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B Need help calculating acceleration out of a gravity well

  1. Feb 24, 2017 #1
    Hey, orbital mechanics!

    I can't find what I need to figure this out on the internet, and I don't do calculus so I don't understand all that I find. Help me make my next sci-fi novel plausible?

    I just did an Oberth maneuver around Sol, 21 radii (.0977 AU, 14,616,000 km) from center. Did I need orbital velocity just before perigee? Or more? Is orbital velocity at that distance 436.7 km/s, or is http://keisan.casio.com/exec/system/1360310353 wrong?

    I had 1.5 gravity of acceleration at perigee, and I gained solar escape velocity ( I need it) and then some? Is that 617.6 km/s+? Or is keisan wrong there, too?
    http://www.calctool.org/CALC/phys/astronomy/escape_velocity says 138.74 km/s V3; keisan says V3 (escape velocity from the solar system) is an impossible 50,570,000 km/s, 167c! Something must be wrong with that calculator.

    Again, I had 1.5 g of thrust at perigee ten solar diameters from the surface. If I had it all through the needed burn, how much velocity would I gain? How long would that burn last? Would I have solar escape velocity? (I must) Or more? Or do I need more thrust?

    If I then lost thrust for three days, and so came around the sun a little behind my desired vector (or is the right term orbit?) for an economical orbit to Jupiter rendevouz; then regained 0.2 g thrust 12 hrs/24 for two months (no thrust half the time); then 1.0 g 24/7 for as long as it takes to get to Jupiter orbit, how long would it take to get to Jupiter, and what would be a reasonable velocity when I got there?

    And if I caught up with Jupiter from behind, and did a gravity sling with 3 g of thrust at that perigee, using the maneuver to angle up above the plane of the ecliptic a little, how much velocity might I reasonably gain?

    If you can show me the algebra, I can tweak the numbers until the math works with the plot.

    Thanx huge!

  2. jcsd
  3. Feb 24, 2017 #2


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    Staff: Mentor

    Hi Jakko,

    Welcome to Physics Forums!

    Perhaps there's a problem with how you're using or interpreting the results of the keisan.casio online calculator? Can you post a screenshot of the calculation you set up?

    At 21 solar radii I find that the circular orbit velocity is 95.3 km/s. The escape velocity is ##\sqrt{2}## times that, or 134.8 km/s. This is using a value of 132712440018*km3*sec-2 for the GM (also called the gravitational parameter μ) for the Sun, and 6.9599x105 km for the radius of the Sun.

    Note that the terms perigee and apogee generally apply to bodies orbiting the Earth. The equivalent terms for bodies orbiting the Sun are perihelion and aphelion.
    Perihelion occurs at a specific instant in time on an orbit, so a 1.5 g acceleration at perihelion is a bit vague. The time duration of the acceleration would depend upon the initial speed going into the maneuver, giving the required Δv to achieve escape velocity. So more information is needed about the initial orbit going in. Where did the ship "fall" from? Assuming that the required Δv is not too large and the burn duration short enough, the radial distance from the Sun won't change greatly over the burn so you can approximate the burn time using ##Δv = a\;Δt##, where ##Δt## is the time of the burn and ##a## is the acceleration (your 1.5 g).
    Constant acceleration orbit trajectories are difficult to handle mathematically. Time on orbit calculations are tricky enough for un-powered trajectories except in special circumstances (for example, see "Hohmann transfer orbit"). When thrust is involved for anything other than very short impulses (so that the ship moves from one conic-section orbit to another in negligible time) the math becomes all but intractable and computer simulations are used instead.

    If you achieved solar escape speed at 21 solar radii, then by the time you reached the orbit of Jupiter on an unpowered orbit your speed would be reduced to the solar escape speed at that distance. Jupiter's orbital radius is about 5.2 AU, or 7.78x108 km. Solar escape speed at that distance is 18.47 km/sec. If you are accelerating along the way though, that speed will be different and would depend on how the acceleration was used (course corrections to maneuver into the right approach trajectory, speeding up or slowing down to meet Jupiter wherever it happens to be on its orbit, and so on). It will be very dependent on details.
  4. Feb 25, 2017 #3
    Hi, gneil.
    I had to add the numbers back into the fields above the line, so it's not quite a screen shot; the numbers are as on the screen.
    the planet Sun
    Mass M 1.989E30 kg
    radius r 6.96E5 km
    Orbit radius R .0977 AU [21 solar radii, 14,616,000 km from surface]
    Galaxy's mass Mc 2.8E41 kg
    Gravity G 6.67438E-11 [the negative sign is on screen. Is it correct? No units given. Gravity that tiny a force? When I cleared the calculator, then reset it to Sun, it cleared the GravityG field permanently. Total lack of anything in that field did not change the outcome, below ]

    the first cosmic velocity V1 436.7 km/s Orbit around the planet
    the second cosmic velocity V2 617.6 km/s Escape velocity from the planet
    orbital speed Vc 35,760,000 km/s
    orbital period Tc 8.138E-8 year
    the third cosmic velocity V3 50,570,000 km/s Escape velocity from the solar system [and that's many times the speed of light--can't be correct, can it?]
    The third cosmic velocities of the Sun and the moon are calculated as the escape velocities from the galaxy and the Earth respectively.
     Earth's mass=5.974E24
     Sun's mass=1.989E30
     Galaxy's mass=2.8E41
    "134.8 km/s." is that escape velocity from the sun (what I need) or from the solar system? http://www.calctool.org/CALC/phys/astronomy/escape_velocity says 138.74 km/s "escape velocity," but it doesn't exactly say whether from sun or solar system, though I gather it means the sun. That's close enough to your number for corroboration. Unfortunately that calculator doesn't tell me anything else.
    "This is using a value of 132712440018*km3*sec-2 for the GM (also called the gravitational parameter μ) for the Sun, and 6.9599x10^5km for the radius of the Sun." Is GM the same as GravityG in the Keisan calculator? When I input that number in that field, the first, second, and third cosmic velocities change to 1.947E+13; 2.754E+13; and 1.617E+13 respectively. That's 1.617 x 10^13? That can't be right
    perihelion and apihelion. I remember now. What do you call closest approach to Jupiter, and to any planet in general?
    "Perihelion occurs at a specific instant in time on an orbit, so a 1.5 g acceleration at perihelion is a bit vague. The time duration of the acceleration would depend upon the initial speed going into the maneuver, giving the required Δv to achieve escape velocity. So more information is needed about the initial orbit going in. Where did the ship "fall" from?"

    It "fell" from Mars orbit, but it had thrust (or rather the cometary body it was hidden in, mounted with a dozen fusion thrusters, had thrust) much of the way down. I haven't specified a velocity at closest approach, because I don't know what's reasonable. The faster you're travelling when you hit perihelion/closest approach, the shallower the change in your course, right? I can work with that. I'd rather it had very high velocity at closest approach, and way more than escape velocity when the Oberth maneuver is complete. I should explain further:

    My crew is stealing Earth's first crude tin-can attempt at a sub-light, oort-cloud grazing "star" ship, hence hiding it in an iceball that was supposed to be dropped into lunar orbit for processing; they overrode the controlls and thrust much of the way to Sol. They want to make rendevous with Jupiter, pretty much on the other side of Sol, to gain whatever velocity they can from a gravity sling, but more importantly to angle up-or down--don't know yet--to leave the plane of the ecliptic. They don't want to go through the asteroid belt too fast--1 million kph should give them time to dodge rocks big enoug to see--but once through they are going to accelerate constantly to a very high fraction of c, on their way to another star.
    They have fusion engines, and "forced compound" (Star Trek universe) coils in the ship's hull that create a field that messes with the Higgs field such that their apparent mass and inertia are 1/10 of what they should be; without the "Inertial Reduction field" they have 1/10 g of thrust; with it on at full normal power their inertia is reduced by a factor of ten, so they and the ship feel 1/10 g while accelerating at 1.0g. (Do you suppose it would work that way--the crew is as effected by the field as the ship, so with their inertia reduced by 9/10, is that what they'd feel--1/10 g--even while actually accelerating at 1.0g?)
    Fuel will be an issue, but for now they have enough to accelerate and decelerate constantly to/from .99c. [I know that's unlikely, but....] They also have to carry only 1/10 the fuel they would need without the IR field. And acceleration at 1 g makes their Odyssey merely a life sentence, and not a multi-generational trip.

    They over drive their engine and IR coils, producing half again their normal thrust, getting themselves through a slolar flare before it cooks them; they leave the ice cloak on the ship for shielding. They fry engine and IR coils just after their perihelion burn, and it takes them three days to re-mount the 12 comet thrusters to the rear of the hull, and begin mining the ice shrouding the ship for fusibles (with which it is enhanced, having just flown through the edge of a CME). Once they have thrust--just over 2/10 g with all comet thrusters firing, but without inertial reduction--they thrust continuously for four days (I have them making Mercury orbit in 3, and I remember doing some math on that--is it plausible?) They then drop to 1/20 g thrust eight hours/24, and coast 16 hrs/24, to save fuel and give them time without thrust to mine the rest of the ice off their hull, while they repair their main drive and IR coils. I want that to take them a couple of months, and then another month or so at full acceleration, 1.0g, to Jupiter orbit. The online relativistic acceleration calculators I've played with make that sound plausible, but I don't know how to deal with the "drag" of Sol's gravity.
    Once at Jupiter the crew can use the still-mounted comet thrusters, with the IR field on full, to accelerate through Jupiter-ogee (perijov-ee?) with three g of thrust (feeling 3/10g?), if that helps them. They then thrust constantly for about a year, at 1 g, right out of the solar system and to .99c. The simple relativistic spaceship calculators online are fine for that bit.

    I can change most of these numbers a fair bit, to make this work, except the factor of ten inertial reduction and 1 g of thrust from the main engine with the IR coils on full.
    "At 21 solar radii I find that the circular orbit velocity is 95.3 km/s. The escape velocity is 2" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.08px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; font-family: "PT Sans", san-serif; position: relative;">√22 times that, or 134.8 km/s."

    Wow. Look what your program does to formatting, simple copy and paste.

    "At 21 solar radii I find that the circular orbit velocity is 95.3 km/s. The escape velocity is √ 2 times that, or 134.8 km/s."

    Average orbit Mercury ~= 59,224,000 km. 3 days x 24 hrs x 3600 sec/hr = 259,200 sec.
    59,224,000 / 259200 = 228.5 km/s average V to go Sol to Mercury in 3 days (straight line; wonder what the curve of the partial elipse adds?), but I was three days without thrust. If I had more than solar orbital velocity--say 120 km/s--at closest approach, and more than escape velocity, after the burn, say 175 km/s, in three days without thrust they'd go 45,360,000km, already most of the way to Mercury--except that Sol's gravity would be a huge, constantly diminishing drag that I don't know how to account for.

    Is that a reasonable Δv--55 km/s-- for a burn at perihelion at 21 solar radii, with 14.7 m/s/s of acceleration? Does that give you enough information to guestimate the length of that burn?
    "...approximate the burn time using Δv=aΔt where Δt is time of burn, a is the acceleration (your 1.5 g)."
    Sure, acceleration x time, and Δv/a =Δt. 55 km/s / .0147 km/s/s = 3741 sec = 1 hour 2 minutes 21.5 seconds?
    Avg. v 147.5 km/s: 147.5 x 3741 sec = 551,795 km travelled in that time. Circumference circular orbit @ 21 r = 2π 14,616,000km = 91,835,036 km; / 551,795 km = 166.43 part of an orbit. 360 / 166.43 = only 2.163° of angle, though I suppose you would ease into and out of that in a pretty shallow arc? You wouldn't get a lot of directional change from that, would you? Would you get more if you had thrust for a longer burn after perihelion, and you applied a vector/angled the ship toward the sun to tighten the arc? I can't do that, though.
    "the math becomes all but intractable and computer simulations are used instead." You mean NASA programs that take a super computer, right? Not open-source, run on Windows 7?
    "Jupiter's orbital radius is about 5.2 AU..." but all space ship trajectories in a stellar system are eliptical orbits, right? That's going to add to the distance travelled, and I have no idea how much. Does it have anything to do with how far Jupiter travells in its orbit in the time it takes to travel to it?

    Thanx huge, bro.
  5. Feb 25, 2017 #4


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    It would appear that when you selected "Sun" for the planet it filled in some numbers having to do with treating the Sun as a "planet" orbiting the galaxy center. I checked and it fills in 1.9E9 AU for the orbit radius, which is far to large to be anything but a galactic scale measurement. Apparently you chose to make that field your ship's orbit radius (the 21 solar radii). Not good. You want to use the "radius r" field for that.

    The Gravity G value is Newton's gravitational constant. It is correct. Yes it's that small. I suppose what you see in that field when it appears is the default value used. Presumably you have an opportunity to play with it. I didn't check that.

    Here's the output I got when I used the "radius r" field for the ship's orbit:

    The orbital speed and orbital period values are consistent with the Sun's speed and period of orbit around the galaxy center. The final escape velocity shown would be for escaping the entire galaxy, not the solar system. The software is not very well done if it doesn't make these things clear.
    I plugged in your ship orbit value into that calculator and it returned:

    I don't know how you obtained 138.74 kps.

    Anyways, 134.8 kps is close enough for what you want as the solar system escape speed. The Sun's mass dominates.
    GM is the product G multiplied by M, where G is Newton's gravitational constant and M is the mass of the Sun. It turns out that this product, GM, is actually a more accurately known constant than is G alone! This may surprise you, but G is a very small constant and notoriously difficult to measure accurately. GM on the other hand is pretty large and can be accurately determined by closely tracking a space probe's orbit over time.
    I've seen perijove used. But I don't think it's common usage. The general terms are periapsis and apoapsis. For a star other than the Sun use periastron and apastron.

    Right. At extreme speeds a ship would barely be deflected. Since your ship has the luxury of continuous propulsion and presumably steering, you can choose any speed within reason for your close approach speed. 1.5 g thrust is significantly larger than the Sun's gravitational acceleration for almost the entire solar system; You need to get within 0.025 AU of the Sun before it's gravitational acceleration matches Earth's 1 g surface gravity.

    Of course, if you're going so fast that perihelion flashes by in an instant then you won't have much time near perihelion to apply thrust and take best advantage of your Oberth maneuver! But the Sun's a fairly big beast, so you've probably got some latitude for your entry speed to allow for several hours of relatively close-in burn. If you've got the fuel, then the Oberth maneuver is really just gravy on top of a continuous burn.

    No idea. Not my universe, not my circus, not my monkeys :smile:
    It'll be negligible given your thrust capability. From your perihelion point to infinity the Sun's gravity is only going to eat the escape speed from that point (134.8 kps). You get to keep everything over that :smile:
    Unpowered, Newton/Kepler rules! For a body in a free-fall orbit the total mechanical energy (sum of kinetic and potential energy) is a constant; Use conservation of energy to determine the speed at any radius of the orbit.
    You should have leeway to use your engines for steering, even at closest approach. After all, the gravitational acceleration due to the Sun at your closest approach (1.462 x 107 km) is only 0.063 g. Your engines swamp that.
    Nah. Your Windows 7 box would have more than enough power. The problem would be finding a code that would let you play with the characteristics you want. Most such simulators would presume short course correction "impulse" type burns with a lot of coasting in between. They might do a search for suitable launch window assuming some maximum available ΔV for the entire trip. You'd have to do some research.
    Most of the planet orbits are pretty close to being circular. You won't go far wrong by considering them to be circles for your estimates. Figure that they travel at an essentially uniform speed throughout their orbits at least for the first approximations.
  6. Mar 9, 2017 #5
    Sorry for not getting back sooner, gneil. Taxes and dentistry and plumbing leaks wait for no man....

    I want to read through this a couple more times, and absorb the numbers. Numbers always did sink into my head very slowly.... But it sounds as if I can almost ignore solar escape velocity, use a relativistic (constant acceleration) spaceship calculator, fudge a little and be close enough to not sound like an idiot--and that you really can go Sol to Jupiter in just a few months, even on just ~1 or 2/10g acceleration. I'll get back again soon.

    But one thing--talking about orbits being elipses, I did not mean planetary orbits, but spaceship orbits. A ship will not go straight from Sol to Jupiter; it will be pulled by sol's gravity into a curved path that will add to the distance travelled, si' no? At these velocities, apparently not a lot? If Jove @ apehelion is 5.46 AU from the hot spot, maybe 8 AU actually travelled, to catch up if I left closest approach to sol a little before I wanted to???

    Thanx Huge!

    Know much about fusion? I'm not understanding why reactions withat produce fairly high MeVs--in this context, temperature?--like D+D once all the side reaction are complete (siomething like 22 MeV?), nonetheless produce too low "energy density" to make good spaceship fuels. Are MeVs additive? D+He3 and He3+He3 look best, once you can ignite them; p+B11 produces less than 8MeV, so I would guess that it would not be such a good fuel mix for a spaceship--great on earth where boron in plentiful and low neutronicity is--well, almost as important as on a spaceship--but maybe this energy density thing I apparently don't understand would nonetheless give you much delta-v on little fuel burning p+B11?

    "Energy density"--energy per unit volume? Unit mass? Of fuel????
  7. Mar 9, 2017 #6


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    When you have significant continuous acceleration available the solar system becomes a lot smaller. At your proposed closest approach to the Sun of about 21 Sun radii the local acceleration due to the Sun's gravity is only about 1/16 g (that's 0.063 g). Your ship's acceleration of about 1 g swamps that. You could drive your ship like a car around the solar system :smile:
    Sol to Jupiter, a distance of about 5.2 AU, could be covered in less than a week, easy. Maybe double that time if you want to come hang around at Jupiter (so accelerate about half the trip, decelerate the rest to match speeds with Jupiter)
    Elliptical trajectories apply to unpowered, fee-fall orbits. With the continuous accelerations available to your ship, except during close proximity to the Sun you could easily mask all evidence of the Sun's influence and fly an essentially straight-line path anywhere you wanted to go in the solar system. At just over 5 solar radii the Sun's acceleration due to gravity drops to less than 1 g.
    In space volume only matters if you need to contain it, mass only matters if you have to move it. There are probably other members here who can go into the specifics of the (currently known) best choices for energy and propellants.

    What matters for rocket fuel is how much mass you can throw and how fast you can throw it. It's a matter momentum (mass times velocity). You need to toss out a lot of light weight particles at very high speeds to match tossing a lump of bulk matter at moderate or low speeds.
  8. Mar 11, 2017 #7
    Sol to Jupiter, a distance of about 5.2 AU, could be covered in less than a week, easy. Maybe double that time if you want to come hang around at Jupiter (so accelerate about half the trip, decelerate the rest to match speeds with Jupiter

    Nope. I'm on the way out of the solar system, don't want to slow down until it's time to decelerate into the first stellar system to visit. I don't really need the gravity assist --or know how to calculate it, either--I'm catching up to Jupiter from behind, and I can have 3gs of thrust at closest approach, if I get the inertial reduction system fixed by then, and use the NOT fuel efficient comet thrusters for--an hour?--at perijove, to supplement the 1g main drive. I want to use Jupiter's gravity to change direction, away from the plane of the ecliptic, and that works with a subterfuge I'm thinking of adding to the plot, too..

    I have one g when the main engine and inertial reduction system are both on line, but they were both damaged producing 1.5 g escaping the solar flare I flew through at closest approach. Without inertial reduction the main drive normally produces 1/10 g. I have fusion thrusters designed to move cometary bodies around--my crew were asteroid miners before they turned pirate--and they mount them on the rear of the ship after the flare, to give themselves some thrust while repairing the mains. But they have a maximum of 2/10 g with those, without the inertial reduction system; and those engines are much more fuel efficient at 1/20 g. AS WRITTEN. I can change any of that I need to, except the 1 g with inertial reduction main drive--that's locked in by the previous book.

    I need way more than a week, to repair both systems and do a bunch of other things in the plot. So I'll drop back to 1/20 g 1/2 the time until I get the main drive fixed. The acceleration calculator at https://www.cthreepo.com/lab/math1/ says 22 days continuous thrust @ 0.05g is 550,407,883 mi; Sol-Jupiter is 483.8 million miles, so that's close enough. Thrusting half the time, that's 44 days, and I can stretch it a bit, not thrusting through the asteroid belt to make it easier to dodge any errant rocks. If I can just use acceleration calculators like this one, fudge a little, and be plausible, that's close enough. I figured sol's gravity would complicate things a lot more than it does.

    So, even 1/20g of thrust--just 0.49 meter/sec^2--is huge, as long as it's continuous. Hard to wrap my brain around.

    You could drive your ship like a car around the solar system cool! Or a motorcycle? Wonder if I can get my Suzuki V-Strom serviced on Mars?

    At just over 5 solar radii the Sun's acceleration due to gravity drops to less than 1 g. That's confusing. If you were just four radii--4 x 695,700 = 2,782,800 km--above the surface of the sun, its pull on you--the Einsteinian "up elevator" you'd feel if you could somehow stand on a non-falling non-orbiting surface?-- would only be 1g? 9.8m/s/s? What is the relationship between acceleration due to gravity and escape velocity, which we decided at 21 radii is 134.8 km/s? That's the velocity needed to leave the solar system with zero velocity--dead stop in the Oort cloud? with no additional thrust, taking forever. If your ship were stopped above a point on the surface of the sun, not in orbit, five radii out, 9.8 m/s^2 is all the acceleration you'd need to hover? Same as at Earth's surface??
    At that distance, would you be subtracting 9.8 m/s from your velocity every second you spent without thrust? And less and less as you get farther away....

    Unpowered, Newton/Kepler rules! For a body in a free-fall orbit the total mechanical energy (sum of kinetic and potential energy) is a constant; Use conservation of energy to determine the speed at any radius of the orbit. Math is not my thing, bro, and it's years since I studied conceptual physics, but if you can show me a formula I can probably learn to use it. Kinetic energy--do I need to estimate the mass of my starship? Three? aircraft carriers.... 3 x 7.3 x 10^7kg..... I know what potential energy is, but not how to figure it here.

    What matters for rocket fuel is how much mass you can throw and how fast you can throw it. It's a matter [of] momentum (mass times velocity). You need to toss out a lot of light weight particles at very high speeds to match tossing a lump of bulk matter at moderate or low speeds. Yep. I'm a "baby ballistician"--I understand more than a bit about ballistics, including recoil calculations, right up to where I'd need calculus to figure a trajectory--but that's what computer ballistics programs are for. I think I understand the tradeoffs between mass and velocity--especially as they relate to recoil--viscerally. But the energies needed are staggering, and that math would likely make me burst an artery in some unfortunate bit of anatomy. Fortunately I don't need to know the foot-pounds needed--let alone the Joules!

    But, yeah, more reaction mass exhausting not-so-fast for lower ship velocity high thrust initial acceleration; less but hotter, faster reaction mass for high exhaust speed at high ship velocity--"high gear"-- which is why I like the basic VASIMR concept. I'm postulating a fusion-driven VASIMR, run on D-T and putting most of the 80% energy wasted accelerating neutrons back to work cooling lithium and water jackets with Brayton cycle generators, putting that electricity to work heating additional reaction mass with microwaves, and accelerating the reaction mass + initial fusion products with electrical fields. In other words, a high-neutron, way- too massive kludge because it uses D-T. The crew will alter it in flight, adding a pair of opposite-direction spiral particle accelerators around the perimeter of the hull, to add the heat of collisions (iron plasma) at 2 x 1/3 c to the fusion chamber--hohlraum?--to make it hot enough to burn much-less-neutronic D-He3 or He3-He3--which should be more available in space than lithium (otherwise needed to make tritium and absorb neutrons).

    Up to 67% of the matter that falls onto a neutron star at .3 c, I've read, is converted to energy (gamma) by the impact; I should get a few TeV smacking a couple of micrograms of iron together at twice that. Should get hot enough to ignite almost any fusion fuels. I think. I'll start a new thread, and ask if there are any "fusion freaks" who can tell me more about it.

    So: I'm going to assume that I leave perihelion, and lose my engines, @300kps (unless you see domething wrong with that number); that I can get away with coasting for three days while I mount the comet thrusters; thrust continuously for three days at 2/10g to gain some distance from Sol; then drop to 1/20g 12 hrs/day for ~40 days while I repair the main drive. Get drive and inertial reduction coils back on line in time for a gravity sling around Jupiter, and with inertial reduction, main engine and comet thrusters all working, I can get 3g all through my burn around Jupiter.

    If it's 57 million km (Mkm) Sol-Mercury (mean orbit; http://www.universetoday.com/15462/how-far-are-the-planets-from-the-sun/ ); / 300km/s x 3600 sec= 52.8 hrs unpowered but also no drag from Sol. If I wanted to subtract out deceleration from Sol's gravity--which apparently lessens--as the square root of distance from source?--rapidly as you gain distance, what would that equation look like? Would that stretch 53 hrs to ~72?
    779Mkm (to Jupiter) less 57 = 722Mkm, 4.8263 AU. Let's see if I can drag and drop this here: Yep!

    U1L6a1.gif http://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations

    I want the first one, don't I? But altered--I know distance, acceleration, initial velocity, want time:
    d = vi t +1/2 a t^2
    d/a = vi t +t^2 / 2
    2d/a - vi t = t^2 (subtracting a term from both sides is a legal operation, right?)
    t = sq rt (2d/a - Vi * total time in flight, right? ????????????? but that leaves me to know what I'm trying to find!, leaves time on both sides of the equation. Where'd I goof?
    easier way: http://nathangeffen.webfactional.com/spacetravel/spacetravel.php :
    Distance meterskilometerslight-secondslight-minutesastronomical unitlight-yearsparsec
    Acceleration m/s^2g (1/20g)
    Maximum velocity 594.7932 kilometers /sec or 2,141,255.53 kph. That's twice as fast as I wanted go through the asteroid belt, but the belt starts at ~2 AU, so I won't be going anything like that fast through most of it; think I'll let it slide.
    Observer time elapsed during journey 28.09875 days
    Traveler time elapsed during journey close enough

    So if I thrust 1/2 time @ 1/20g I have about 56 days to get my systems up and running--except that doesn't account for the other 1/2 the time, when I'm not accelerating but moving very fast, does it. And I'm moving faster at every thrust break; now I'm getting a headache.

    I could refine that further: the first 3 days without thrust @ 300 km/s I'd go 77,760,000 km: in 3 days @ 2/10g (1.96 m/s^2) I'd go 32,920,467.69 km + initial velocity x time--but I should assume I lost some velocity without thrust--average 250 km/s? = 64,800,000 km +32,920,468 + plus the first three days 77,760,000 km = 175,480,468, 23 Mkm, past Earth orbit, or maybe just about there, given some curve to the ship's orbit. Really? Already?
    779Mkm -175.48Mkm = 681 Mkm @ .49 km/s/s = 27.3 days--at constant thrust, not taking into account time/distance travelled coasting. I'm stuck.


  9. Mar 11, 2017 #8
    Maybe I'll have them not thrust through the asteroid belt, 2-4 AU, to make it easier to dodge any rocks.
  10. Mar 11, 2017 #9


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    At one solar radius (i.e. on the Sun's "surface") the acceleration due to gravity is about 274 m/s2, or 27.9 g. Since all the scenarios discussed are a bit hand-wavy, call it 28 g. The gravitational acceleration drops as the square of the distance, so if n is the number of solar radii distance from the Sun's center, then the Sun's gravity produces acceleration:

    ##a = \frac{28}{n^2}## in g's.

    As long as the mass of the orbiting body is insignificant with respect to what it orbits (the Sun in this case), you can ignore the mass and work with what is called specific energy, which is energy per unit mass. Just about all calculations for astrodynamics are done using specific energy, specific angular momentum, etc., since in variably the masses of everything from spacecraft to asteroids is insignificant when compared to the Sun.

    For unpowered trajectories, if you know the velocity and distance from the Sun at some instant then you can determine the specific total mechanical energy for the orbit:

    ##ξ = \frac{v^2}{2} - \frac{μ}{r}##
    ##μ = 1.327\times 10^{20}~m^3 s^{-2}~~## is the Sun's gravitational parameter (μ = GMSun)
    ##v~~## is the velocity
    ##r~~## is the radial distance from the Sun's center

    For bound orbits ξ will be negative and the orbit will be a circle or ellipse (or degenerate straight line fall into the Sun!). For escape trajectories ξ ≥ 0. For exactly escape velocity, ξ = 0, and the trajectory is parabolic.

    Once you know the specific mechanical energy for an orbit you can predict the velocity for any distance:

    ##v(r) = \sqrt{2 \left( ξ + \frac{μ}{r} \right)}##

    The problem is, the basic kinematic equations above only apply when the acceleration is constant. For unpowered orbits the acceleration depends upon distance, and if distance is changing with time then so is the acceleration. If your ship's engine acceleration completely dominates then you can use them to get a good approximation.

    The time on orbit (or "time of flight") between two locations on an unpowered orbit, except for some particularly simple situations, is not a trivial problem. Kepler, who was no slouch mathematically, found it to be difficult. There is no exact (algebraic) solution to the general problem, and either iterative or series solutions must be used. If you're interested, look up "The Kepler Problem".

    With the speeds you're using in your scenarios you can probably ball-park the results using the starting velocity and final velocity at the destination and assume an average velocity for the trip. Use the specific mechanical energy and the v(r) formula to find the final velocity.
  11. Mar 11, 2017 #10
    Distance meterskilometerslight-secondslight-minutesastronomical unitlight-yearsparsec -- this computer program messes with formatting when you post. And when I tried http://nathangeffen.webfactional.com/spacetravel/spacetravel.php :, it took me nowhere.
    So from another relativistic starship calculator:
    Distance 4.522 AU
    Acceleration .05 Earth g
    Time on board 27.28 days
    maximum speed 0.0019272726c should ~= 577.782 km/s~= 2,080,014.5 km/h

    Here's another constraint, that I was going to "write around," but maybe it will help refine all this. My ship "fell" from Mars orbit--but again, it had an unspecified smallish amount of thrust--1/20, 1/50, even 1/100g would work--for a time unspecified except as "days" "adding thousands of kph to the [artificial] comet's velocity". It had that thrust early in its trajectory, between Mars and Earth orbits, when it was not yet already falling very fast. Thus it went into perihelion 21 solar radii out travelling faster than it would have on an unpowered trajectory, burned through its Oberth maneuver at 1.5g, and (hopefully) left with something well over solar escape velocity.

    I can change thousands of kph to hundreds if that helps.

    The ship fell past Earth--L5, actually--now unpowered, on Christmas Day 2060. I need it to be on the other side of the sun, and back out somewhere near Venus or Earth or even Mars orbit, 97 days later, on April 2, 2061, the date of the nuclear war that is part of the Star Trek universe. And I need something like a couple of months to go Sol to Jupiter.

    Right now I have the ship's drive shut down entirely for three days after its Oberth maneuver burn; then I have three days constant burn at 2/10g; then burn 12 hrs a day @ 1/20g, and coast 12 hrs a day, for a month or two. If I need more time I can coast all the way through the asteroid belt, 2 to 4 AU. Does that help?
  12. Mar 11, 2017 #11


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    As soon as you go un-powered the orbit is fixed by the conditions when the thrust is turned off. The instantaneous velocity and position define the entire orbit, including where perihelion will occur. In order to "fall" inwards from an outer orbit you need to "kill" a lot of the speed associated with its original (presumably somewhat circular) orbit. If you have thrust then you can "drive" into any position you want.

    A true Oberth maneuver just coasts along its orbit (whatever it may be) until it gets near perihelion where it applies thrust. It's efficient because you get more ΔV for your craft for a given amount of fuel that way. If you've got significant amounts of thrust already and a seemingly endless fuel supply, then it becomes harder to justify the maneuver. Unless of course you are just taking a shortcut across the solar system and taking advantage of the close flyby. Perhaps your crew needs the time to prepare the high thrust engines, to avoid other ships, just be generally sneaky, or other story specific reasons.

    If the ship actually fell totally un-powered from near Mars orbit, then with a perihelion distance of 21 solar radii it would have a perihelion speed of about 131 kps. Escape velocity at that location is just about 135 kps, so you only need a ΔV of about 4 kps to achieve an escape speed. Of course it would be a long, slow coast out of the system. The infall itself would have taken just over 133 days. To cut that time you'll need to use thrust.

    If you use your available thrust to slowly "bend" your infall path to place it on target for a 21 radii passage, then you could presumably already have more than escape speed by the time you reach perihelion.

    The faster you go, the more "straight line" the trajectories will be, and the better you can approximate the times for reaching various distances by just using average velocities and distances.
  13. Mar 15, 2017 #12
    Thanks for the equations, gneil. I'll play with them and see what they teach me. I think I understand this well enough, now, to keep my timeline reasonable.

    You understand this stuff rather well. What do you do professionally?

    Thanx huge!

  14. Mar 15, 2017 #13


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    Staff: Mentor

    I do consulting work in High Performance Computing (HPC). I've a background in Pure & Applied sciences and a degree in Computer Science & Electronic Systems. Astrodynamics just happens to be a bit of a hobby of mine.
    You're welcome!
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