Lesson 12: Gravity

Aristotle

From the time of Aristotle (384-322 BC) until the late 1500’s, gravity was believed to act differently on different objects.

• Drop a metal bar and a feather at the same time… which one hits the ground first?
• Obviously, common sense will tell you that the bar will hit first, while the feather slowly flutters to the ground.
• In Aristotle’s view, this was because the bar was being pulled harder (and faster) by gravity because of its physical properties.
• Because everyone sees this when they drop different objects, it wasn’t questioned for almost 2000 years.

Galileo

Galileo Galilei was the first major scientist to refute (prove wrong) Aristotle’s theories.

• In his famous (at least to Physicists!) experiment, Galileo went to the top of the leaning tower of Pisa and dropped a wooden ball and a lead ball, both the same size, but different masses.
• They both hit the ground at the same time, even though Aristotle would say that the heavier metal ball should hit first.
• Galileo had shown that the different rates at which some objects fall is due to air resistance, a type of friction.
• Get rid of friction (air resistance) and all objects will fall at the same rate.
• Galileo said that the acceleration of any object (in the absence of air resistance) is the same.
• To this day we follow the model that Galileo created.

ag = g = 9.81m/s2

ag = g = acceleration due to gravity

Since gravity is just an acceleration like any other, it can be used in any of the formulas that we have used so far.

• Just be careful about using the correct sign (positive or negative) depending on the problem.

Examples of Calculations with Gravity

Example 1: A ball is thrown up into the air at an initial velocity of 56.3m/s. Determine its velocity after 4.52s have passed.

In the question the velocity upwards is positive, and I’ll keep it that way. That just means that I have to make sure that I use gravity as a negative number, since gravity always acts down.

vf = vi + at

= 56.3m/s + (-9.81m/s2)(4.52s)

vf = 12.0 m/s

This value is still positive, but smaller. The ball is slowing down as it rises into the air.

Example 2: I throw a ball down off the top of a cliff so that it leaves my hand at 12m/s. Determine how fast is it going 3.47 seconds later.

In this question I gave a downward velocity as positive. I might as well stick with this, but that means I have defined down as positive. That means gravity will be positive as well.

vf = vi + at

= 12m/s + (9.81m/s2)(3.47s)

vf = 46 m/s

Here the number is getting bigger. It’s positive, but in this question I’ve defined down as positive, so it’s speeding up in the positive direction.

Example 3: I throw up a ball at 56.3 m/s again. Determine how fast is it going after 8.0s.

We’re defining up as positive again.

vf = vi + at

= 56.3m/s + (-9.81m/s2)(8.0s)

vf = -22 m/s

Why did I get a negative answer?

• The ball reached its maximum height, where it stopped, and then started to fall down.
• Falling down means a negative velocity.

The Rules

There’s a few rules that you have to keep track of. Let’s look at the way an object thrown up into the air moves.

As the ball is going up…

• It starts at the bottom at the maximum speed.
• As it rises, it slows down.
• It finally reaches it’s maximum height, where for a moment its velocity is zero.
• This is exactly half ways through the flight time.

As the ball is coming down…

• The ball begins to speed up, but downwards.
• When it reaches the same height that it started from, it will be going at the same speed as it was originally moving at.
• It takes just as long to go up as it takes to come down.

Example 4: I throw my ball up into the (again) at a velocity of 56.3 m/s.

a) Determine how much time does it take to reach its maximum height.

• It reaches its maximum height when its velocity is zero. We’ll use that as the final velocity.
• Also, if we define up as positive, we need to remember to define down (like gravity) as negative.

a = (vf - vi) / t

t = (vf - vi) / a

= (0 - 56.3m/s) / -9.81m/s2

t = 5.74s

b) Determine how high it goes.

• It’s best to try to avoid using the number you calculated in part (a), since if you made a mistake, this answer will be wrong also.
• If you can’t avoid it, then go ahead and use it.

d = (vf2 = vi2) / 2a

= (0 - 56.32) / 2(-9.81m/s2)

d = 1.62e2 m

c) Determine how fast is it going when it reaches my hand again.

• Ignoring air resistance, it will be going as fast coming down as it was going up.

Gee's

You might have heard people in movies say how many "gee’s" they were feeling.

• All this means is that they are comparing the acceleration they are feeling to regular gravity.
• So, right now, you are experiencing 1g… regular gravity.
• During lift-off the astronauts in the space shuttle experience about 4g’s.
• That works out to about 39m/s2.
• Gravity on the moon is about 1.7m/s2 = 0.17g