the other way around. An acceleration will produce a speed (on a resting object). Newton's 2nd law. On 12/8/2023 at 8:56 PM, swansont said: Acceleration is a change in velocity. Yes. On 12/8/2023 at 8:56 PM, swansont said: It works in more than one dimension. Yes, but NOT always. Sometimes it works in ONLY 1 direction. But yes 3D automatically includes all 3 dimensions (for any "thing", in the universe). On 12/8/2023 at 8:56 PM, swansont said: Velocity is a vector; it has a magnitude (the speed) and a direction =angle. Let us say wrt the x_axis, e.g. (x,y,z)=(1,0,0). But that still (also) needs y, too. ? On 12/8/2023 at 8:56 PM, swansont said: . If you change direction, there is an acceleration, even if speed is constant. Yes. (Good example (last phrase).) On 12/8/2023 at 8:56 PM, swansont said: If the discussion is about circular orbits, you can't be looking at this in one dimension. Yes. On 12/8/2023 at 8:56 PM, swansont said: A circle has two dimensions . Yes. Edited December 9, 2023 by Capiert
swansont Posted December 9, 2023 Posted December 9, 2023 2 hours ago, Capiert said: That'( i)s new to me. What then is your definition of motion? Mine is, a change of position with time. A change in position with respect to time is velocity. 2 hours ago, Capiert said: Please explain. I can NOT imagine acceleration without a change of position (wrt time). Please fill me in, there. Toss something up in the air. It's moving up. Later, it moves down. There will be a point in time where it is at the top of its arc, and is motionless. v=0 2 hours ago, Capiert said: (=That (statement) does NOT make sense to me. E.g. I can NOT have a chicken without the (prerequisite=)egg, 1st. I can NOT build(=continue) upon NO foundation.) You can tell if you are accelerating. Even blindfolded. You don't need a point of reference. That's not true of velocity, which is relative to some frame of reference. 2 hours ago, Capiert said: As I said (=implied), I need help there. So it (=g) is wrt radius r. (y/n?) g is only an acceleration (it's called free_fall acceleration), & it is vertical. It's vertical in a cartesian coordinate system, where you have a y axis and an x axis. 2 hours ago, Capiert said: (That means:) It has magnitude, & direction. (So it must (also) be a vector, too. ?) It depends. g can be a scalar, 9.8 m/s^2, or you can use it as a vector if you acknowledge the direction, toward the center of the earth 2 hours ago, Capiert said: That sentence is a head_twister for me. Would you mind explaining a bit better. I assume your 1st "g" is a symbol, but your 2nd "g" makes NO sense to me. g is 9.8 m/s^2 If you are not at the surface of the earth, your gravitational acceleration will be different than this value. It's sloppy to use g as a generic acceleration 2 hours ago, Capiert said: My examples stated how (my) g varied, e.g. wrt height, because they were wrt to a reference. At least I knew (exactly) what & where I was dealing with. But with your definitions I go off into nirvana (because they seem not_founded e.g. NOT specified enough, or arbitrarily ambiguous). That's why one should use GM/r^2 2 hours ago, Capiert said: I DON'T mind an extra complication if it helps me understand clearly, instead of get lost. The problem being that you are posting this for others to read, and it just begs the question of why you would do calculations one way when you can do it in far fewer steps. It's confusing, and points to a lack of understanding of the concepts. 2 hours ago, Capiert said: Unnecessary? complicated? maybe for you; but NOT for me.] Seeing as how often you get a wrong answer, that's not readily apparent 2 hours ago, Capiert said: Considering it (=my syntax) is only algebra I should NOT (even) need to state "average_" every time; NOR delta, etc. for simple differences. But I do (have to) (just to make the (algebraic) distinction.) It'( i)s that simple! but has become that complicated. Most of the time you use average it's because you've decided to make things more complicated by insisting on using an average value. I might have something to do with your disdain of calculus. 2 hours ago, Capiert said: Naturally, NOT falling. Standing, was (ONLY) an analogy (to an orbit equivalent). Take it or leave it. I find it useful. Orbit defines a specific set of conditions. Standing on the earth is not an orbit. If you make up your own definitions of terminology you can't communicate with others 2 hours ago, Capiert said: Yes. Ellipse, etc. You miss the point. A circular orbit has more conditions than moving in a circle. 2 hours ago, Capiert said: That'( i)s why I kept it simple KIS to start with ONLY circular_motion. Once the basics have been established correctly, the (complexity) details may be (included later &) improved upon. Then get the basics right. An orbit means the gravitational force is what is keeping it moving in a circle. Not standing on the ground. 2 hours ago, Capiert said: There you go preferring a (particular, alternative) coordinate system; when I simply convert (if needed) as option. The choice of coordinate system seems logical given the problem. You can use x and y if you want, but then there is motion in both the x and y directions, which varies over time. The speed remains the same; v^2 =vx^2 + vy^2 (gosh, that's the equation of a circle!) 2 hours ago, Capiert said: Why is that (suppose to be) better? I assume you mean simpler.? Simpler and more descriptive. If you are trying t describe a satellite's orbit, vertical only works when it's directly overhead. At any other time "vertical" doesn't work. 2 hours ago, Capiert said: Thus it can NOT fall.? What does that mean (more exactly, please)? "free" and "bound" have definitions in physics. You would do well to learn such terminology. A free particle is one that could get infinitely far away without being subject to new force. A bound one cannot. In terms of energy, KE + PE < 0 for a bound particle 2 hours ago, Capiert said: Yes. Faster (orbit_speed) is a larger radius. No, it's not. For a circular orbit, v = sqrt(GM/r) As r gets larger, v decreases 2 hours ago, Capiert said: So if things like a sattelite go faster, then they automatically (ascend) & go up to a higher radius. The dynamics of it is more complicated than that, actually. You add energy, but this goes into increasing your potential energy (it's negative, but gets smaller in magnitude) and your speed decreases. KE goes down, but PE increases twice as fast. 2 hours ago, Capiert said: accelerating(?)(y/n?) ? I assume you mean to say: Freefall just means you are accelerating at the local gravitational acceleration. Yes 2 hours ago, Capiert said: But I am still NOT clear on what you mean by local & how you measure that acceleration, e.g. what reference do "you" use (to measure, with). How is such a measurement done=performed? Local means where you are. You can measure it with a spring scale. If you know your mass, the spring scale tells you the apparent weight, and from that you can get the acceleration. If you want to know the rotation (which is another form of acceleration), you can use a Foucalt pendulum. 2 hours ago, Capiert said: But its units are so. The SI units of acceleration are meters/ second^2 There's no area 2 hours ago, Capiert said: Energy too. (Kilogram) Meters_squared per second_squared. The m^2 in Joules does not represent an area 2 hours ago, Capiert said: I'm still trying to picture that. E.g. From what perspective. You talk about the (circle's) center.(?) A satellite is oin circular motion. The center of that circle is the center of the object it's orbiting. 2 hours ago, Capiert said: The 2nd law is the law of acceleration. Particularly linear_acceleration. I interpret it (=The 2nd law) as the average_momentum squared. Force is tied in with momentum, not momentum squared. 2 hours ago, Capiert said: Then I must conclude (=interpret, centripetal_acceleration) zero_seeking is (instead) deceleration. The acceleration in circular motion does not change the speed, it only changes the direction (velocity is a vector. Changing the direction of motion requires an acceleration) 2 hours ago, Capiert said: Disclaimer: I can NOT explain it (=that (radial), vertical orbit position, constant height) any other way. How otherwise can you get more from less? There is no more, no less. The speed is constant. 2 hours ago, Capiert said: How (else) can you get a (linear, vertical) "acceleration" from a constant (circular_)"speed" vc=cir/T, cir=2*Pi*r. You don't. It's not a linear (i.e. one-dimensional) system, and the direction is not vertical. The direction is "toward the center of the circle" (i.e., it's radial) 2 hours ago, Capiert said: Normally (=Typically) it is the other way around. An acceleration will produce a speed (on a resting object). Newton's 2nd law. If the force is perpendicular to the velocity, it will only change the direction. No work is done, so there is no change in KE
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