A small planet having a radius of 1000 km exerts a gravitational force of 100 N on an object that is 500 km above its surface. If this objec

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A small planet having a radius of 1000 km exerts a gravitational force of 100 N on an object that is 500 km above its surface. If this object is moved 500 km farther from the planet, the gravitational force on it will be

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Delwyn 6 months 2021-07-19T08:46:52+00:00 1 Answers 68 views 0

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    2021-07-19T08:48:32+00:00

    Answer:

    56.25 N

    Explanation:

    Given,

    Radius of the planet, r = 1000 km

    Gravitational force, F = 100 N

    Initial distance from the planet, d = 500 km

    Final distance from the planet, D = 1000 km

    Lets assume, mass of planet = M

    mass of the object = m.

    As per the gravitational law, the force due to gravity is given as

    F = \frac{G \times M \times m}{R^{2}}

    where, G = gravitational constant

    R =  distance between the two objects.

    Case 1: when the object is at a distance of 500 km from the planet,

    F = \frac{G \times M \times m}{R^{2}} = 100

    Here, R = 1000 + 500 = 1500 km, so

    \frac{G \times M \times m}{1500^{2}} = 100

    G \times M \times m = 100 \times {1500^{2}

    Case 2: when the object is at a distance of 1000 km from the planet, so

    now, R = 1000 + 1000 = 2000 km

    Now the gravitational force, F’ will be

    F' = \frac{G \times M \times m}{R^{2}}

    F' = \frac{100 \times 1500^{2} }{2000^{2}}

    F' = 56.25 N

    Thus, the gravitational force will now be reduced to 56.25 N.

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