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A newly-discovered plant, "Cosmo", has a mass that is 4 times the mass of the Earth. The radius of the Earth is Re. The gravitational field strength at the surface of Cosmo is equal to that at the surface of the Earth if the radius of Cosmo is equal to...

I guess I need some kind of relationship between mass, radius, and gravity. Originally I thought that since Fg=ma and a=v^2/r for circular motion (IS this circular motion???) then Fg=mv^2/r. But then if m is multiplied by 4, so is r. But that's wrong...

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Hannah

  1. You need Fg = G m M2/r^2
    What the problem says is that the force on a sample mass m is the same at the surface.

    Fg earth = G m Mearth/Rearth^2
    Fg cosmo = G m 4 Mearth/ Rcosmo^2
    they are the same for the same little m
    so

    4 /Rcosmo^2 = 1/Rearth^2

    Rcosmo = 2 Rearth

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    Damon

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