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Suppose the Earth was only half the size it is now (half the mass and half the radius), what would "g" be?

  • Physics -

    Newtons law:

    g= G Me/r^2

    Now, if one halves the radius, mass is not half.

    Mass=density*volume=density 4/3 PI r^3 so if radius goes down by 1/2, then mass goes to 1/8 Me

    So your question is really confusing.

  • Physics -

    Newton's law of universal gravitation:

    where force equals the universal constant of gravity multiplied by mass of the earth and mass of an object divided by the raduis squared

    F = G((m1*m2)/(r^2))

    Force equals the mass of an object multiplied by the acceleration or 'g'

    mg = G((m1*m2)/(r^2))

    Mass of the object would cancel out on both sides since infact we do not have a second mass

    g = G((m1)/(r^2))

    The universal constant of gravity as Newton discovered is:

    G = 6.67 * 10^-11 m^3/kg*s^2

    The mass of the Earth is:

    mass of the earth = 5.98 * 10^24 kg

    The radius of the Earth is:

    radius of the earth = 6.38 * 10^6 m

    The mass then needs to be divided in half:

    5.98 * 10^24 kg/2 = 2.94 * 10^24 kg

    The radius then needs to be divided in half:

    6.38 * 10^6 m/2 = 3.19 * 10^6 m

    The equation for the acceleration or 'g' is:

    g = 6.67 * 10^-11 m^3/kg*s^2((2.94 * 10^24 kg)/(3.19 * 10^6 m^2))

    Leaving 'g' as:

    g = 19.27 m/s^2

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