Re: If you kick a brick on the moon or in outer space, will it hurt your foot?

The short answer is "Yes, kicking a brick will always hurt your foot!!"

It is because of what physicists call "inertia". There are, by the way, many previous answers in our archives dealing with inertia, which you can find by using our search engine. Having said that, I will briefly describe the topic.

All real objects are massive. That is, they are made of stuff that has mass. The constituent particles that all real objects are made of are primarily protons, neutrons, and electrons, and physicists have measured how much mass each has.

Mass affects us in two major ways: gravity and inertia.

All massive objects interact will all other massive objects by way of the gravitational force. That is, there is a force that exists between all massive objects, and it can be calculated by the equation
F = Gm₁m₂/r²
where F is the gravitational force, G is the Gravitational constant, m₁ is the mass of object number one, m₂ is the mass of object number 2, and r is the distance between the two objects. Our main experience with gravity on earth is the force that exists between the earth and us.

The other way mass affects us is inertia. We know that all massive objects "want" to stay at whatever speed they have. That is, an object in motion tends to stay in motion, and an object not moving tends to stay motionless. This effect can be calculated with the equation
F = ma
where F is a force (possibly gravitational but not always), m is the mass of the object, and a is the acceleration that the object experiences due to the force F. Acceleration is how quickly an object changes its speed.

Physicists have measured the acceleration that objects obtain near the surface of the earth, and we call this the "gravitational acceleration", g. The magnitude of g is about 32.2 feet per second per second, or 9.8 meters per second per second. What we call weight is the strength of the gravitational force near the earth's surface, or W = mg, where W is the weight. So you can see that weight is a force. If we equate the gravitational force equation with this (W = mg) equation we get
F = Gm₁m₂/r² = m₁g
and if we divide through by m₁ we obtain
g = Gm₂/r².
If we call m₂ the mass of the earth and use the radius of the earth at its surface for r, we can calculate the gravitational acceleration. It really does work out; I've done the calculation!

By the way, dividing through by m₁ really bothered Albert Einstein. It is not intuitively obvious that the gravitational mass (the mass on the left side of the equation
Gm₁m₂/r² = m₁g )
is the same as the inertial mass (the mass on the right side of the equation). That is, the mass m₁ on the left side is used to calculate a gravitational force, and the mass m₁ on the right side calculates an inertial force, and there is no particular reason why the magnitudes of these masses should be the same. It is one of the outstanding questions of physics to figure out why this is so, but the equivalence of the two masses has been measured to be true to great precision.

Now, finally, to the kicking part. When you kick the brick it wants to remain at rest (inertia), and it takes a certain amount of force to give it a certain speed. You can calculate that force if you know the mass of the brick and the acceleration it obtains when you kick it. And this force does not depend on gravity's being around. That is, the inertia of the brick is there whether it is on earth or in outer space or on the moon.

In "collisions" like kicking a brick, it is often not possible to measure the actual acceleration, so we use another concept which is "impulse". Impulse is force multiplied by a time interval, and equals the change in momentum of the object. That is,
I = FDt = mDv
where I is impulse, F is force, Dt is the time interval over which the force acts, m is the mass, and Dv is the change in velocity of the object. So if we know the brick's mass and how much velocity it obtains we can calculate the change in momentum (which is mDv), and then if we know about how long the force (your foot!) is in contact with the brick we can calculate the (average) force our foot has applied to the brick to make it move. And this is all independent of gravity!!

Incidentally, if in the equation
FDt = mDv
we divide both sides by Dt we get
F = mDv / Dt
and since Dv / Dt is just acceleration, we are back to the equation
F = ma.
The form of the equation
F = mDv / Dt
is the form that Sir Isaac Newton originally published (many many years ago) as the now famous F = ma that we attribute to Sir Isaac.

John Link, MadSci Physicist