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Falling objects: help an ignorant fool

  1. Jan 16, 2007 #1
    First of all, let me start by saying that I am quite uneducated when it comes to physics (not even as much as a high school physics course), so please be gentle with me. I wouldn't even be at your lovely forum, but this question has been driving me nuts, and no one else seems to answer it to my satisfaction. Thanks in advance for any help you provide.

    Do all objects fall at the same speed or not? I've been deeply confused over this issue for some time. The conventional answer is that, yes indeed, all objects do fall at the same speed. The acceleration of an object depends upon the force of gravity and the mass of an object. A more massive object will require more force to accelerate at the same speed as a less massive object, and gravity will act with greater force upon a more massive object. The conflicting points exactly cancel each other out, such that all object accelerate at the same speed.

    But if I accept this, I am confronted with a wide range of paradoxes I can't explain. I'll list three of them:

    1) If I were to drop a bowling ball toward the Earth, it would fall at a certain rate. If I were to drop the same ball toward the Moon, it would fall at a slower rate. Correct? The more massive the object (Earth vs. Moon), the greater the force of gravity and the greater the rate an object will accelerate tworads it due to gravity. Correct? But why should I use the Earth or the Moon as my frame of reference? Why can't I use the bowling ball as my frame of reference, and drop the Earth and the Moon towards the surface? In that case, we already know the Earth would fall faster twoard the bowling ball, and we know this is *because* of the greater mass of the Earth. So "heavier" or more massive objects do fall faster, don't they?

    2) I saw on a physics website an equation that is used to determine the mass of a planet by setting an object in orbit around it. The speed with which this object orbits the planet determines the mass of the planet, but cannot (according to the site) determine the mass of the orbiting body- to do that you would have to set some other object in orbit around the first object. Why? Because, says the site, all objects, no matter the mass, will orbit at the same speed. So let's assume that we are sitting on an object in a universe devoid of anything other than the object we are sitting on and a body orbiting us/our object. We can use this body to determine the mass of the object we are on (pretending we are devoid of mass ourselves). Correct? So what if I said the object we are on is a satalite used by NASA and the orbiting body is a planet? Again, what object is orbiting what is simply an arbitrary frame of reference. I can just as easily choose the satalite as the object being orbited as I can the planet! And in doing so, I would come to the exact conclusion that the site says I can't- determine the weight of the object doing the orbiting. What has gone wrong here?

    3) There was a debate for a long time regarding whether or not the universe would continue to expand forever or would collapse back in on itself. It was argued that if there was enough mass in the universe, it would collapse, and otherwise it would continue to expand. If mass is not important to the acceleration of two objects towards each other, why would it matter how much mass is in the universe at all? It seems the argument requires that more massive objects accelerate faster than less massive objects.

    Is the whole idea of "all objects fall at the same speed" just bogus? Are the intro physics books simply wrong or oversimplifying the issue? What am I missing here?
  2. jcsd
  3. Jan 16, 2007 #2
    1) The trick is that you have to use a non-accelerating frame of reference. In this frame you'll find that all other bodies accelerate towards the bowling ball at the same rate. The reason we can normally use the earth as our frame of reference is that it's acceleration is very small (because the ball has such a weak gravitational pull).

    2) Same problem, the equation uses approximations only valid if the first object is much more massive (eg. assumes that the center of mass is at the first object). From the surface, without reference to other bodies, I'm not sure how to distinguish a moon from a planet.

    3) While it's true the mass of a projectile does not determine whether it falls back down again, the mass-density of the universe does nonetheless determine the curvature of space-time. This is in accordance with a theory that is of a higher mathematical level and which turns out to be surprisingly accurate.

    In your attempt to list paradoxes you seem to obviously realise that a cannon ball and a fishing weight, if dropped from a house, will accelerate down at the same rate. What you're missing is how importantly this contradicted prior common belief, and thus initiated the scientific study of gravity.
    Last edited: Jan 16, 2007
  4. Jan 16, 2007 #3


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    When an object falls towards the earth. It is being influenced by two factors - gravity and air resistance. It is because of air resistance that a bowling ball would hit the floor faster than a sheet a paper. Take air resistance out and both objects hot the ground at the same time.

    Sounds good to me.

    Actually, I think you are confused. When you drop an object to the earth, both the earth and the object move towards each other to conserve momentum! The movement of the earth is negligible becuase it is massive. If you where to somehow drop the earth to the ball. Both would move towards each other, but the movement of the ball would be more noticeable becuase of its size.

    Consider this, in a vacuum, the only force acting on an object falling towards a surface is gravity. The acceleration can be found by:
    [tex]a = \frac {F}{m}[/tex]
    But since gravity is the only force, F in the above equation can be replaced by the objects weight, W.
    [tex] W = mg[/tex]
    where m is the mass of the object and g is acceleration due to gravity.
    [tex]a = \frac {F}{m} = \frac {W} {m} = \frac {mg}{m} = g[/tex]
    Thus the acceleration a is equal to g.
    Last edited: Jan 16, 2007
  5. Jan 17, 2007 #4
    Are you nuts? How can the earth fall towards the object? According to you, if we drop two objects, at exactly opposite sides of the earth, and at exactly the same time, the earth would have to split in two to reach each object, considering the earth's atmosphere was large enough to keep the object in the air for an exceptionally long time.

    If you're right, please explain your statement to help me understand.
  6. Jan 17, 2007 #5


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    No I am not nuts :rolleyes: This is exactly what happens. When you drop a ball, the action is the ball being pulled towards the earth, the reaction force is the earth being pulled by the ball. Because the earth is so massive, it seems unaffected by this. Please read up on the law of conservation of momentum.

    Consider the following example:
    A 5 kg ball is dropped [from rest] and allowed to fall for one second. The force pulling on the ball is 49N:
    [tex]F = ma = 5 * 9.8 = 49[/tex]
    The acceleration is 9.8 m/s/s
    [tex]a = \frac {F}{m} = \frac {49}{5} = 9.8[/tex]
    Since the ball is allowed to fall for one second, its velocity is also 9.8 m/s.
    This gives the ball a momentum of 49 kg m/s:
    [tex]p = mv = 5 * 9.8 = 49[/tex]

    Now remember I said that the earth is also falling towards the object. The force on the earth is 49N. The acceleration of the earth is 8.16*10^-24:
    [tex]a = \frac {F}{m} = \frac {49}{6*10^{24}} = 8.16*10^{-24}[/tex]
    Thus after one second the earth will be moving up at a velocity of 8.16*10^-24 m/s. Now the momentum is 49 kg m/s in the opposite direction!
    [tex]p = mv = 6*10^{24} * 8.16*10^{-24} = 49[/tex]

    Thus momentum is conserved!
  7. Jan 18, 2007 #6


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    It's not acceptable to go about calling people "nuts", especially when they're right and you don't know they're wrong. Please show a little more respect to people, particularly those that you're asking to help you understand better.
    Last edited: Jan 18, 2007
  8. Jan 18, 2007 #7
    The earth's mass has something to do with its overall kinetic energy around the sun, which would effect it's gravity. And it's speed around the sun has something to do with the mass of the sun, which has something to do with ITS momentum relative to everything else.
  9. Jan 18, 2007 #8
    Thanks for the replies everyone. But I'm still confused.

    Will I? I thought the Earth would accelerate towards the bowling ball faster than the Moon. In fact, it seems quite obvious from the videos I've seen of people in space that a marble will have almost zero acceleration towards the bowling ball compared to the Earth- when starting from the same distance apart. Your answer just seems so obviously wrong. But everyone always gives it, so I must still be obviously missing something.

    That aside, what exactly does it mean to use an "accelerating frame of reference" and how do you know I doing that? Isn't every object at rest with respect to itself? Only the Moon and Earth were accelerating in my example, as far as anyone can prove. In a universe consisting of only two objects, what reference frame isn't accelerating?

    Again, if this was a properly controlled experiment with a universe consisting of only two objects, you would have no way of knowing which object's movement is "more noticable". Forget "falling" objects, where one object is much more massive than another. Is it true or not that the relative acceleration of two objects twoards each other is dependant upon the combined mass of both objects? Won't a bowling ball and a marble accelerate towards each other more slowly than a bowling ball and the sun- when starting from the same distance apart?

    I don't understand this at all. Isn't g a constant? But acceleration isn't constant- it depends on distance. An object dropped on a mountain top will accelerate more slowly than an object dropped at sea level. Correct? And we are talking about two objects accelerating towards each other. Why is there only a single variable for mass in the equation?
  10. Jan 18, 2007 #9


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    I see the point you are trying to make. But this is strictly from an observable view point. If you do the mathematics of it (my second response), it will become clear as to which object is falling towards the other at a faster rate. Just in case you are wondering how I got that force acting on the earth, its newtons 3rd law.

    Every mass has a gravitational field that will exert an attractive force on another mass. The strength of the field in proportional to the mass of the object.

    I simply did that to show you that the only force acting on a mass in free fall [ignoring air resistance] is gravity itself. In other words, the claim made by the book isnt bogus. Note that g is measured from the center of the earth to sea level. So 9.8 m/s/s is simply the force experienced by an object near the surface. You can find g for the different distances by varying r in following formula:
    [tex]g = G \frac{m_1}{r^2}[/tex]
    [tex]g = 6.67 * 10^{-11} \frac{5.9 * 10^{24}}{(6.3 * 10^6)^2} = 9.8[/tex] <--near the surface.
    Last edited: Jan 18, 2007
  11. Jan 18, 2007 #10
    Thinking in terms of accelerating reference frames is the correct answer.
    You know you are in one because you will observe an inertial force.
    For example, if you place a marble in your hand while accelerating then it will roll in response to this acceleration. If you are simply moving but not accelerating then there lacks any measurable quantity that makes your frame unique.
  12. Jan 18, 2007 #11
    It is actually true that the marble will accelerate towards the earth more than it does towards the bowling ball, and you'll note this is exactly the reason why your original argument 3 fails.

    If you use the appropriate equation (supplied by ranger), you will find the marble accelerates toward the bowling ball at exactly the same extremely slow rate that the earth accelerates toward the bowling ball. However, the bowling ball accelerates towards the earth much faster than that ball accelerates toward the marble.

    The (gradient of the) rate* at which the distance closes, between the bowling ball and the earth, is calculated by adding the two accelerations: producing the quantity that conforms with your intuitions (being much faster than the equivalent quantity for just the ball and marble).

    (*Really need another word that nicely expresses second derivative.)

    In this case this is untrue, and this is what makes scifell's questions so difficult to answer. (According to relativity theory, motion under the force of gravity still counts as inertial.) In a satellite, your hand will fall towards the planet at the same rate that a marble in your hand does, so you can't easily detect the fact that the satellite is circling the planet (or more likely, that they are both circling a point in between, somewhere closer to the bigger one).
    Last edited: Jan 18, 2007
  13. Jan 18, 2007 #12
    I don't want to be rude scifell but the formulas that are being introduced, are very simple algebra formulas in which he is merely substituting physical values, in order to give you a physical understanding of what he is explaining.

    If you are not able to grasp the mathematics, perhaps you should first develop an understanding of the physics concepts as that might help you more. At the very least, don't denounce the answers given to you, if you don't understand the math that is being introduced in connection with the physics.
  14. Jan 18, 2007 #13
    Thanks everyone- I think I understand now. Basically, the answer to my original question ("do objects fall at the same speed?") depends entirely upon what is meant by "fall". If it means the time taken for an object, A, to travel some distance, d, as a result of gravity from another object, B, then yes, all objects fall the same speed. However, if what is meant is how long it takes for objects A and B to come into contact as a result of gravity when starting distance d apart, then the answer is no (due to the fact that object B will also travel some portion of distance d).

  15. Jan 18, 2007 #14


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    Right now, you are being pulled toward the earth, the sun, the moon, Jupiter......

    Are you being split into 4+ pieces?

    Have you taken geometry yet and learned of the concept of vectors? You can take multiple forces, accelerations, or other vector quantities up and add up the component pieces to make calculating them easier, then add them all together to get a resultant.

    Your attitude is really hindering your ability to learn, linux kid. Don't think you already know everything - if someone says something that seems wrong to your preconceptions, try to learn what the explanation is. You need to accept that you need to learn.
    Last edited: Jan 18, 2007
  16. Jan 19, 2007 #15
    ranger, I must commend you on your point about the earth moving towards the object in response to the object's movement to the earth as an act to conserve momentum.

    I've always merely thought about it in newton's third law. Eg. Force of object on earth = force of earth on object.

    Edit: Actually conservation of momentum is derived from Newton's third so ok, but still i've never thought of it that way. THanks!
  17. Jan 23, 2007 #16
    Cesium, you bring up a good point (this isn't the first time you have corrected an oversight on my part.):blushing:
    I was wondering though, if you cannot feel an inertial force when all parts accelerate simultanously (as in gravity) then does this imply that some accelerating reference frames are unmeasurable. Specifically, what if our entire universe had a translational acceleration. Would there ever be any hope of proving this?
  18. Jan 23, 2007 #17


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    Are you asking how to prove all objects fall at the same rate? With the caveat already discussed about the earth being pulled toward the object in freefall, plugging Newton's gravity equation into f=ma yields a=f/m=9.8.
  19. Jan 23, 2007 #18
    That wasn't quite what I was driving at. I wasn't really clear, the entire subthread related to my question is as follows:

    What caught my attention is that I was taught to consider accelerating reference frames as measurable. This not only resolved problems like the twin paradox (the twin that ages is the one in the rocket ship, and the rocket ship is the one that accelerated because of a measurable inertial force), but also placed accelerating reference frames outside of the reach of special relativity, (ie: since everyone can agree on the magnitude of an inertial force due to an acceleration, this implied that the frame was not relative to another.) Is all of this correct?

    Anyways all this made me wonder if acceleration was always a measurable quantity. Cesium's post implies that you cannot alway count on inertial forces as your "yard stick". So in that case what do you rely on. For instance, if the entire universe where undergoing a translational acceleration, would we be able to determine this?

    Hope this is now more answerable.:smile:
  20. Jan 23, 2007 #19
    No one mentioned a simplification used on the original question.
    Newton's gravitational law states that the force(s) between two punctual masses is directly proportional to each mass an inversely proportional to the square of the distance separating them.
    The gravitational force produced by any object depends therefore on where it is exerted (what's the distance between the ball and the center of the Earth)

    In "free fall" problems some assumptions are commonly made:
    1) No air resistance is taken into account. Otherwise the acceleration wouldn't be determined by gravity alone.
    2) The frame of reference is taken to be inertial, that is, it's imposed that the Earth has a fixed position in space and the Newton laws are valid in a frame attached to it.
    3) The value of the gravitational force doesn't change sensibly within all the range of the fall. It will take a free fall of several kilometers to be able to notice a change on the acceleration.

    In the satellital problem (two bodies alone in deep space) the approach is usually to take as reference frame one in which the center of mass is at rest. Since there are not external forces on the system it is supposed to be an inertial frame. Each body attracts the other with the same force, producing different accelerations if the masses are unequal.

    In the mentioned site in which the planet mass alone is obtained, probably they have assumed that the planet has a mass so much greater than the satellite's that the center of mass almost overlaps with the (center) of the planet at any given time, allowing to say that the planet is stationary (in the sense that one can attach an inertial frame to it).

    Note on the math: you can 'translate' some formulas to 'daily language' thinking in the products as direct proportionalities and divisions as inverse ones.
  21. Jan 23, 2007 #20

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    Ignoring air resistance, would two objects with different masses (e.g., a bowling ball and a marble) released from the same height hit the Earth at the same time? Mathematically, the answer is no. The Earth does indeed fall toward the bowling ball or marble while the bowling ball (or marble) is falling toward the Earth. The bowling ball will hit the Earth first.

    Physically, the answer to the question is yes. The time difference is immeasurably small.

    Ignore everything but the Earth and the bowling ball or marble. Mathematically, the center of the Earth is not an inertial frame because the Earth accelerates toward the bowling ball (or marble). The mathematically correct inertial reference frame is the Earth/bowling ball center of mass frame. This frame is not measurably different from the Earth center of mass frame.

    One difference between physics and mathematics is that physicists know to ignore things that don't matter. In this case, even though the Earth is accelerating toward the bowling ball, we can safely ignore that acceleration and treat the center of the Earth as an inertial frame.
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