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Distance from earth an object would have to fall from to reach speed x

  1. May 11, 2013 #1
    big G = G = 6.674E-11 N(m/Kg)^2

    Mass of earth = Me = 5.972E24 Kg

    Radius of earth = Rsurface = Rs = 6.371E6 metres

    THIS IS NOT HOMEWORK

    Okay so I was interested to know how you would work out the distance you'd need drop an object from (above the earths surface), for it to reach a given speed (neglecting air resistance).

    Here's what I've got so far:

    We know Force = F = G *(Mass of object 1 * Mass of object 2)/radius^2 = G * Mm/r^2

    F = mass * acceleration = m*a

    F/m = a

    a = G * m/r^2

    This is an equation for the acceleration of the object at any distance away from the centre of the earth (considering earth as a point mass for distances < Rs), taking into account the inverse square law of gravity.

    Right now it would be handy to have a in terms of t, so we could integrate to get an equation for v = velocity (units m/s). But instead we have it in terms of r (units metres).

    So I thought I might be able to integrate a in terms of r away to get something with the units (m/s)^2, which I thought might give us v^2. which we could square root to get an equation for v in terms of r? But I wasn't sure if that is correct/if I had overlooked some concept which means it isn't giving me what I wanted.

    Here's the integration I did:

    a = G * Me/r^2

    v^2 = G * (integralOf(m/r^2)dr for the range lower limit = Rs to upper limit = r) = G * ([-(Me/r)] for the range Rs to r)

    so v^2 = G * (-Me/r - (Me/Rs) = G * Me*(r - Rs)/(r * Rs)

    so v = Sqroot( G * Me *(r - Rs)/(r * Rs) )

    with a rearrangement

    ( (r*Rs) * v^2 )/(G*Me) - r = -Rs

    r * ( (Rs*v^2)/(G*Me) ) - 1 ) = -Rs

    r = -Rs/( ( (Rs*v^2)/(G*Me) ) - 1 ) )

    Now I have tested this out and it wasn't correct (too far out to be correct when I compared it for the constant acceleration model given by the suvat equations).

    But I did notice that if you do this (using suvat equations):

    a = 9.81 acceleration

    s = r = displacement in metres

    u - start velocity = 0 m/s (metres per second)

    v^2 = 0^2 + 2*a*s

    v = Sqroot(2*a*s) you obviously get the right answer for constant acceleration, but what they're doing here is multiplying a by a distance vector to get v^2, which makes me think my decision to integrated above should have been okay (since integration find the area under a curve, which is like multiplication of the axes).

    Then I noticed that if you divide the right hand side of that suvate equation by Sqroot(2), you get extremely similar answers to my previous equation:

    v = Sqroot( G * Me *(r - Rs)/(r * Rs) )

    So I multiplied the RHS of my variable acceleration equation by Sqroot(2) and it seems to be right. But I'm not sure where I went wrong in the first place!

    New equation:

    v = Sqroot(2 * G * Me *(r - Rs)/(r * Rs) )

    So I do the rearrangement for r again and get (thinking it will work this time):

    r = -Rs/( ( (Rs*v^2)/(2*G*Me) ) - 1 ) )

    Now it works! That value of r is including the earths radius, so you need to minus 6.371E6 to get the distance about the earth surface :).

    What I actually want help with READ :P:

    BUT I'm still not sure where that factor of 2 came from. I think it may have come from the integration, but the power of r in my initial equation for acceleration, was -2, which you increase to -1, then divide by the new power (being -1, not 2). Any clarification will be appreciated. Thanks!

    EDIT:

    Okay so I thought I'd also have a go at trying to find out the true gravitational potential energy, considering the inverse square nature of gravity.

    I just subbed the equation for v^2 into KE = 1/2 * m * v^2 :).

    So true GPE = 1/2 * m * (2 * G * Me *(r - Rs)/(r * Rs) ). If you want it relative to the centre of the earth make Rs 0, or if just for when the object hits the earths surface make Rs = 6.371E6 :).
     
    Last edited: May 11, 2013
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  3. May 11, 2013 #2

    arildno

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    "So I thought I might be able to integrate a in terms of r away to get something with the units (m/s)^2, which I thought might give us v^2. which we could square root to get an equation for v in terms of r? But I wasn't sure if that is correct/if I had overlooked some concept which means it isn't giving me what I wanted."

    Sure. But remember that for EVERY dimensionless constant "k", the quantity k*v^2 has the right dimensions!

    Thus, without knowing which "k" is the right one, you won't get the right answer.


    The factor 1/2 is the one appearing in the quantity called "kinetic energy",
     
  4. May 11, 2013 #3
    Thanks for the reply. I'm still not sure where I went wrong. I'd have thought that is my acceleration equation was correct, then as long as I integrated everything properly my v^2 equation should still be correct, because I'd have though that the division part of integrating would have sorted out any constant factors?

    Also where is the 1/2 that is appearing you speed of? It is a factor of 2 that appears, which is the reciprocal of 1/2. But I haven't used the KE equation. I know that KE = 1/2 m v^2 can be derived using F = ma, which I did use.

    Any more explanation would be appreciated :).
     
  5. May 11, 2013 #4
    Step by step

    a = GM/r[itex]^{2}[/itex]

    a dr = GM dr / r[itex]^{2}[/itex]

    a = dv/dt

    Subst and re-arrange

    vdv = GM/r[itex]^{2}[/itex] dr

    Integrate, yielding

    v[itex]^{2}[/itex]/2 = -GM/r

    v[itex]^{2}[/itex] = -2GM/r


    Now you have the 2 ( and the square root of 2 )

    Take your limits or r from r1 to r2 where r1>r2 and you should come up to your good equation.
     
  6. May 12, 2013 #5
    Thank you very much! So my mistake was to presume that I could jump from:

    a dr ----> v^2

    because the units matched up :P. Well this was fun and definitely helped me brush up on my maths :)!!!!!

    Regards, Greg.
     
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