1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Space Elevator: Physics Challenge

  1. May 4, 2009 #1
    Space Elevator

    Your first job after graduation you are hired by a engineering firm that just got a contract to build the first space elevator. The space elevator consists of a cable hanging from geo-stationary satellite to an anchor point on earth’s surface. Once installed this equipment can be raised and lowered on the elevator, saving the expense and risk associated with rocket launches and reentry. In this problem the radius of the earth will be taken as Re, mass of earth is Me, rotation of the earth (months per day) given by angular frequency is ω (e). The satellite has mass Ms, and system circular orbit of Rs. The elevator cable has length Rs-Re to reach the surface of the earth, and a constant mass per unit length λ.

    A) Where are on earth should the anchor station be located in order for the cable to extend vertically upwards from the surface of the earth? And Why?
    B) Let the tension on the cable where it attaches to the satellite be T (some variable). What is the Ms of the satellite that the satellite must have to maintain in stable circular orbit? (What mass must the satellite have in order to maintain in stable circular orbit in the presence of the weight of the cable and its own weight?)
    C) At what radius Rs in the presence of centripetal force of the weight of the cable in addition to its own weight must the counterweight be? What mass must the satellite have in order to maintain in stable circular orbit in the presence of the weight of the cable and its own weight.
    D) What is the tension on the cable as a function of the height? Consider only the weight of the cable and neglect the tension due to external load hanging on the cable.
    E) What is the tension at the top of the cable (where it connects to the satellite)? The cable is made of a material with mass density ρ.
    F) What’s the minimum force per unit cross sectional area required to maintain its own weight?
    G) Both the climbers ascend on the stationary cable at a fixed # of meters of cable per second. What are the radial and tangential components of the velocity of the climber as a function of height? (since the cable is rotating with the earth it has a tangential velocity)
    H) What is the force by the cable on the climber? (has to satisfy Newton’s second law)
    Do Not Use mg! (Surface of earth diff from orbit)
    A given amount of cable will weigh different half way up.

    Basically the questions are:

    A) Where does the anchor have to be located?
    B) What should the mass of the satellite be?
    C) What is the minimum theoretical value for the radius of the satellite?
    D) What’s the tension of the cable as a function of height?
    E) Evaluate that at the top of the cable.
    F) How strong does the cable have to be in terms of force per unit area?
    G) What’s the velocity of the climber, radial and tangential?
    H) What is the force by the cable on the climber?
  2. jcsd
  3. May 5, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi jones590! Welcome to PF! :wink:

    Try A B and C first.

    Show us how far you get, and where you're stuck, and then we'll know how to help! :smile:
  4. May 5, 2009 #3
    A) The anchor station should be located at the equator. This is due to the gravitational force being as close to g = 9.8 as possible. The farther you go from the equator the gravitational force slightly changes up and down. Making the best location for the anchor, on the equator.

    B) T = m(ω^2)R so m = (T)/((ω^2)R)

    C) This is where I get confused.....but I think it might be

    (m(ω^2)R)+(1/3λ(ω^2)(R^2)) = ((MmG)/(r+R)^2) - ((GMλ)/R) + ((GMλ)/r)

    Where if you solve for R you get the theoretical radius/length of the cable. Where (1/3λ(ω^2)(R^2)) is the solution to the integral for the centripetal force.
  5. May 5, 2009 #4


    User Avatar
    Homework Helper

    The satellite is supposedly in geosynchronous orbit. To do this it must remain directly overhead as the earth rotates. There is only 1 orbit aligned with the equator that will meet this condition.
  6. May 6, 2009 #5
    I'm very confused with part D, can someone help?
  7. May 6, 2009 #6
    I have the same challenge problem in my physics class. Your choice for the anchor point is correct, but for the wrong reason. You place it at the equator not because of the value of g, but because if you place it anywhere else on the planet it wouldnt be in the same place at all times.

    Its period might be the same as earths, but it would bob through the equator. The earth rotates on a given axis with a given period. For a satellite to be geostationary it must have the same period and rotate on the same axis. Otherwise the area underneath the sattelite would move.

    I agree with your answer for B.

    However I disagree on C. The solution I arrived at is as follows:

    I said that the centripetal force of the sattelite must equal the force of gravity on the sattelite plus the total force of gravity on the cable.In this expression R is the radius from the center of the earth to the sattelite, and r is the raidus of the earth so:

    (m)(w^2)(r) = [ (G)(mass sattelite)(mass earth)/(r^2) ] + [Integral_of: (-G)(Lamda)(r)(mass earth)/(r^2)]
    The integral in this expression should be evaluated from r to R

    this gives you:
    (m)(w^2)(r) = [ (G)(mass sattelite)(mass earth)/(r^2) ] + [ (-G)(Lambda)(mass earth)( ln (R) ) - (-G)(Lambda)(mass earth)( ln (r) ) ]

    And then you would solve for R
    Last edited: May 7, 2009
  8. May 7, 2009 #7
    Can anyone shed some light on C?
  9. May 7, 2009 #8
    jones or Richard jones? That is the question
  10. May 7, 2009 #9
    How would you get started on part F or H? I'm kinda lost on this too
  11. May 7, 2009 #10
    What about this for C:

    (mass sat)(w^2)(R sat) = (mass sat)(w^2)(R sat - R earth) + [ (G)(Mass Earth)(Mass Sat)/(R sat)^2 ]

    What this expression says is that the centripetal force of the sattelite must be equal to the tension in the rope plus the force of gravity on the sattelite. The radius in the tension of the rope/cable is (R sat - R earth) because it foes from the sattelite to the surface of the earth. Solving for R sat you get:

    R sat = [ [ (G)(Mass earth) ] / [ (w^2)(Radius earth) ] ]^(1/2)
  12. May 7, 2009 #11


    User Avatar
    Science Advisor
    Homework Helper

    Hi fogel1497! Thanks for your PM. :smile:
    Well, suppose the satellite were at the usual height for geostationary orbit …

    then the centripetal acceleration it needs would be exactly provided by the gravitational force on it …

    and so any extra force (from the tension in the cable) would start to pull it down …

    but if it were slightly further out, some tension would be needed to help out the gravitational force. :wink:

    (and btw, you're right about A, of course … the satellite would "bob through the equator" if it started either side of it … it has to go in a circle around the centre of the Earth, so it can't "hover" anywhere other than over the equator)
  13. May 7, 2009 #12
    So by what your saying:

    Tension (directed downwards) + Gravity (also directed downwards) = Centripetal Force


    1. Centripetal Force = Tension + Gravity
    2. (mass sattelite)(v^2)/(radius sattelite) = Tension + (G)(Mass sattelite)(Mass earth)/((radius sattelite)^2)
    3. Sub in v = (GM/R)^(1/2)
    4. (mass sattelite)(G)(mass earth) / ((radius sattelite)^2) = Tension + (G)(Mass sattelite)(Mass earth)/((radius sattelite)^2)
    5. Tension = (2)(G)(Mass sattelite)(Mass earth)/((radius sattelite)^2)
    6. Mass sattelite = [(Tension) ((Radius Sattelite)^2)] / (2G)(Mass earth)

    Is that correct?
    Last edited: May 7, 2009
  14. May 7, 2009 #13


    User Avatar
    Science Advisor
    Homework Helper

    Hi fogel1497! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    You're using the usual formula v2/r.

    But v has nothing to do with G … it's fixed (depending on distance) because ω = v/r is 2π/24 hours-1

    So it's easier to use ω2r :wink:

    (and i'm going to bed :zzz: …)
  15. May 7, 2009 #14
    by equation number 4, tension would = 0
  16. May 7, 2009 #15
    also, in the integral should not contain (lambda)(R), it should be (lambda)(R-R_e)
    of course, this makes it very hard to work with
    after the integration, i was not able to isolate r_s
  17. May 7, 2009 #16
    any 1 figure out part G or H ?
  18. May 8, 2009 #17
    need help with this ASAP
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook