We can calculate the weight of an object using gravitational acceleration and Newton’s second law of motion. Near the surface of the earth, the pull of gravity is strong enough to accelerate objects at 9.8m/s^{2}, so a 10 kg rock weighs 98 newtons when located near the earth’s surface:

force = mass × acceleration

force = 10 kg × 9.8m/s^{2}

force = 98 kg m/s^{2}

force = 98 N

force = 10 kg × 9.8m/s

force = 98 kg m/s

force = 98 N

As Galileo had discovered, heavy objects and light objects fall to the ground at the same rate, provided there is little air resistance. Near the surface of the earth, free-falling objects accelerate at a nearly constant rate of approximately 9.8m/s^{2}. In other words, the velocity of a free-falling object increases by 9.8 m/s every second. This gravitational acceleration is often denoted with the lowercase letter “*g*”:
*g* = 9.8m/s^{2}

However, this value only holds for free-falling objects near the surface of the earth (gravitational acceleration near the surface of the moon is much slower, for instance), and the acceleration only remains constant so long as air resistance is negligible. As an object falls through the air, the force of air resistance increases as the velocity increases, reducing the rate of acceleration until eventually the downward force of gravity is exactly matched by the upward force of air resistance. At that point, the object continues to fall at a constant velocity called its terminal velocity. That’s why a feather falls more slowly than a rock: it quickly reaches terminal velocity, because it has a large surface area relative to its weight. In a vacuum (with no air resistance), a rock and a feather would fall at the same rate.

What would happen if raindrops fell with no air resistance? Cumulonimbus clouds—storm clouds—typically form between 2,000 and 16,000 meters, sometimes much higher. So, how fast would a raindrop be moving if it fell from (say) 10,000 meters without air resistance? First, let’s figure out how long it takes for the raindrop to reach the ground:

distance = ½ × acceleration × time^{2}

10,000 m = ½ × 9.8m/s^{2} × t^{2}

2041 s^{2} = t^{2}

45 seconds = t

So, the raindrop falls for 45 seconds with no air resistance. Since it is accelerating at 9.8m/s10,000 m = ½ × 9.8m/s

2041 s

45 seconds = t

Thanks to air resistance, fortunately, the terminal velocity of raindrops is in fact much lower—about 9 m/s (20 miles per hour) for the largest and fastest raindrops. Large hailstones fall slightly faster, with a terminal velocity of about 20 m/s (44 miles per hour).

But *why* do heavy objects and light ones fall at the same rate? Why do they fall at all? Galileo effectively refuted Aristotle’s theory, but proposed no alternative in its stead. As we’ll see in what follows, Newton’s laws provided at least a partial answer to these questions.