img_3675-1If you drop two very different objects from the same height will they land at the same time? And if you repeat this in vacuum will the result be different? Click through to find out the answer with Newton’s laws of motion and universal gravitation!



I came across the video below a couple of days ago, in which British physicist Brian Cox visits the world’s biggest vacuum chamber to show you that two objects dropped in vacuum fall at the same rate, regardless of their mass. So today I am going to use Newton’s laws of motion and universal gravitation to explain why that happens :)

First of all, let’s talk about gravity. Newton published his book “Newton’s Principia” in 1686, in which he described the inverse square law relationship between the gravitational force between two bodies and their distance. Newton’s law of universal gravitation states that two bodies in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The mathematical expression for this relationship is as follows:

F = G \cdot \dfrac{M_1 M_2}{R^2}

Where F is the force between the masses, G is the gravitational constant, M1 is the mass of the first body, M2 is the mass of the second body, and R is the distance between the centres of mass of the two bodies. If we are calculating the force due to the Earth’s mass on an object that sits on it’s surface (a person for example) then the mass of the first body is the mass of the Earth, the mass of the second body is the mass of the object, and the distance between them is the radius of the Earth:

M_1 = M_{Earth} = 5.972 \cdot 10^{24} kg

M_2 = M_{person} = 60 kg

R = R_{Earth} = 6371 km = 6.371 \cdot 10^6 m

G = 6.67384 \cdot 10^{-11} Nm^2/kg^2


Introducing the numbers into Newton’s law of Universal Gravitation we obtain:

F = 6.67384 \cdot 10^{-11} Nm^2/kg^2 \cdot \dfrac{5.972 \cdot 10^{24}kg}{\left( 6.371 \cdot 10^6 m \right) ^2} \cdot M_{person} \rightarrow

F = 9.8 N/kg \cdot M_{person} \rightarrow

F = 9.8 N/kg \cdot 60 kg = 588 N

From now on, we will call g = 9.8 N/kg = 9.8 m/s2 the acceleration due to the Earth’s gravity. (NOTE: the units of N/kg are equivalent to those of m/s2 due to Newton’s Second Law.)

Now that we know what the gravitational force is on an object on the surface of the Earth, let’s calculate the time the object takes to reach the ground. In order to do this we are going to use Newton’s Laws of Motion, which were also published in 1686 in “Newton’s Principia” and state the following:

1) First Law: When viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by an external force.

2) Second law: The vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration vector a of the object

3) Third Law: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

For our purposes right now we are only concerned with the second law:

\Sigma F = m \cdot a

If we consider a ball of mass M = 5kg on the surface of the Earth, being dropped from a height H = 10 m while being in vacuum the only force acting on it is the gravitational force due to the mass of the Earth. If we combine this with Newton’s second law it follows that:

F_{gravity} = m \cdot a \rightarrow m \cdot g = m \cdot a \rightarrow

g = a

So, the acceleration of the ball does not depend on it’s mass!!


The equation of motion of a ball being dropped from rest from a height H = 10 m is:

h_{end} = h_{start} + v_{start} \cdot t + \dfrac{1}{2} \cdot a \cdot t^2

In our case h-end = 0 m (because it ends on the ground), h-start = H (the height from which we drop it), v-start = 0 m/s (we drop it from rest) and a = – g (due to the Earth’s gravity):

h_{end} = 0 m

h_{start} = H = 10 m

v_{start} = 0 m/s

a = - g = - 9.8 m/s^2

(NOTE: the negative sign on the acceleration due to the Earth’s gravity is because of the fact that we have chosen the height to be positive in the upwards direction and gravity goes downwards.)

This gives:

0 = H - \dfrac{1}{2} \cdot g \cdot t^2

So the time it takes for the ball to reach the ground is:

t = \sqrt{2 \cdot \dfrac{H}{g}}

The time required for the ball to reach the ground does not depend on it’s mass!! It depends only on the height the ball is dropped from and the acceleration due to gravity. So if we were to drop any other object from the same height it would take the same amount of time to reach the ground, regardless of the object’s mass, size, shape…

For our chosen height of H = 10 m the fall time is 1.42 seconds.

t = \sqrt{2 \cdot \dfrac{10 m}{9.8 m/s^2}} = 1.42 s

However, we all know that if we drop a ball and a feather in air from the same height, they feather will take much longer to reach the ground than the ball. Why is this? Well, we have to consider the effect the air has on the two objects.


The air offers some resistance to the falling objects, more scientifically known as “drag”. The drag depends on the properties of the fluid and on the size, shape, and speed of the object. It can be expressed by means of the drag equation:

F_D = \dfrac{1}{2} \cdot \rho \cdot v^2 \cdot C_D \cdot A

Where F-D is the drag force, rho is the mass density of the fluid (in our case air), v is the velocity of the object relative to the fluid, A is the cross section area and C-D is the drag coefficient, a dimensionless coefficient related to the object’s geometry and taking into account both skin friction and form drag. The expressions for the drag coefficient and the drag equation are very complicated (to be honest I don’t understand them very well myself hehe, I have never really liked fluid dynamics much), so I won’t go into any detail. I will just point out that in the case of a bowling ball and a feather, the ball experiences very little drag (air resistance) but the feather experiences a very high drag. Hence the ball will fall much faster than the feather when both dropped from the same height in air.

To finish off, here is a quick photo tip: when taking photos of people jumping, make sure you compose from as low as you can, this way it will seem your subject is jumping even higher. They will seem to be defying gravity!! Here are a couple of examples:

Nic Douglass from Adventures of a Sailor Girl jumps over Sydney at sunset!
Nic Douglass from Adventures of a Sailor Girl jumps over Sydney at sunset! Photo by Ana Andres-Arroyo Photography:
Bouncy Ball at Bondi Beach
Bouncy Ball at Bondi Beach. Photo by Ana Andres-Arroyo Photography:

Oh, and one very last thing, I promise hehe. I saw this the other day and I thought it was really funny :) It’s created by Dan Piraro from Bizarro Comics and it fits really well with the topic of today’s post hehehe.

Comic by
Comic by Dan Piraro from Bizarro Comics:

I hope you liked this post and enjoyed the video :)

Thanks for reading!!

xx Ana :)


4 thoughts on “Gravity

  1. That was very well explained. It is actually very simple once it is explained in this fashion. The difference is air resistance which we do not take into account in elementary physics. The techniques regarding the jumping shots are quite practical. You have posted some unusually interesting stuff.


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