1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Energy to escape gravitational pull

  1. Jul 11, 2011 #1
    forgetting the need to enter an orbit, is there a difference in the energy required to escape the earths gravity by firing a rocket straight up against firing a rocket on a trajectory? Straight up would be a shorter distance.
     
  2. jcsd
  3. Jul 11, 2011 #2
    The difference in gravitational potential energy in going from an initial to a final separation is independant of the route taken.There are,however,other considerations to be taken into account,such as if a craft is launched vertically upwards it takes the shortest route out of the atmosphere and suffers less resistive losses.
     
  4. Jul 11, 2011 #3

    rcgldr

    User Avatar
    Homework Helper

    Ignoring the effect of the atmosphere, what's required to escape gravity from a planet is to accelerate an object to escape velocity. If the acceleration is directly "outwards" (up) from the center of the planet, then that rate of acceleration is reduced by the pull of gravity. At the earth surface, the acceleration would be reduced by 9.8 m / s2. It would be better to continously reorient the accelerating force so that it is always perpendicular to the force of gravity.

    In the earth's case, the atomspheric drag also resists acceleration, and also puts a straing on a space craft during launch, but the goal is to eventually get the space craft's velocity so that it's nearly perpendicular to the force of gravity, so the path used balances out these conflicting aspects of an earth based launch of a space craft.
     
  5. Jul 11, 2011 #4
    I don't believe that is true. It is taking the full retarding force of gravity into account that the equations of motion predict the 11.2 km/s escape velocity. There is no further retardation to impose after that. The calculation of this figure assumes an object escaping in a direction in full opposition to the force of gravity.

    That actual trajectory the rocket takes might be away from the vertical, to allow it to reach its place in orbit in the intended orientation, I suppose.

    Ranjit Konkar
     
  6. Jul 11, 2011 #5

    Filip Larsen

    User Avatar
    Gold Member

    However, thrusting against gravity will incur gravitational losses. Just imagine the rocket thrusting 1G vertically and its obvious that all reaction mass of the rocket in this case will be "wasted" just to counter gravity. So the trick is to get quickly out of the atmosphere, but at the same time quickly get the velocity vector down near horizontal, assuming here that thrusting due to aerodynamic constraints is restricted to being mainly opposite the velocity vector.
     
  7. Jul 11, 2011 #6
    When I said no further retardation to impose, I meant gravitational retardation. Air resistance and any other resisting forces would certainly slow it down, increasing the force required to overcome these.

    Ranjit Konkar
     
  8. Jul 11, 2011 #7

    rcgldr

    User Avatar
    Homework Helper

    Excape speed depends on the gravitational potential at some point from the center of a planet. That 11.2 km/s speed is based on an object on the surface of the earth or at that distance from the center of the earth, at a radius of about 6370 km.

    Take the case of a stationary object 6370 km from a point source in outer space (vacuum) with the same mass as the earth. As posted above, you could exert 9.8 netwons of force on a 1 kg object directly away from from the that point mass and the object would not accelerate. However you could apply a very tiny force to that object as long as that force was always oriented to be perpendicular to the force of gravity and the object would accelerate, and it's speed would increase. Initially the object would spiral inwards (somewhat elliptically) until v2 / r exceeded G M / r (were M is the mass of the Earth), after which the object would start to spiral outwards (again elliptically), eventually acheiving a parobolic path once it's velocity equaled escape velocity for it's current position (in which case the acceleration force would no longer be needed). As long as the accelerating force is always oriented perpendicular to gravity, then the energy required or the work done would be equal to 1/2 m escape_speed2, regardless of the amount of force (it would just take more distance and time with a smaller force).
     
    Last edited: Jul 11, 2011
  9. Jul 11, 2011 #8
    I don't think he's disputing that. Vertically or horizontally, you need 11.2 km/s....the point he was making is that it will be harder to achieve this 11.2 if you do go vertically instead of horizontally.
     
  10. Jul 11, 2011 #9
    But the work done per unit mass in reaching a certain height is the same whether you go vertically upwards or at an angle.It's analogous to lifting a mass through a certain height by sliding it up a slope rather than by lifting it vertically.In both cases the work done against gravity is the same.
     
  11. Jul 11, 2011 #10

    Filip Larsen

    User Avatar
    Gold Member

    When a rocket engine is thrusting "against" gravity, only a part of this thrusting force end up accelerating the mass, and thus only a part of it can be contributed as work done on the mass which you speak of. Again, having a rocket just hoovering at the same height means that the mechanical work on the rocket is zero even though the engine is clearly producing a lot of work.
     
  12. Jul 11, 2011 #11

    russ_watters

    User Avatar

    Staff: Mentor

    You can't eliminate the upward component of force, otherwise the rocket will crash into the earth before it reaches orbital, much less escape velocity!

    The escape velocity is the same but the energy required of a real rocket is lower if you go vertical(neglecting rotation of the planet). (Edit.....ehh not sure if it is equal or lower)
     
    Last edited: Jul 11, 2011
  13. Jul 11, 2011 #12
    You seem to be considering the thrust only and not the distance moved.Look at my analogy above and imagine a mass m lifted vertically through a height h in a uniform field of strength g.The work done against gravity =force times distance=mgh
    Now consider the same mass being lifted a height h by sliding up a slope which is at an angle X to the horizontal.This time the force is smaller and equal to mgsinX but the distance moved is bigger being equal to h/sinX.The work done against gravity is again mgh.The point is that the work done against gravity in taking a certain mass from one height to another is constant regardless of what route is taken.Generally speaking the shortest route is the best because the mass arrives at its height in a shorter time and there are less resistive losses.Of course my analogy is simplistic but I am using it to illustrate the principle.
     
  14. Jul 11, 2011 #13

    Filip Larsen

    User Avatar
    Gold Member

    I do not agree. For a rocket to obtain a given amount of orbital energy (without considering atmosphere or ground impacts) it can do so with a shorter burn if it is thrusting horizontally than if it is thrusting upward.

    A practical rocket would of course, as already mentioned in this thread, have to consider both atmosphere and avoid impacting with the ground.

    (I have a feeling of deja vu here; didn't we had this discussion in another thread where DH came by and cleared it up? If so, perhaps he could do so again, because my explanations apparently do not impress much :rolleyes: )
     
  15. Jul 11, 2011 #14
    I think we didn't come to any firm agreed conclusions because there are so many other factors to account for such as the rotation of the earth,the variation of g,the rapidly reducing mass of the rocket as it burns fuel and so on.It ain't rocket science to work out the ideal trajectory.Whoops of course it is.:biggrin:

    (where's the rocket people?)
     
  16. Jul 11, 2011 #15

    rcgldr

    User Avatar
    Homework Helper

    I was assuming the earth was replaced by a point mass and that the space craft stated at 6370 km from the point mass. In this case you don't need an upwards component of force, just an increase in velocity perpendicular to the direction of gravity, while letting the component of velocity in the direction of gravity cycle due to the elliptical like component of the spiral path of the object. As long as there are no collisions, it's not an issue.

    If you assume a frictionless earth and no atmosphere, then the rocket just accelerates along the surface of the earth (horizontally) until v2 / r is greater than gravitational acceleration at the surface of the earth.

    If you take the real world situation with atmosphere, a rotating earth, fuel that gets depleted while thrust is created, constraints on maximum drag force, ..., then you get the more complicated "rocket science" stuff.
     
  17. Jul 11, 2011 #16

    cjl

    User Avatar

    Trajectory optimization is complicated :smile:

    As for gravity losses, those depend on the trajectory taken and the burntime. Ignoring the atmosphere for a moment here, it turns out that the faster something is accelerated, the less loss there is, and the often quoted escape velocity is the required delta V if the acceleration is instantaneous (or if you could accelerate horizontally with no atmosphere). If your acceleration is vertical and non-instant, you need additional delta V capability approximately equal to the burntime multiplied by the acceleration of gravity (so a 10 second burn in earth gravity would need about 100 m/s more delta V capability than an instant burn would for the same actual velocity gain). Of course, there are limits to how much acceleration spacecraft (and people) can take, so extremely fast burns aren't really practical. In addition, very high acceleration means that you are going faster in the lower atmosphere, which increases the energy lost to aerodynamic drag. So, the loss due to a longer burntime needs to be balanced against the loss due to going fast in the lower atmosphere and the loss due to needing more structure to survive the high acceleration loads.
     
  18. Jul 11, 2011 #17

    russ_watters

    User Avatar

    Staff: Mentor

    I guess it depends on how far from reality one wants to go with the scenario. I assumed, at least, we had an earth.
     
  19. Jul 11, 2011 #18

    rcgldr

    User Avatar
    Homework Helper

    In that case the Apollo missions come close. They went in to a normal orbit using a path that started off vertical, then transitioned into horizontal, similar to the shuttle. After 2 or 3 orbits, they waited for the proper position to fire the engines again to achieve enough velocity to leave the earth and get captured by the moon (called translunar injection), with a constraint on the time it would take to reach the moon. I would assume that the firing of the engines while in orbit to achieve translunar injection path and velocity was nearly perpendicular to the force of gravity (about 8 degrees from "horizontal" based on the "flight path" numbers included in the article's links). Link to an article I found:

    http://www.braeunig.us/apollo/apollo11-TLI.htm

    A somewhat related link to Hohmann transfer orbit, which includes the math and efficiency factors to transition from one circular orbital path to another with two impulse cycles perpendicular to the direction of gravity:

    http://en.wikipedia.org/wiki/Hohmann_transfer_orbit

    Attached is an image of an example Apollo path:
     

    Attached Files:

    Last edited: Jul 11, 2011
  20. Jul 11, 2011 #19

    cjl

    User Avatar

    Yes. Once you're in orbit, you pretty much always fire along your current flight path. Since you are already in orbit, you don't have to worry about countering gravity, and the most efficient way to add or subtract velocity will always be to thrust parallel to your flight path. That isn't possible during initial ascent though (due to gravity).
     
  21. Jul 11, 2011 #20

    rcgldr

    User Avatar
    Homework Helper

    True, but the original post was asking about the difference between a continously "vertical" ascent, versus one that transitions into nearly "horizontal". The point I was trying to get at is it's more efficient to transition into a near "horizontal" path to avoid the rocket engine's power being wasted on opposing gravity versus increasing velocity.
     
  22. Jul 12, 2011 #21
    I don't understand these comments about it not being necessary to counter gravity.Gravity doesn't just disappear when in orbit.In fact to calculate escape velocity we calculate the work done against gravity in taking an object to infinity and use KE lost =PE gained.Gravity is the main factor to consider throughout the journey,eg on a journey to the moon the gravitational pull of the moon becomes stronger than that of the earth only after about 9/10 of the distance between earth and moon has been covered.
     
  23. Jul 12, 2011 #22

    rcgldr

    User Avatar
    Homework Helper

    What was missing in the previous comments is you need to consider the work done on the ejected mass (spent fuel) as well as the work done on the rocket when considering the efficiency of a particular path. The example of inefficiency previously mentioned is one where a rocket wastes power by hovering and using it's fuel to oppose gravity without achieving any acceleration. No work is done on the rocket in this case, but there is work done accelerating the spent fuel towards the earth.

    The goal is to increase the total energy, KE + PE, but because of the way rocket engines work, it's more efficient to have most (or all?) of the acceleration of rocket and fuel perpendicular to the force of gravity to increase velocity and KE, than to directly oppose gravity to increase distance and PE.

    In the real world, there would be compromises if there are time and/or position constraints on the target path of the space craft, but the original post was just asking about achieving escape velocity via any path and if a radial acceleration would be less efficient than a tangental acceleration (the tangental acceleration is more efficient), and I'm assuming that a conventional rocket engine and the onboard fuel would be the source of power.
     
    Last edited: Jul 12, 2011
  24. Jul 12, 2011 #23
    I looked again at the original question and I think we have strayed slightly from it.The question enquired about the energy required to escape the earths gravity.The energy needed for a mass m to escape from the earth of mass M to an infinite distance is given by:

    W=GMm/r (r=radius of earth)

    This energy is independant of the route taken and the method of taking that route so the main goal is to increase the PE.The OP made no reference to escape velocity and started the question by stating "forgetting the need to enter an orbit"
    Taking the craft to escape velocity early in the flight may well be an efficient way to overcome gravity but whatever method is used work must be done against gravity.
     
  25. Jul 12, 2011 #24

    rcgldr

    User Avatar
    Homework Helper

    I was assuming energy required for firing a rocket would include the energy spent on the ejected mass as well as the rocket (the total chemical energy required to achieve escape velocity). In this case the path does matter.

    It would be possible to always orient the rocket so that acceleration of fuel and rocket are always perpendicular to the direction of gravity. In this case, at any moment in time, the rocket engine does no work in the direction of gravity on either the rocket or the spent fuel. Over time, gravity is constantly reducing the velocity and KE achieved by the rocket's engine, so gravity is doing work against any work previously done by the rocket engine. Also, even though the rocket engine does no work against gravity at any specific moment in time, that work is done in what will be the direction of gravity at some later time, and in that sense, work is done against gravity.

    I'm not sure what the optimal orientation is for a continous burn. The Apollo missions used an upwards pitch of about 8 degrees from tangental to transition from earth orbit to lunar injection, but I don't know if this was the optimal orientation or one compromised by path, time, and velocity constraints to intercept the moon.
     
  26. Jul 12, 2011 #25
    Well I agree that we have to consider energy given to the ejected fuel.This is just one of several factors to account for in real systems.It certainly is a thorny problem.

    Where's Werner and the other rocketeers?:smile:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...