Is it possible to determine the percentage of the gravitational force at a specified distance using only the geometry of the planet?

Example: The ISS at ~420km altitude “weighs” about 90% of what it would on the Earth’s surface.

Is there an equation using only geometrical values that would give you this info?

  • mozz@mander.xyzEnglish
    11·
    10 months ago

    Force is proportional to m1 * m2 / r^2, so your weight as a percentage of your surface weight is equal to (radius of earth / (radius of earth + height above surface))^2.

  • nooneescapesthelaw@mander.xyz
    3·
    10 months ago

    Mass weight of object on earth = g(mass of object * mass of earth)/(earth radius)^2

    Weight of object in orbit= g(mass of object * mass of earth)/(earth radius + distance from earth)^2

    Dividing the two equations gives us:

    (1/(earth radius + distance from earth)^2)
 ---------------------------------------
(1/earth radius)^2

    Which yields:

    (Earth radius)^2
-----------------------------------------
(earth radius + distance from earth)^2

    Multiplying the fraction above by 100 will give a percentage.

    To check if the equation makes sense imagine if the object was very far from earth. Then the weight (earths gravity) would be small and thus the percentage would be small. This checks with our equation

    We can try it for the ISS and see we get:

    ((6378)^2) ÷(408+6378)^2 = .88

    Or 88%