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Acceleration due to gravity of a rocket

  1. Jan 16, 2015 #1
    If an object say a rocket is thrown up in the sky with an additional acceleration say ' x ' , then why do we add the value of acceleration due to gravity i.e 9.8 m/s^2 to the acceleration ' x' in order to find the total acceleration.......
    Since vector of acceleration due to gravity is directed downwards...so we consider it - 9.8 m/s^2 . So in accordance to this, if we find total acceleration then we should subtract 9.8 m/s^2 from 'x ' m/s^2 because...firstly, gravity is -9.8 m/s^2 and secondly, acceleration due to gravity will constantly decrease if the rocket or object will continue going up int he sky with that X acceleration. So why should 9.8 m/s^2 be added to X m/s^2 and not subtracted ?
  2. jcsd
  3. Jan 16, 2015 #2


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    If you add a negative value you're subtracting it, no?
  4. Jan 16, 2015 #3

    But as the bookish explanation says, we add the positive value of acceleration due to gravity and not the negative one....
    So here adding it means adding it only....
    That's why I am confused...
    Need some help and guidance
  5. Jan 16, 2015 #4


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    That doesn't seem right. Can you provide more context for the question? Perhaps a verbatim description of the problem as stated in your book? (a picture will do)
  6. Jan 16, 2015 #5

    The example goes on like this :

    A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m/s^2 . Find the initial thrust(force) of the blast.
    Take g=9.8 m/s^2

    Solution (as given in my book) :
    The rocket moves up against gravity with an acceleration of 5.0 m/s^2 .
    Hence the blast produces a total acceleration of =
    a=9.8+ 5.0 =14.8 m/s^2
    By Newton's II law, the initial thrust(force) of the blast is =
    F=ma = 20,000 kg x 14.8 m/s^2 =2.96 x 10^5 N.
    Therefore, answer is 2.96 x 10^5 N
    Last edited: Jan 16, 2015
  7. Jan 16, 2015 #6

    My doubt is the underlined portion only. If the rocket moves against gravity then why should acceleration due to gravity be added to the initial acceleration?????
  8. Jan 16, 2015 #7


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    That could be written more clearly to show what's going on.

    Try this:
    where a is the net acceleration (5m/s^2, positive is up), g is the gravitational acceleration (-9.8m/s^2) and t is the thrust you want to find. Here, the signs show you the directions of the vectors.

    If you rearrange the above to solve for t, you get the equation in your book.
    The book simply went straight for the rearranged equation.

    Note that the acceleration due to gravity is not
    but is deducted from the net (initial) acceleration. ##-(-9.8) = +9.8##
    Last edited: Jan 16, 2015
  9. Jan 16, 2015 #8


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    To get the proper acceleration, relative to free fall. That's what the engine must provide to cancel gravity and accelerate up.
  10. Jan 16, 2015 #9

    Oh....THANKYOU soooooooooooooooo much....
    I got it!
    I can now see it clearly....yaa......I've undersood it now....
    Once again...THANKS A LOT !
    It was really very sweet of you to guide me .....
  11. Jan 16, 2015 #10
    5 - (-9.8)=14.8

  12. Jan 16, 2015 #11

    Thank you so much .. Sir....
    I got it !...
    It was really simple and I was stuck at it like anything...
    Thank you so much for helping me out.
  13. Jan 16, 2015 #12
    It would have been much easier if you had drawn a free body diagram of the rocket, and then, based on the free body diagram, written the force balance:

    F - mg = ma

  14. Jan 16, 2015 #13


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    Perhaps think like this...

    How much thrust would the rocket have to produce so that it only just manages to lift off the ground without accelerating upwards?

    Then how much more does it have to produce to accelerate upwards at "a"?

    What's the total?
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