What can be concluded between acceleration due to gravity and mass of an object under free fall

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Key concepts Physics Free fall Forces Gravity Mass

Inertia

Introduction
Have you ever wondered how fast a heavy object falls compared with a lighter one? Imagine if you dropped both of them at the same time. Which would hit the ground first? Would it be the heavier one because it weighs more? Or would they hit the ground at the same time? In the late 1500s in Italy the famous scientist Galileo was asking some of these same questions. And he did some experiments to answer them. In this activity you'll do some of your own tests to determine whether heavier objects fall faster than lighter ones.

Background
In fourth-century B.C. Greece the philosopher Aristotle theorized that the speed at which an object falls is probably relative to its mass. In other words, if two objects are the same size but one is heavier, the heavier one has greater density than the lighter object. Therefore, when both objects are dropped from the same height and at the same time, the heavier object should hit the ground before the lighter one. Is this true?

Some 1,800 years later, in late 16th-century Italy, the young scientist and mathematician Galileo Galilei questioned Aristotle's theories of falling objects. He even performed several experiments to test Aristotle's theories. As legend has it, in 1589 Galileo stood on a balcony near the top of the Tower of Pisa and dropped two balls that were the same size but had different densities. Although there is debate about whether this actually happened, the story emphasizes the importance of using experimentation to test scientific theories, even ones that had been accepted for nearly 2,000 years.

Materials • Two balls of the same size, but different mass. For example, you could use a metal and a rubber ball or a wooden and a plastic ball, as long as the two balls are about the same size. If two spherical balls like this are unavailable, you could try something like an apple and a similar-size round rock. • A ladder or step stool 

• A video camera and a helper (optional)

Preparation • You will be dropping the two balls from the same height, at the same time. Set up the ladder or step stool where you will do your test. If you are using a heavy ball, be sure to find a testing area where the ball will not hurt the floor or ground when it lands. • If you are using a video camera to record the experiment, set up the camera now and have your helper get ready to record.

• Be careful when using the step stool or ladder.

Procedure • Carefully climb the ladder or step stool with the two balls. • Drop both balls at the same time, from the same height. If you are using a video camera, be sure to have your helper record the balls falling and hitting the ground.

• Did one ball hit the ground before the other or did both balls hit the ground at the same time?

• Repeat the experiment at least two more times. Are your results consistent? Did one ball consistently hit the ground before the other or did both balls always hit the ground at the same time? • If you videotaped your experiments, you can watch the recordings to verify your results.

• Can you explain your results?

• Extra: Try this experiment again but this time use balls that have the same mass but are different sizes. Does one ball hit the ground before the other or do they hit it at the same time?
• Extra: Try testing two objects that have the same mass, but are different shapes. For example, you could try a large feather and a very small ball. Does one object hit the ground before the other or do they hit it at the same time?
• Extra: You could try this experiment again but record it using a camera that lets you play back the recording in slow motion. If you watch the balls falling in slow motion, what do you notice about how they are falling over time? Are both objects always falling at the same speed or is one falling faster than the other at certain points in time?

Observations and results

Did both balls hit the ground at the same time?

You should have found that both balls hit the ground at roughly the same time. According to legend, this is what Galileo showed in 1589 from his Tower of Pisa experiment but, again, it's debated whether this actually happened. If you neglect air resistance, objects falling near Earth’s surface fall with the same approximate acceleration 9.8 meters per second squared (9.8 m/s2, or g) due to Earth's gravity. So the acceleration is the same for the objects, and consequently their velocity is also increasing at a constant rate. Because the downward force on an object is equal to its mass multiplied by g, heavier objects have a greater downward force. Heavier objects, however, also have more inertia, which means they resist moving more than lighter objects do, and so heaver objects need more force to get them going at the same rate.

More to explore
Elephant and Feather—Free Fall , from The Physics Classroom
Engines of Our Ingenuity: No. 166: Galileo's Experiment , from John H. H. Lienhard, University of Houston
Video: Fall of 2 Balls of Different Weights , from Matthias Liepe, Cornell University
What Goes Up, Must Come Down: Conduct Galileo's Famous Falling Objects Experiment , from Science Buddies

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Take something in your hand and toss it down. Its speed is zero when you free it from your grip. Its pace rises as it descends. It flies faster the longer it goes. This sounds like acceleration. 

Acceleration, on the other hand, implies more than just rising speed. Pick up the same object and throw it into the air vertically. It will slow down on the way up until it comes to a halt and reverses course. Acceleration may also be described as a decrease in speed. However, acceleration is more than just a change of height. 

Pick up your battered object and give it one last launch. This time, toss it horizontally and see how the horizontal velocity increases as the vertical velocity decrease. This change of direction is often called acceleration since acceleration is the rate of change in velocity with time and velocity is a vector quantity. The acceleration in both of these cases was caused by gravity. Since gravity was dragging the object down, it was speeding up. Even the straight-up object is dropping — and it starts falling the moment it leaves your side. If it hadn’t been, it would have proceeded in a straight line away from you. This is the acceleration due to gravity. 

Now, Let’s discuss some basics before understanding the factors affecting acceleration due to gravity. 

The universal force of attraction among all the entities or matter in this universe is also known as gravity. It can be considered as the driving force which pulls together all the matter. Gravity is measured in terms of the acceleration or movement that it gives to freely falling objects. At Earth’s surface, the value of the acceleration of gravity is about 9.8 m/s2. Thus, for every second an object is in free fall, its speed increases by about 9.8 m/s2.

Some important that one must learn before going on further are: 

• There exist a direct correlation between mass and gravity, that is, mass is directly proportional to gravity. The heavier the object, the greater the intensity of pull. For instance, stars and sun, have greater gravity.
• Direct relation also exists between the mass of the object and gravity, that is increase in mass leads to increase in the pull due to gravity.
• Gravity is also inversely proportional to the distance between two objects.

Acceleration due to Gravity

The Earth pulls any particle lying on its surface towards its center with a force known as gravitational pull of gravity. When a force acts on a body, it causes acceleration, and in the case of gravitation, this acceleration caused by gravity is referred to as acceleration due to gravity

Acceleration due to gravity, usually referred by the symbol ‘g‘ is the acceleration attained by any object in the universe due to gravitational force.

The SI unit of acceleration due to gravity is m/s2. 

The value of g in SI system is 9.806 ms-2. And the value of g in CGS system is 980 cm s-2.

The Dimensional Formula for g is [ M0 L1 T-2 ].

It has a magnitude as well as direction. Thus it is a vector quantity.

Derivation of the formula for Acceleration due to Gravity

Near the surface of Earth, the acceleration due to gravity is approximately constant. But, at large distances from the Earth, or around other planets or moons, it is varying. The acceleration due to gravity depends on the terms as the following:

• Mass of the body,
• Distance from the center of mass,
• Constant G i.e. Universal gravitational constant.

Let’s consider an object of mass m, on which the acceleration due to gravity g is acting, now suppose F is the force acting on it and is given by:

F = mg                                                                                                                                                                                  ……(1)

where F is the force acting on the object, 

g is the acceleration due to gravity and 

m is the mass of the object.

According to the universal law of gravitation, the attractive gravitational force is given as:

F = (G × m × M) / (r+h)2

where F is the force between two objects, 

G is the universal gravitational constant (6.67 × 10-11 Nm2 / kg2), 

M is the mass of the earth, 

r is the radius of the earth, and 

h is the height at which the body is from the surface of the earth.

Since the height is negligibly small compared to the radius of the earth, so the last equation can be rewritten as:

F = (G × m × M) / r2                                                                                                                                                          ……(2)

Lets equate equation (1) and (2) and solve to evaluate the expression for g as:

mg = (G × m × M) / r2

g = GM / r2

Hence, the formula of acceleration due to gravity is evaluated as shown above.

g is majorly affected by the following four factors:

1. The shape of the Earth.
2. Rotational motion of the Earth.
3. Altitude above the Earth’s surface.
4. Depth below the Earth’s surface.

Variation of g due to shape of the Earth

The earth is not perfectly spherical, but is an oblate spheroid. The polar radius (radius near poles) of the earth is 21 km smaller than its equatorial radius (near the equator).  As per the formula derived, the acceleration due to gravity is inversely proportional to the square of the radius of the earth

Where, ge and gp are the accelerations due to gravity assumed at the equator and poles, Re and Rp are the radii of earth near the equator and poles, respectively. 

From the above equation, it can be easily derived that acceleration due to gravity is more at poles and less at the equator, that is ge < gp . Thus, acceleration due to gravity increases when one moves from the equator to the pole.

Rotational motion of the Earth.

Let us assume,  ω is the angular velocity of rotation of earth about its own axis, then the acceleration due to gravity at a place having latitude λ is shown as

g′ = g – Rω² cos² λ

Here, λ at poles is 90° and λ at the equator is 0°.

g’ = g

Thus, there is no effect of rotation of earth about its own axis at poles.
At the equator,

 λ = 0° and g’ = g – Rω²The value of g is minimum at the equator.

If earth stops its rotation about its own axis, then g will remain unchanged at poles but increases by Rω² at the equator.

Hence, we can conclude that;

Acceleration due to gravity is minimum at the equator. 

At poles, θ=90°

Acceleration due to gravity is maximum at poles.

Altitude above the Earth’s surface

Assume a mass ‘m’ that is under the effect of earth’s gravity at a height ‘h’ from earth’s surface. The force due to gravitational pull acting on the object is;

Acceleration due to gravity at the height ‘h’ from the Earth’s surface

 ……..(i)

Here, 

M is the mass of earth

R is the radius of the earth. 

The acceleration due to gravity at a certain height is given by ‘h’ 

Therefore,

  ……(ii)    

Then, the acceleration due to gravity on the surface of the earth is shown by;

  ………..(iii)

After dividing equation (iii) and (ii) we will get,

 …………(iv)

It is clear that the value of g decreases with an increase in height of an object. Hence, the value of g becomes zero at infinite distance from the earth.

Depth below the Earth’s surface.

The acceleration due to gravity on the surface of the earth is given by

Assume 

 be the density of the material of the Earth

As we know

Mass = volume × density

  …….(i)

Thus, if we put the value of (i) in g, we will get

 

Also, let us assume the body to be taken to the depth ‘d’ below the surface of the earth. Then acceleration due to gravity gd at the depth ‘d’ below the Earth’s surface is shown as;

   ……..(iii)

Now, divide the equation (iii) by (ii), we will get

    

Here, equation (iv) is an expression for the acceleration due to gravity at the depth ‘d’ below the surface of the earth.

From equation (iv) we get to know that acceleration due to gravity decreases as we go down into the Earth. 

At the Centre of earth d = R, 

Thus, acceleration due to gravity at the Centre of the earth is 0.

Sample Problems

Problem 1: At what height above the surface of the earth value of acceleration due to gravity is reduced to one by ninth of its value on the surface of the earth?  

Solution:

From the formula

 gh = 

Here gh = acceleration due to gravity at height ‘h’

g = acceleration due to gravity

R = radius of the earth

Thus, 

From the problem

Hence,

At twice the height of the radius of the Earth value of acceleration due to gravity is reduced to one by ninth of its value on the surface of the earth.

Problem 2. If the radius of the Earth is decreased by 2% while keeping its mass the same, how will the acceleration due to gravity change?

Solution: 

As we know that

 

Where,

g = acceleration due to gravity

R = radius of the Earth

M = mass of the Earth

If R is decreased by 2% it will become 

 

Therefore,

Therefore,

g’ increases by 0.04 

 therefore increases by 4%.

Problem 3. At what height from the surface of the earth will the value of ‘g’ be reduced by 19% of its value at the surface of the earth.

Solution:

g’ = 81% of g

g’ = 

 

Therefore,

At the height of 

 from the Earth surface the value of ‘g’ will be reduced to 19% of its value at the surface of earth.

Problem 4. How many times is the acceleration due to gravity on the surface of Jupiter as compared to Earth’s surface?

(Use the following data: mEarth = 5.98 × 1024 kg, rEarth = 6.4 × 106 m, mJupiter = 1.9 × 1027 kg, rJupiter = 6.99 × 107 m)

Solution:

To solve this example we will need to find the ratio of acceleration due to gravity of Jupiter to acceleration due to gravity of Earth

By the relation;

Here,

m = mass of planet

r = radius of planet

Thus, the acceleration due to gravity on Jupiter surface is 2.66 times more than that of Earth’s surface.

Problem 5. Explain acceleration due to gravity increase or decrease with the increase in the altitude?

Solution:

As we know that acceleration due to gravity at the depth of h is shown as;

Here,

Re = Radius of the Earth

g = Acceleration due to gravity on the surface of the Earth

Thus, we can clearly see that from the equation that acceleration due to gravity increases with decrease in height and decreases with increase in height.

Problem 6. Explain that acceleration due to gravity is independent of the mass of earth as well as the mass of the body?

Solution:

As we know that acceleration due to gravity of body of mass m is shown as;

Here,

G = Is the universal gravitational constant

M = Mass of the Earth

R = Radius of the Earth

Thus, we can clearly see here that acceleration due to gravity is independent of the mass.

Problem 7. Explain that acceleration due to gravity is increased or decreases with the increase in depth?

Solution:

As we know that acceleration due to gravity is shown as;

Thus, here we can clearly see that the acceleration due to gravity increases with decrease in depth and decreases with increase in depth.