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The reason why objects fall at the same speed on the moon, with detail:

  1. Oct 9, 2012 #1
    This WILL sound sketchy and far-fetched, but read every bit of information carefully, and you will hopefully understand, I know overall the reason why the objects fall together at the same rate is because there is no air-resistance, but working with formulas it shows how some things are inter-linked and how the mass is irrelevant due to the cancellation of mass from the basic GCSE Physics formulas of:

    "Kinetic energy gained = Gravitational potential energy lost."

    Copied from my facebook thread:

    "
    Students taking the Physics-2 Exam in November 2012:

    The topic of Kinetic-Energy and Gravitational-Potential-Energy:

    You are required to learn as part of the Syllabus:

    The ability to work out Speed of a falling object being given the Mass and the Height of the object from where it starts to fall (assuming the object is falling to the ground on Earth (meaning g = 10N/Kg)).

    The example that follows shows you how to do this:

    Kinetic Energy Gained = Gravitational Potential Energy lost.

    mgh = 1/2mv^2

    Whereby:

    m = Mass in Kilograms (Kg)
    g = Gravitational Field Strength (Earth being 10N/Kg & The Moon being 1.6N/Kg)
    h = Height of the object (the height of where the object is falling from to the ground) in meters
    v = Velocity (in meters per second).

    =======================================================

    Example:

    An apple of mass 140grams has fell from a tree of height 1.7 meters. Calculate the speed as it hits the floor.

    First off, the speed "as it hits the floor" is the overall speed the apple is traveling at as it is falling, so lets not get confused by that.

    Find G.P.E lost: G.P.E = mgh = 0.14 x 10 x 1.7 = 2.38J (J = Joules). This must also be the K.E gained.

    Equate the number of joules of K.E. gained to the K.E. formula with v in: 2.38 = 1/2mv^2

    Stick the numbers in:
    2.38 = 1/2 x 0.14 x v^2 or 2.38 = 0.07 x v^2
    2.38 / 0.07 = v^2 so v^2 = 34
    v = √34 = 5.83 m/s

    =======================================================

    This seems all rather complicated no? This is what the revision book has told me to do (which I have added a couple of sentences here just to clarify what is actually going on, as by the "First off" sentence).

    It has just dawned to me that if you look at the Height of the object, and times it by 20, that is the Velocity squared.

    Proof below:

    Look at the formulas, it is basic cancelling and changing the subject:

    mgh = 1/2mv^2

    Divide both by m (which is mass), because what you do to one side, you MUST do to the other, therefore:

    gh = 1/2v^2

    'g' does not change, and neither does the height, however:

    g = 10N/Kg (ONLY on the Earth, the question WILL state if the object is falling on the Moon, if so, then the question will tell you that).

    Therefore:

    10h = 1/2v^2

    Therefore:

    20h = v^2

    √(20h) = v

    This means, no matter what the mass of the object is, it could be 5 Kilograms, it could be 9.9 x 10^4 Kilograms, if you have the height of the object, then you times it by 20 and you then square-root that answer, to find the speed of the object as it is falling on Earth.

    If it states the "Moon" rather than the "Earth" then you follow these steps:

    gh = 1/2v^2

    Replace g with 1.6 (rather than 10, as 'g' on the moon is 1.6N/Kg)

    1.6h = 1/2v^2

    3.2h = v^2

    √(3.2h) = V

    Times the Height by 3.2 and then square-root the overall answer to get the Velocity in meters per second.

    No matter the mass, times the height by 20 (for the Earth) or 1.6 (for the Moon) and you then follow the last couple steps, to get the velocity squared.

    Overall if I said to you:

    Mass = 7.65635252345 x 10^14 and Height = 500 meters, would you wnt to write down all the calculations of the first way (which is how they expect you to learn it), or would you just times '500' by 10, and then 2 (to make 1/2v^2 into v^2), and then square-root that to find the speed of the falling object?

    It works regardless, it just saves you having to mess around with the mass, which could, in any case in real life, be a really confusing decimal number.

    I found this out about 25 minutes ago (so, about 8:10pm).

    Joshua Brazier.
    "

    Now how it is inter-linked to the falling of the Hammer & Feather on the moon:

    "
    This is also the reason why two objects with different masses hit the ground of the moon together:

    If there is no air-resistance on the moon, which there isn't, mass doesn't have a say in the speed of the object, because the resistance isn't there to slow the object down, therefore just like my formula, mass is irrelevant, meaning:

    You have two objects with different masses, and drop them at the same time on the moon, they WILL hit the ground together.

    Also:

    Speed = Distance / Time

    This means if you have the height of the object which you do, and you follow my steps, or the steps given by the book, you can work out how long it takes for the object to hit the ground.

    Meaning: Just like with the Hammer & Feather experiment on the moon:

    If the Astronaut dropped the object from just above waist height, the height would be 83cm (on average), therefore:

    Speed = 83cm / Time

    Time = 83cm / speed

    Therefore:

    speed = √3.2h

    √(3.2x0.83) (0.83 Because the height is in Meters, and not Centimeters).

    Meaning: √(3.2x0.83) = √2.656 = 1.63 meters per second,

    Therefore:

    Time (seconds) = 83cm / 1.63 meters per second

    Therefore: 0.83 meters (because overall we are using meters per second with speed, and you must then convert the centimeters into meters).

    0.83 / 1.63 = 0.51 Seconds (rounded to 3.S.F):

    Meaning:

    If you dropped a hammer (or any object for that matter) at 83 Centimeters on the moon (waist height for the average person), then it will take 0.51 Seconds to hit the ground.

    This is what happened when the Hammer and Feather were dropped together by the Astronaut on the moon.

    The mass of the objects is irrelevant.

    Overall, 0.51 Seconds is how long it takes for any mass to fall from 83 Centimeters on the moon.
    "

    If anyone has anything to prove me wrong, please tell me, and I will agree and say that I was wrong, although, please read the entire two threads first and then tell me, before making any accusations that I am wrong without fully reading my theory and examples.

    Thank you.
    Eggman100 / Joshua Brazier.
     
  2. jcsd
  3. Oct 9, 2012 #2

    sophiecentaur

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    I didn't read anything like all of this; I recommend editing your posts more heavily in future. But you seem to have got the right idea yet you seem a bit surprised (?).
    However, the beauty of Physics is that it produces equations and formulae that produce the right answer, whatever particular values you insert into them. It can hardly be surprising that, if you equate Gravitational Potential Energy lost in falling (wherever you are) to the Kinetic Energy gained you will find that the feather and hammer experiment on the Moon will give the same basic results as Galileo obtained all those years ago in Pisa. There is no need for 'instances' in order to make a point and you could have done it all in two or three lines.
     
  4. Oct 9, 2012 #3
    No I was in the end basically stating that, if you use Distance = Speed / Time, that is okay if you have two of those values, if you have one, for example, the height of the object at the start, and you do not have the speed or the time, then how could you work it out?

    Simply by just using the two formulas and working out the speed, because the distance is also the height of the object in the formula, I was just simply stating that it has surprised me because you can still only know one value out of the three, and work out the time it takes to fall that distance.

    However I understand your point.

    Also because they don't tell you what I said (about how it is always √(20xHeight) to find the velocity of the falling object, if the object that is falling, is falling from somewhere on the earth (so g = 10N/Kg)).

    They expect you to learn the entire formula, when you can simply times the height by 20 and then square root the answer for the velocity, you don't even need the mass stated in the question, besides you also said you didn't read the entire thread, therefore, I don't wish to be rude, but you cannot comprehend what I said overall, without doing so.

    Eggman100.
     
  5. Oct 9, 2012 #4

    ZapperZ

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    Have you checked the FAQ subforum?

    Zz.
     
  6. Oct 9, 2012 #5

    Doc Al

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    Rather than cutting and pasting a discussion from another site and expecting folks to wade through it, why not just state your point directly?
     
  7. Oct 9, 2012 #6

    sophiecentaur

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    @eggman
    Be careful: distance = speed TIMES time!!
    Your Maths needs to be 100% right if you want to prove anything.

    I'm not sure what point you are trying to make. You seem to be surprised that Maths can reliably give you answers in Physics. The formulae for motion under constant acceleration (not just for falling objects) are correct and, let's face it, the few 'SUVAT' equations are not hard to learn or to understand. What more do you want? You must expect to learn a few things and have them available from memory if you plan ever to get this topic straight.
     
  8. Oct 9, 2012 #7
    However said that I copied the threads from another site obviously didn't read what I said. I said to you that I copied it from MY facebook thread, that was made by ME.

    Read first please, then post.

    No, I was just saying if you have the distance from S=D/T, then you are missing a value, and you can use those physics formulae to work it out, however they didn't tell you that in the book you have to read, that is what I'm saying.

    Rather than having to use the mass in mgh=1/2mv^2 you can just do √20h and then you have the velocity, and how it can just be applied to things like the hammer&feather, and how to work out the time it would take for the impact of an object to happen.

    I first off do not expect to get slated by someone who didn't read my comment and calls me a leecher of another thread, that is actually my work: reference: https://www.facebook.com/RxB0t.Lulzsec.Cybergate/posts/3667009961776?notif_t=like [Broken]

    To the others, simply, I am just surprised at how you can use basic formulae to work out lots of other questions, like for e.g. the time it would take for any object to fall at a given height.

    I don't see them teaching you that in class or at-least referencing it somewhere,

    I take everyone's point of view seriously, but as far as I am concerned, I have been slammed and not understood.

    Mod please close.
     
    Last edited by a moderator: May 6, 2017
  9. Oct 9, 2012 #8

    Doc Al

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    Yes, you cut and pasted it from another site. I realize that it was from your Facebook page. And it seemed like you were presenting an argument for someone else. (References to "prove me wrong" and "my theory".)

    Better for you to state your point right up front.

    Didn't mean to imply that it wasn't yours. (Just that it wasn't clear what you point was. And that it was written for another venue and pasted in here.)

    Sorry to hear that. Basic stuff. (But good to figure out on your own.) What book are you using?

    Was there another point of view that you were arguing against?
     
    Last edited by a moderator: May 6, 2017
  10. Oct 9, 2012 #9
    @Doc Al -

    a.) Agreed I suppose, I was providing an arguement in the terms of, rather than following the procedure in the book, why not say that you can do Sqrt(20h) and then you have the velocity, why do you need to provide all the calculations the book gives when that is simply what you need to do to find the speed of a falling object on the earth, or Sqrt(3.2h) for the moon.

    b.) I did explain that through the thread really, overall how you can use multiple different formulae to get from something as simple as mgh=1/2mv^2 and how it can be applied to finding out the time it takes also for the falling object, just assuming you know the height.

    c.) It's okay, I should have explained myself better. Yeah, but it was just to show quickly what I did to get to my conclusion.

    d.) It's okay, I personally use CGP's books, for additional science, have some AS ones somewhere, but for these exams, the Add-Sci book. We have a new teacher that cannot control the class properly, it is slowly changing, but we have so little time, and we have to take 3 biology classes in one because the school is running out of time for the science exams, so I will fail, because of these kids *sighs*, the kids that don't want to learn will fail and then not have any jobs, but will make those who want to do well also not have any jobs because they are too busy messing around. =/

    e.) No, I just mean, I feel as if I got slammed just because of how I tried to show my work, and that it doesn't make me feel better just to get slated rather than constructive criticism. =$

    It's cool dude, one day, I'll have a degree, and I'll be some really good physicist. Oh, no, wait, that's a dream because I will fail my exams. =(
     
    Last edited: Oct 9, 2012
  11. Oct 10, 2012 #10

    sophiecentaur

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    @eggman
    We all have to deal with adverse teaching and learning conditions to some degree. You have my sympathy in that respect. Have you considered making a (polite) complaint about your situation - just to get it on record?

    But that is an entirely separate issue from your apparent problem with what is basically the way that all Physics tries to treat the World. It's all about making the simplest models possible to describe a process (hence my objection to your first rambling post). Rather than complaining about it, you should be overjoyed that it is possible to cover so many problems in Dynamics with just a handful of simple formulae (basically just four). All you need to do is to learn them and get used to applying the right one to the situation you are given. Forget learning to use √20 for falling objects - that will only confuse you when an object is rolling down a slope or under the influence of brakes. The CGP books are fine and have all those formulae in them with worked examples.
    Don't get mad - get informed.
     
  12. Oct 10, 2012 #11
    First of all: mass is essential for such kinetic energy calculations. However:
    That immediately reduces to gh = 1/2v^2.
    And if I understand you correctly, that is the essence of what you are saying here; the calculation example that they next give by including multiplication with the mass is unnecessarily cumbersome. Thus:
    Exactly. It looks as if you are supposed to be able to plug in numbers, but not supposed to be able to do basic math. :grumpy:
     
    Last edited: Oct 10, 2012
  13. Oct 10, 2012 #12

    sophiecentaur

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    @eggman
    If you are doing AS or A2 you should be using 9.8 or even 9.81 ms^-2.
     
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