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Change in acceleration

  1. Jan 7, 2015 #1
    Does the change in acceleration along the path of a space flight to the moon, for example, get taken into account when planning initial velocity?

    I would imagine that the initial velocity leaving earth will be decelerating and the deceleration will increase the further it gets until another bodies gravitation compels it?

    How is this change in acceleration taken into account? I am especially wondering if it is an average value of the acceleration function over the radius, or if an average doesn't work in this case?

    Any input would be appreciated

    Thanks!
     
  2. jcsd
  3. Jan 8, 2015 #2

    Matterwave

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    The important quantity that we look at is the acceleration because that is the quantity which corresponds to forces through Newton's second law F=ma. Of course, when we work with planning any kinds of flight paths, we will look at an acceleration which will be generally a function of time a=a(t) and not a constant one. This is definitely taken into account.
     
  4. Jan 8, 2015 #3
    Is it taken as an average when its increasing and decreasing throughout a trajectory path?
     
  5. Jan 8, 2015 #4

    Matterwave

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    No, when you are solving equations for these problems, you usually solve for r(t) which is the position as a function of time. This function determines uniquely all relevant quantities as functions of time. The velocity is v(t)=r'(t) and the acceleration a(t)=r''(t).

    We rarely work with averaging since the instantaneous quantities tell us a great deal more information and it is very necessary to be precise.
     
  6. Jan 8, 2015 #5
    when a rocket leaves earth it's not decelerating , rather it is accelerating with a decreasing acceleration ..... and this acceleration has different equation regarding the path you choose and will be a function of time and mass
     
  7. Jan 8, 2015 #6
    Ok, I was thinking an average like this

    b
    ∫ G*M/r2 * 1/(b-a)
    a

    So, in that case, how would you keep track of the change in acceleration between upper interval b and lower interval a? Do you use the interval as a function of time? and if so what equation?
     
  8. Jan 8, 2015 #7
    Yes the rocket accelerates, but it accelerates against gravity accelerating back at it, so eventually you will come back down if you don't have escape velocity...
     
  9. Jan 8, 2015 #8

    Matterwave

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    That's quite a peculiar equation...I am pretty baffled why you think an "averaging" of this kind is desirable or necessary when doing calculations (I am assuming you are integrating there with respect to r?).

    Simpler would be to look at the instantaneous forces on this rocket (definitely not a simple equation to solve though): $$\vec{F}_{net}(t)=\vec{F}_{thrust}(t)-\frac{GM_E m_r(t)}{|\vec{r}(t)|^2}\hat{r}(t)=m_r(t)a(t)$$
     
    Last edited: Jan 8, 2015
  10. Jan 8, 2015 #9
    averaging is not possible here ... its not a linear equation .. and accelaration changes every second and moreover , a rocket travels in a parabolic path amd when it's not so close to earth the thrust equation also changes ( we have to take the gravitational force of moon into account )
     
  11. Jan 8, 2015 #10
    I'm was specifically thinking about once its left earth, say the lower bound is the Karman line, and you need to propel your craft with enough velocity to the moon, if you make your main shot with enough velocity to get to the point where G ⋅Me/r2 = G ⋅Mm/r2, the moon would pull you the rest of the way, so the main consideration would be to make sure that you don't decelerate and be attracted back to earth before you reach where G ⋅Mm/r2 > G ⋅Me/r2 So then, if this is the case, the vi= vf - a ⋅ t
     
  12. Jan 8, 2015 #11
    once you start a rocket that has enough fuel you will never decelerate until your fuel is over ... you can prove it with this
     
  13. Jan 8, 2015 #12
    But what about the poor man's space craft, which does not have a big enough tank to burn gas the whole way there, only one controlled burn to get a burst of velocity, the only thing that will slow you down is the gravity force attracting to earth, so if you get a good enough impulse from LEO in the right direction you will make TLI?
     
  14. Jan 8, 2015 #13
    that poor man has to be richer than the rocket owner .... within a very little time he has to put in a great enough force to get that spacecraft the escape velocity ... that's an impossible case you are talking about... but yes if you can get that escape velocity you will leave the earth's gravitational field
     
  15. Jan 8, 2015 #14

    Drakkith

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    I'm not sure what you mean by this. You will certainly cease accelerating when you stop your rocket engine. Turning around and activating it again will cause you to decelerate.
     
  16. Jan 8, 2015 #15

    Drakkith

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    That's correct. That's how all of the Apollo missions worked. A short burn near Earth, and then they coasted the rest of the way, with perhaps a few course corrections if needed.
     
  17. Jan 8, 2015 #16
    ...i didn't mean it will still accelerate when the engine is stopped ... i just tried to say that when the engine is on it always accelerates
     
  18. Jan 8, 2015 #17
    If centre to centre distance = d, and the distance from earth where the moon's FG > earth' FG = a

    How do you calculate the loss of velocity required to get to the point where → G ⋅Me/ a2 = G ⋅ Mm/ (d-a)2

    I thought if I summed the acceleration outcomes in the interval and averaged them like this it would work but I guess the values don't add up because they're based on time and not just the radius... Can you think of anything else. I don't see how the Fnet(t)=Fthrust(t)... equation gets me what I'm looking for but I'm still learning about it so I need time on that.

    I'm probably missing something, so if you can lead me in the right direction I'd be grateful.
     
    Last edited: Jan 8, 2015
  19. Jan 8, 2015 #18

    Drakkith

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    Sorry, no idea! :s
     
  20. Jan 8, 2015 #19

    jbriggs444

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    You could use potential energy to get to that number.

    Let's take a simplified scenario where we imagine that Earth and Moon are stationary and that we launch with an initial burn and coast to the mid-point between Earth and Moon where their gravitational attraction cancels out.

    The gravitational potential energy of the spacecraft at its initial location at the surface of the earth will be the sum of its potential energy with respect to the Earth and its potential energy with respect to the moon. That's ##\frac{-GmM_e}{R_e} + \frac{-GmM_m}{d}##

    The gravitational potential energy of the spacecraft at its midpoint location at distance "a" from Earth is ##\frac{-gmM_e}{a} + \frac{-GmM_m}{d-a}##

    By conservation energy, the change in kinetic energy on a coasting trajectory is equal and opposite to the change in potential energy. Use ##KE=\frac{1}{2}mv^2## to work out the change in velocity required to match the change in potential energy computed as above.
     
  21. Jan 8, 2015 #20
    Yes sorry, the integral is with respect to r. Anyway's, your equation is a little above my paygrade atm, I wouldn't know where to start with the matrices I believe it requires and so far my thought problem is 1 dimensional. Why is this "simpler" than calculating the loss of velocity? Can you think of any other ways I could do this, maybe with a derivative for the rate of change?
     
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