Jace asked in Science & MathematicsPhysics · 6 months ago

an object has a weight of 15N on the surface of the earth, what would it weight if object is tripped its distance from center of the earth?

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  • 6 months ago

    In this situation, the weight is equal to the Universal gravitational force.

    Fg = G * M * m ÷ d^2

    According to this equation, the force is inversely proportional to the square of the distance from the center of the earth. Since this distance is tripled, the new force is be one ninth of the original force.

    F = 15 ÷ 9 = = 1⅔ N

    In the following work, I will use the distances to check this answer. The initial distance is the radius of the earth. This is 6.38 * 10^6 meters.

    New distance = 3 * 6.38 * 10^6 = 1.914 * 10^7

    At a distance of 6.38 * 10^6 meters, the force is 15 meter

    15 = k ÷ (6.38 * 10^6)^2

    F = k ÷ (1.914 * 10^7)^2

    F = 15 * (6.38 * 10^6)^2 ÷ (1.914 * 10^7)^2 = 15 ÷ 9 = 1⅔ N

    This is the correct answer. I hope this is helpful for you.

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  • 6 months ago

    Weight is inversely proportional to r^2, the distance to the center of the Earth squared.

    So if 15 N = G*M*m / r^2,

    then the weight of that object when the distance that previously was r has become 3*r, the new weight is

    W_new = G*M*m / (3*r)^2

    W_new = G*M*m / (9*r^2)

    W_new = (1/9)*G*M*m / r^2 = (1/9)*15 N = 1.66667 N =~ 1.7 N

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  • 6 months ago

    I am sure that you mean "tripled"

    As F = GMm/ r^2 and the only thing that you have changed on the right is R

    F2/F1 = (r1/r2)^2 = (1/3)^2 = 1/9

    ie the force at 3 times the distance is 1/9 that on the surface of the earth.

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  • 15/9 =>

    5/3 =>

    1.666666...

    1.7 N

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