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Cake day: July 11th, 2023
  • stembolts@programming.devtoPeople Twitter@sh.itjust.works•Whoa
    5·
    20 days ago

    Disclaimer Edit: I realize there are some semantics between force, weight, mass, etc. The purpose of this post is not the rigor of the mathematics but the enjoy-ability of being able to abstract interesting measurements. Sorry if I made any mistakes physicists! Always open to corrections :)

    Of course we can weigh the earth. In your example the earth weighs nothing because you didn’t define a reference frame. Since earth is the usual reference frame we use to weigh objects, choosing the earth itself as an object to weigh does stretch our minds a bit, but we can do it.

    In order to weigh something, first we must understand what weight is expressing. Weight is the measure of the gravitational pull of an object A with mass M_a on a second object M_b summed with the gravitational pull of object B on A.

    Weight = G_a-on-b*M_b + G_b-on-a*M_a

    In our scenario, if A is the earth and B is a non-planetary-sized object, then the pull G_a-on-b is so much larger than G_b-on-a that we can set G_b-on-a to zero for convenience.

    Weight = G_a-on-b*M_b + 0

    However if we are weighing an earth-sized object we will not be able to do that ;-)

    Now, lets talk about the reference frame for this question.

    On earth, the weight of some object B is recorded in a reference frame 6,367km from the center of a 5.97E24 kg mass object (object A).

    With M(ass) and G(ravity) now defined, we can work toward W. G is easy since we are working with earth-like objects.

    Back to the question “how heavy is earth”. There is only one earth, so the answer is, “In what reference frame?” Without defining a reference frame, no conclusion of any significance can be drawn.

    If we want to weigh the earth in the reference frame of the surface of another earth, we would need two earths. Place a 5.97E24 mass object 6,367km from the center (radius of earth R=6.367E6m) of another 5.97E24 mass object, we will arbitrarily assign a diameter of 1,000km to each, this assignment has no bearing on the result since density is irrelevant in this context.

    Where G = 9.81m/s², and the weights/masses of object A wrt B and B wrt A are W_ab/M_a W_ba/M_b respectively. In this case, M_a and M_b have the same value so we’ll reduce to M.

    W_ab = G*M + G*M = 2GM

    …where M is the mass of earth and G is gravitational acceleration. Voila, our answer.

    I’ll leave you to do the math as that task is trivial.

    I digress.

    I must insist that the real question is what is the weight of a third object C (the husband of A) at a distance 8R (the distance from the bedroom door to the foot of AC’s bed) when it catches A and B (at relative distances R_ab=2Ra, assuming R_a=R_b) in AC’s bedroom when A said it was picking up their son D at a distance 2,700R (soccer practice) and B came home early from work to find them entangled in the passionate embrace of ecstasy when A told C they were too young to get married and have kids but they had a surprise pregnancy and did the best they could. Calculate the weight of C wrt AB and C wrt D assuming the positions of AB, C, and D form a straight line.