What is the magnitude of the gravitational force of attraction between two 0.425-kilogram soccer balls when the distance between their centers is 0.500 meter?

(1) 2.41 x 10-1l N
(2) 4.82 x 10-11 N
(3) 5.67 x 10-11 N
(4) 1.13 x 10-10 N

Respuesta :

In the Newtonian theory of gravitation, the effects of gravity are always attractive, and the resulting force is calculated with respect to the center of gravity of both objects. The law of universal gravitation formulated by Isaac Newton postulates that the force exerted by a point particle with mass M on another with mass m is directly proportional to the product of the masses (and the Universal Gravitation Constant), and inversely proportional to the square of the distance (r) that separates them:

[tex]F = \frac{GMm}{r^2}[/tex]

[tex]F = \frac{(6.67*10^{-11})(0.425)(0.425)}{0.5^2}[/tex]

[tex]F = 4.819*10^{-11}N[/tex]

[tex]F \approx 4.82*10^{-11}N[/tex]

Therefore the correct answer is 2.

The magnitude of the gravitational force of attraction will be "[tex]4.82\times 10^{11} \ N[/tex]".

According to the question,

  • Mass, [tex]m = 0.425 \ kg[/tex]
  • Distance, [tex]d = 0.500 \ m[/tex]

We know that,

→ [tex]F = \frac{GMm}{r^2}[/tex]

By putting the values, we get

       [tex]= \frac{(6.67\times 10^{11})(0.425)(0.425)}{(0.5)^2}[/tex]

       [tex]= 4.82\times 10^{-11} \ N[/tex]

Thus the response above i.e., "option 2" is correct.  

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