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Homework Help: Estimating convergence of GRACE twin-satellites due to gravitational mass

  1. Nov 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Hello noble physicists,

    I am struggling to solve a problem with any sort of confidence whatsoever, so to you I turn in the hopes of guidance.

    The problem refers to GRACE twin satellite convergence due to gravitational anomalies.

    I’d like to estimate the convergence of the two satellites due to the gravitational mass of Mt. Everest, and subsequently estimate the size of the smallest mountain detectable by GRACE.

    Twin trapezoid satellites: 3.1 m × 0.8 m × (1.9–0.7) m each
    460kg each

    500km orbital height
    220km apart
    Can measure their separation to within 10 microns
    16 orbits/day

    It is assumed Everest is a pyramid
    Length: 8.8km
    Width: 8.8km
    Height: 8.8km
    Rock Density: 3000kgm^3

    2. Relevant equations


    I think:

    F = Gm/R^2
    a = F/m
    s = at^2
    s = Vi t + 1/2at^2

    3. The attempt at a solution

    Time spent above Everest = 1.2 seconds

    Circumference of Earth = 40075km

    40 075 000m x 16 orbits = 6.412x10^8 m
    Speed = 6.412x10^8 m / 86 400 secs = 7421m/s

    Width of Everest = 8800m
    Time spend over Everest = 8800/7421 = 1.2 secs

    Mass of Everest = 6.815x10^5 kg

    V = 1/3 (base area) height
    V = 227.157m^3

    Mass = Vol x Density = 6.815x10^5 kg

    Radius of satellite from centre of earth = 6.88x10^6m

    Radius of Earth + Orbit height
    6.38x10^6 + 500000m = 6.88x10^6m

    Now I’m unsure of which steps to take next to calculate the convergence of the two satellites, I hope someone would be able to offer some guidance.

    Many thanks in advance.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 27, 2012 #2


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    Staff: Mentor

    The time of influence is longer - of the order of the flight time for the orbital height (550km).

    You could assume that one satellite passes 110km "left" of the mountain (relative to the orbits) and another one passes 110km "right" of it. Calculate the sidewards acceleration for both and integrate over a fly-by (or use some approximation for the integral). Everest can be approximated as point-mass. This gives the change in relative velocity during a central fly-by.
  4. Nov 27, 2012 #3
    Thanks for your help,

    though could you explain how to get the correct time of influence?
  5. Nov 27, 2012 #4


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    Staff: Mentor

    Well, the range of gravity is infinite, but the minimal distance is about 550km - if the satellites are >>550km away, the remaining influence does not matter any more. There is no precise number for the time of influence. If you are just interested in an order of magnitude approximation, you can take the maximal force and use 550km flight distance as range of influence with that maximal force.
  6. Nov 27, 2012 #5
  7. Nov 29, 2012 #6
    I think the problem has been over complicated.

    Here is a massively simplicated solution which seems to present a reasonable answer.

    Though if there was one thing I were not too sure sure about it would be whether to include in the time of influence calculations the actual distance between the satelites - 220km - or half of that as they are equal distances below...

    G 6.67E-11 ms^-2
    Mass of EARTH 5.98E+24 kg
    Radius Earth 6400000 m
    Orbital Height 500000 m
    Orbital Radius 6900000 m
    Separation 220000 m

    Vol = 1/3(base area) height
    Everest Base Width 8800 m
    Everest Height 8800 m
    Everest Density 3.00E+03 kgm^-3
    Volume Everest 2.27E+11 m^3
    M = V D
    Mass of Everest 6.81472E+14 kg

    V^2 = GM/r (r = orbital radius = 6900km)
    Satellite Velocity 7603.0695555589 ms^-1 V^2 = GM/r

    F = (G.Meverest)/ r^2 (r = 500000)
    Acceleration 1.818167296E-07 ms^-2

    T = D/S = 220000/Satellite Velocity
    Time 28.93568162 secs

    s =( Vi t) + 0.5(a t^2)
    s 220000.000076115 m
    s - 220km 0.0000761152 m
    x2 Satellites 0.00015223 m

    Convergence in Microns 152.23 microns


    38.06 microns using 110km equal separation.
  8. Nov 30, 2012 #7


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    Staff: Mentor

    That is a temporary distance modification during a single fly-by if those satellites follow each other - after they passed Mount Everest (and ignoring other mountains), their distance is the same as before.
  9. Nov 13, 2013 #8
    Hi I'm working on a similar question and am trying to understand the benefit of interpreting the convergence of acting across the two satallites at 220km spacing or two 110km equally spaced. Depending on interpretation the convergance is drastically different.

    Based on the fact that GRACE can only detect convergance of 10 microns or greater - the interpretation of the convergance gives differing impressions of the detectable mass
    Last edited: Nov 13, 2013
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