Newton's third law states that the force object A exerts on object B is equal to the force object B exerts on object A (in the opposite direction). So if, for example, a 5 kg ball is dropped from a plane, the force of the ball is 50 (F = 5*10), so if the ball is exerting a force of 50N on the air molecules around it (even if it is one air molecule), shouldn't those molecules exert 50N on the ball too? And so shouldn't that ball always be in terminal velocity?
2 Answers
The force on the ball from the Earth is 50N. Newton's 3rd Law then states that the force on the Earth from the ball is 50N. The forces that the air molecules place on the ball, and that the ball places on the air molecules are dependent on the relative velocity of the two.
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1$\begingroup$ So basically the force applied by the air molecules will be the 'action' and the 'reaction' will be the same force being applied by the ball on the air molecules $\endgroup$– De SithCommented Mar 1, 2020 at 12:42
The 50 Newtons of weight is due to gravity. Its third law pair is that the earth is attracted toward the ball with an equal force (you don't see the acceleration because of the mass of the earth being very large). There is a different force that the air exerts opposing the object's motion, and the object exerts an equal and opposite force on the air. Just like when it lands, the gravitational force is not third law pairs with the normal force.The normal force is third law pairs with a compression force pushing the ground down. That understanding will become necessary when dealing with an object being accelerated upward by an elevator for example.
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$\begingroup$ Oh, right. So the force the ball exerts on the air molecules, is the 'reaction' force to the 'action' force the air molecules exert on the ball? $\endgroup$– De SithCommented Mar 1, 2020 at 13:52