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The Skyhook

  1. Apr 5, 2012 #1
    Several science fiction writers have proposed a simple “skyhook” satellite. This
    would consist of a (very) long rope placed in a geostationary orbit at a point directly
    above the equator, and aligned along the radial direction from the centre of the
    earth. The bottom end of the rope would thus remain just above a fixed point on the
    equator and the rope would extend vertically upwards. Find the length of rope which
    would be needed for such a device. Express your answer in units of the earth’s
    radius RE

    You can assume that the rope would have a uniform mass density and be strong
    enough not to break. You can also ignore the mass of any actual satellite at the end
    of the rope.

    So i assumed straight away that this question is all about infinitesimals. I thought the approach was, if you worked out what forces were acting on the rope, these could be summed over the entire length of the rope, as if they are broken small enough, these forces would essentially be constant. From this it would then be possible to solve for 'L', which is the length of the rope.

    Firstly i equated the forces;

    Gravity acting towards earth and 'centrifugal' force acting out due to the rotation of the earth.

    => mg = (mv^2)/r
    => mGM/r^2 = mw^2r

    These i thought would be integrated from RE to RE+L (where L is the length of the rope).

    However, i couldn't actually get to the point where you would solve for L so have became a bit lost.

    Any help would be appreciated. Thanks.
  2. jcsd
  3. Apr 5, 2012 #2


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    I can't help you with the math, but I can tell you that Space Elevators, as they are called, are no longer science fiction. There are pilot projects under way now. I'm sure a Google search will turn up lots of info.
  4. Apr 5, 2012 #3


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    Staff: Mentor

    The question asks for length, not force. Yiu need to calculate the altitude of geostationary orbit.
  5. Apr 6, 2012 #4
    Yes, but the force acting on the rope is related by r^2, which in this case, will be the length of the rope no? So by summing the forces, you can also find the length of the rope. At least this is how i approached it. Not sure where you would even begin with calculating the altitude.
  6. Apr 6, 2012 #5
    You need to integrate, both the gravity and the centrifugal force ove the entire length of the rope starting at the surface to the altidude of the top of the rope h, and equate both of them

    [tex] \int_R^{R+h} {\rho g \frac { R^2} {(R+r)^2}} dr = \int_R^{R+h} {\rho \omega^2 r dr} [/tex]

    where [itex] \rho [/itex] is the mass per length of the rope (wich will fall out) and [itex] \omega [/itex] = 2 pi /86400 = the angular velocity of the earth and R is the radius of the earth.
  7. Apr 6, 2012 #6
    Aha, this is what i did, but i set my first integral up incorrectly. So if i'm not mistaken, i solve these integrals then isolate h, which will be the length of the rope, right?

    Thanks A LOT for the help mate.
  8. Apr 6, 2012 #7


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    Staff: Mentor

    I see what you are doing and agree that it is an answer, but I think the question is too poorly worded to be certain that that is the answer the teacher is looking for.

    A space elevator concept typically utilizes a large, weighted satellite as ballasted counterweight. Your calculation will yield the length of a self-balancing rope with no counterweight. When the teacher says to ignore the weight of the satellite, it isn't clear to me if s/he is saying that the counterweight is arbitrarily small (your assumption) or arbitrarily large (my assumption). If it is arbitrarily small, why not just say that it isn't there at all? Your formulation does not require it to be there. If we assume that it is arbitrarily large, then it is in geostationary orbit itself and its altitude can be found by equating the centripetal acceleration equation to the acceleration due to gravity equation and solving for r.
  9. Apr 6, 2012 #8
    i see your point Russ and thanks again for replying. It really is much appreciated. I actually found the question in a book called 'problems in physics' written by Abhay Singh.
    I'm assuming from what you said, it ends up working out to be the same though? As if i'm not mistaken you are essentially treating the satellite (at the end of the rope) as almost like a dead weight attached to it right? and in this case are equating the forces acting on the satellite and thus how far it must be from the earth.

    I may have misunderstood, but i think this is what you mean. It is worded fairly poorly as i seem to find the problem with a lot of the questions i do.
  10. Apr 6, 2012 #9
    also i know i being very stupid here but i can't quite see why its (R/r+R)^2 - how comes you dont just have r^2 integrated? but you do just have r^2 for the centrifugal - thanks again guys!
  11. Apr 6, 2012 #10


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    No, they aren't the same. Your version has the rope acting as its own counterweight and in order to do that it has to extend far beyond the altitude of GSO.
  12. Apr 6, 2012 #11

    D H

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    Russ: There is no counterweight here. It's just a long piece of rope.

    Here are a couple of ways to look at this problem. The two approaches will yield the same answer.

    One approach is to look at things from the perspective of a rotating frame, one in which the rope is stationary. This is essentially your gravitational plus centrifugal concept. The net external force on the rope must be identically zero for the rope to be stationary. There are two external forces acting on the rope, gravitational and centrifugal. Note that gravitational force is directed inward while centrifugal is outward. You had the wrong sign on the gravitational force, but then you put the total forces on opposite sides of the equal sign. This yields the right answer, but more by luck than smarts.

    Another approach is to look at things from the perspective of an inertial frame. The net external force on an object is equal to the product of the object's mass and the acceleration of the object's center of mass. The only external force from this perspective is gravitation. Integrate to get the total. The location and acceleration of the object's center of mass are implied by the problem statement. Hint: The object's center of mass is in uniform circular motion.
  13. Apr 6, 2012 #12
    Thanks for the response DH.

    Why exactly in an inertial frame does the centrifugal force no longer have an impact?

    Still not entirely sure why its R^2/(R+r)^2 for the force due to gravity. I can see its trigonometry, is this due to only a component of the force due to gravity being taken into account? i.e mgcos(theta) => pglcos(theta),
  14. Apr 6, 2012 #13

    D H

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    Centrifugal force is a fictitious force. There are no fictitious forces in an inertial frame.

    Here's a way to envision this. The rope is stationary from the perspective of a frame that rotates with the orbiting rope. The net apparent force on the rope must be identically zero in this frame. The rope is orbiting the Earth from the perspective of an inertial frame. The net apparent force cannot be zero in this frame. If it was zero the rope would have to be moving at a constant velocity (stationarity is a special case of constant velocity motion).

    It's not. willem2's post is erroneous. Try writing the correct integral for the total gravitational force on the rope.
  15. Apr 6, 2012 #14
    Ok, i think i see what you mean now.

    The integral for the gravitational force would be

    The integral from R+h to R (this will account for the negative sign, that you spoke about, as i flipped the integral and added a negative which now makes it positive)

    mg dr ;

    where dm = pdr
    and g = GMe/r^2

    Thus, you are integrating pGMe/r^2 between the limits above.

    When integrated this yields;

    -((pGme/R) - pGMe/R+h))

    Does this look at all correct?

    Thanks for sticking with me!
  16. Apr 6, 2012 #15

    D H

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    A suggestion: The problem asked for the length of the rope in terms of Earth radii. You are using length in units of meters. It will be easier if from the onset you use a unitless parameter, call it l. Then your integral becomes
    [tex]F_{\text{grav}} = \int_{R_e}^{R_e(1+l)} -\frac {GM_e \rho}{r^2}\, dr[/tex]

    That should be good enough for your level of work. As an aside, you'll often see the product [itex]GM_e[/itex] replaced by a single parameter [itex]\mu_e[/itex], for example,
    [tex]F_{\text{grav}} = \int_{R_e}^{R_e(1+l)} -\frac {\mu_e \rho}{r^2}\, dr[/tex]
    This latter change from [itex]GM_e[/itex] to [itex]\mu_e[/itex] is perhaps a bit pedantic, but it is quite important if you want high accuracy. The universal gravitational constant is only known to 4 decimal places. It is arguably the least certain physical constant. The same goes for the mass of the Earth. The product [itex]GM_e[/itex] is independently measurable, and to a very degree of precision. The Earth's standard gravitational parameter is [itex]\mu_e = 398600.4418\pm 0.009\,\text{km}^3/\text{s}^2[/itex].
    Last edited: Apr 6, 2012
  17. Apr 6, 2012 #16
    One last question, do i not need to worry about the centrifugal force anymore, using the integral you displayed (as in, is this the way of considering the inertial frame).

    Thank you so much for talking me through it, you really were an amazing help.

    It's always good to be accurate, so i will keep that in mind for future questions.

    I hope i can be of some help at some point in the future...

  18. Apr 6, 2012 #17

    D H

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    Real forces such as gravitation are frame invariant. The gravitational force on the rope is the same in the inertial and rotating frames. The difference between the two perspectives is that in the rotating frame you are trying to solve [itex]F_{\text{gravity,tot}} + F_{\text{centrifugal,tot}} = 0[/itex] while in the inertial frame you are trying to solve [itex]F_{\text{gravity,tot}} = F_{\text{centripetal,CoM}}[/itex] . You already have the total gravitational force [itex]F_{\text{gravity,tot}}[/itex]. The total centrifugal force [itex]F_{\text{centrifugal,tot}}[/itex] is another integral. The effective centripetal force [itex]F_{\text{centripetal,CoM}}[/itex] is the centripetal force needed to make a point mass undergo uniform circular motion at one revolution per sidereal day such that the mass of the point mass is that of the rope and the location of the point mass is the rope's center of mass.
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