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A hypothetical spherical planet consists entirely of iron (p=7860 kg/m^3). Calclate the period of a satellite that orbits just above its surface.
I started out with the density formula P=M/V, but i kept subbing in other formulas, but i get stuck..can someone help me? thanks.
I posted this before, but i made a mistake in the question and I am now fearing that whoever helped me will not go 3 pages into the history just to find this question and help me again. Thanks =P

  • Physics -

    Also, teacher gave me a hint that "r(radius)" was suppose to cancel during the formula manipulation process..

  • Physics -

    Time required for 1 revolution,
    T= (2πr^(3/2))/√(G∙M) (p. 146 of 7th ed Cutnell & Johnson)

    Volume of a sphere,
    V=4πr^3 (book inside cover)

    Formula for mass density,
    M=ρ∙V (p.321)

    T: the time period
    G: universal gravitational constant, 6.673 E-11 (N∙m^2)/kg^2
    note: N=(kg∙m)/s
    r: distance from center of planet to satellite, aka planet’s radius
    ρ: mass density of planet.
    Iron density = 7860 kg/m^3
    V: Volume of planet

    Substitute formula for volume of a sphere into equation for mass:
    M = ρ∙ 4πr^3
    Then substitute this into the equation for time period
    T= (2πr^(3/2))/√(G∙ρ∙ 4πr^3)
    The r^(3/2) in the numerator cancels the √(r^3) in the denominator,
    So the equation simplifies to
    T= 2π/√(G∙ρ∙4π)
    Plug in the known value of G and given value of ρ.
    For a planet made of iron, satellite period
    T ≈ 2447.39 seconds


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