# Why does a force not do any work if it's perpendicular to the motion?

I have a book that says the Moon's orbit is [in this context assumed to be] circular. The Earth does no work on the moon. The gravitational force is perpendicular to the motion. Why is there no work done if support force is perpendicular to the motion?

• The perpendicular (to veloicity) component provide change in direction only, not the magnitude in velocity that in turns the kinetic energy. See another answer that has some sorts of linkage. Feb 6, 2017 at 15:21
• The Earth does do work on the moon because in fact the force is not perpendicular. The Earth is rotating faster than the moon is going around the earth, and so the tidal bulge on the side closest to the moon is slightly in front of the moon. Therefore the force is not perpendicular, and the moon is slightly accelerated by the Earth, and therefore the Earth is slightly decelerated by the moon. Thus our day is getting slightly longer all the time, and the moon is getting faster. Exercise: when will this process stop? Exercise: if the moon is getting faster, is the lunar month getting shorter? Feb 6, 2017 at 18:06
• The moon is being accelerated by the tidal forces but it's not getting faster because the speed it gains is quickly traded for height. Welcome to the strange world of orbital mechanics. Feb 7, 2017 at 3:30
• Exactly; and so the lunar month is getting longer, and the day is getting longer. So when does the process end? Original poster, see if you can work it out. Also: what relevant quantities are conserved by this exchange between the Earth and the moon? Feb 7, 2017 at 15:40
• Because the resultant of the triangle of forces in the direction of the motion is zero. Feb 7, 2017 at 17:10

As explained by SchrodingersCat, mathematically work is proportional to the scalar product of force and line element. Therefore any forces acting perpendicular to the path do not contribute to the work.

Now you might want to ask why work is defined like this. I would like to justify this definition taking your example of the moon.

In physics work is intimately related to energy: basically if you want to change the energy of an object you need to do work on it. Now in the case of the moon there are two relevant energies, (1) kinetic energy of the moon related to the magnitude (but not direction) of the moon's velocity, i.e. its speed; and (2) gravitational energy related to the position of the moon in the earth's gravitational field; this one depends on the distance moon-earth.

For (1), since perpendicular forces do not change the magnitude of velocity (only their direction), the perpendicular force should not enter into the equation of work (since it does not contribute to the energy change).

For (2) if you displace the moon always perpendicular to the direction of the gravitational force, you stay at the same distance, i.e. at the same gravitational potential energy. Therefore such perpendicular displacements do not change the energy and should not enter in the expression for work.

• Work is the change in kinetic energy, so gravitational potential energy (2) shouldn't be included here. Indeed if the Earth's gravity is the only force acting on an object, then (1)+(2) will be constant for that object, so this reasoning would lead to the conclusion that gravity never does work on the object. Feb 7, 2017 at 1:51
• @stewbasic: In my understanding one can also speak of work when e.g. lifting an object in gravity, i.e. when increasing its potential energy. I cannot follow your second argument. In the example Earth's gravity is the only force acting on the object/moon. What do you mean by (1)+(2) will be constant? Feb 7, 2017 at 2:13
• Suppose I drop a ball from rest (in vacuum). During its descent it gains kinetic energy (1) and loses the same amount of potential energy (2), so the total energy (1)+(2) is unchanged. Gravity has done positive work on the ball, and the amount is the change in kinetic energy (1). Feb 7, 2017 at 5:31
• @stewbasic If I lift a box from the floor ontop a table I do not change the kinetic energy but I increase its potential energy. And I also do a work on it. Actually, I increase the kinetic energy from zero to some arbitrary value and then I decrease it back to zero while I increase the potential energy all the time. Feb 7, 2017 at 13:04
• @Crowley in that scenario the positive work done by the force you apply cancels with the negative work done by the force of gravity. You could arrive at the same answer by ignoring the work done by gravity, instead including change in gravitational potential energy (which effectively captures the work done by gravity). But this alternate method won't work for the OP's question which is specifically asking about work done by gravity. Feb 7, 2017 at 22:24

Actually, work or, "work" as we mean in physics, has been mathematically defined as $$W=\int_{x_1}^{x_2}\vec F\cdot d\vec{s}$$ So if $\vec{F}$ and $d\vec{s}$ are orthogonal vectors, that is, the angle between the vectors is $90^\circ$, then according to the above formula, no work is done, physically; in this case, by the earth on the moon.

• The larger question is "why is work defined using that scaler product?" Is there physical behavior reasoning behind that definition? Yes, there is. Feb 7, 2017 at 18:35
• @BillN It is the same sort of questions like "Why is blue colour called blue?" He who defined the work found that the scalar product of force and displacement is useful and named it. Story ends. Feb 8, 2017 at 9:53

As @SchrodingersCat explained, there will be no work on the system if the force is orthogonal to the displacement. However, I would like to elaborate the answer a little further.

What is the physical meaning of representing work done on a body by the dot product of the force and displacement of the object?

An object can move even without a force (Newton's first law says about this), the motion however will be an unaccelerated one. So a body could displace even if there is no force. If you have studied classical mechanics, you may have heard that it is the linear momentum which is the generator of translation, not force.
To say a work is to be done by a force on the object, it should have some effect ..... on the object, right? Any dynamical property (like in this case, the effect of a force) is represented by a change in position coordinate of the object (the order of which varies according to the dynamical quantity) w.r.t time, because position is something very fundamental in dynamics.
If the force has some effect on the object (which is acceleration, of course), then this force contributes some displacement along the direction of applied force (even if there is motion already in some other direction). In such a case, the effect of force on the body can be measured by taking the component of the resultant (or net, if you insist) displacement along the direction of the applied force. So, work done by a force is defined as the product of the applied force with the displacement component caused by this force. This can be achieved by taking the dot product of the the two vectors-Force and the net displacement.

So, what does it mean no work is done if force and displacement are orthogonal?

In Euclidean geometry, orthogonal vectors implies mutually perpendicular vectors. However, the actual sense is that the two vectors are independent. This means that one has no common component with the other, which according to the above discussions states that one vector has no effect on the other. So, the displacement occurred is not due to the force given. Geometrically, that is possible only if force and displacement are perpendicular so that their dot product vanishes. This is why there is no work done on the system if the applied force and the resultant displacement are perpendicular.

But, that perpendicular force could affect the direction of motion of the body (since a force on a body should accelerate it somehow). So, no work is needed to change the direction of a body, even though it happens only by a force. In such a case, there is no displacement due to the force applied, only a change in direction- the effect of which is defined by the torque on the body (the rotational analogue of force).

: "effect" in the present context is used to imply anything that can contribute to work. We cannot say that the force has no effect on the object. It could accelerate the body, even if no displacement has happened due to that force, which is by changing the direction of motion.

## Background

The basic physics principle regarding the energy of a system is that the energy changes when work is done on the system, or the system does work on a different system. The work can either increase (positive work) or decrease (negative work) the total energy of the system.

Forces are the agents of work. Only external forces can change the total energy of a system. Internal forces cause exchanges of energy between pieces of the system.

Let's consider a system, the Moon (only). The gravitational pull of the Earth can do work on the system. That would change the total energy of the Moon, which in this case would simply be a change in kinetic energy. What does a change of kinetic energy mean? It means that the speed of the object has changed.

## The Moon

If the Moon is moving in a circular orbit, then the instantaneous velocity of the Moon is always perpendicular to the instantaneous acceleration, so according to the (correct) definition of work in other answers, $$W=\int \vec{F}\cdot\mathrm{d}\vec{s},$$ the work is zero.

Why is there no work done if support force is perpendicular to the motion?

That is, what is the physics behavior that says the perpendicular force doesn't change the energy?

Because the change in energy (by work being done) is a change in the kinetic energy, the force must change the speed of the objects in the system. Accelerations which are perpendicular to the instantneous velocity only change the direction, not the speed. In order to change the speed there must be a component of acceleration which is not perpendicular.

## An aside on potential energy

Potential energy is a system energy. If your system is the Moon only, there is no gravitational potential energy. If your system is the Earth and Moon, then one can consider the gravitational energy due to the interaction of the two. But when accounting for work, it's an either/or situation: Either you measure energy as the sum of kinetic and potential, or you consider kinetic only, modified by the work done by the gravitational interaction. You cannot count both simultaneously, because a potential energy change is defined as the negative of the work done by a conservative force (in this case, gravitational), when relative positions of the interacting objects change.

As others have mentioned, work (in 3D) on a system in defined as the dot product between the force on the system and its displacement. If they are perpendicular, no work is done.

In terms of intuition, what it means is that the force "redirects" the system (the moon, in your case) but in such as way that some of the particles are sped up and some are slowed down in the redirecting such that the net result is that the total energy of the moon is the same as it was earlier. The moon is forced to move in a circular, but in such a way that its energy remains the same at every point (at least, in the extremely simple model that you are assuming).

I see that most of the posters have answered this question in terms of the usual definition of work. That's fine as far as it goes, but to many people the definition of work seems somewhat arbitrary in the first place.

An alternative would be to treat the work-energy theorem $$W_\text{net} = \Delta T \;,$$ (with $T$ the kinetic energy) as an expected behavior (because it is integral in building the conservation law from Newtonian principles and the conservation principle is so useful) and use that to deduce the form that work must take and thus show the reason for the scalar product.

What follows is just an outline.

• The straight-line version gives us $W_\parallel = F_\text{net} \,\Delta x$.
• Uniform circular motion shows us that speed (and therefore kinetic energy) are not changed by forces perpendicular to the direction of motion. That is $W_\perp = 0$.
• We may then break any net force into its parallel and perpendicular components and note that the work comes wholly from the parallel one so that writing the work in terms of the scalar product becomes a natural step.

This also leads us to the main physical content of that definition: forces applied perpendicularly to the direction of motion don't chnage the speed of the object and are in that way different from forces applied along (or against) the direction of motion.

At a higher level of sophistication one would employ Noether's theorem as a postulate and work from there.

• I noticed no one brought up Stokes' theorem either, since any field that can be reduced to $\oint_{C} \ \mathbf{F} \cdot d\mathbf{l} = 0$ is by definition conservative (i.e., path independent). Feb 8, 2017 at 0:37
– user143115
Feb 9, 2017 at 6:41
• @sofky's question: physics.stackexchange.com/questions/310620/… Apr 5, 2017 at 6:44

I think you are right @Shrodingers cat. It would be better to clarify the answer this way.

Work done (w) = F . d

                         = F d Cos θ


At 90 degrees, θ = 90 and Cos 90 = 0,

W = F x d x Cos 90

= F x d x 0 = 0 Joule.


So, when the force is applied perpendicularly to the surface, the work done will be zero.

• Only that you don't apply it perpendicular to any surface but perpendicular to the line element, perpendicular to the path the moon takes. Feb 6, 2017 at 15:18

As others have already explained. Well its because there is no displacement in the direction of the gravitational force. It is assumed that the orbit stays the same during our observation. The Earth is pulling the moon towards the centre but the moon is moving in a circular orbit, with no displacement towards the centre.

A simple analogy is a block released from a particular height. Now as it falls down gravitational force does work on it, by pulling it down through a distance and raising its kinetic energy. Now suppose as it moves down you apply a constant horizontal force to the right, now the block is moving slanted path, because of the resultant force due to the two forces. Now if we consider the horizontal motion, gravitational force is not the cause of that, it is simply your hand. So your hand does work for the horizontal displacement, not the vertical one which is done by gravity.

So, in the case of the moon, Earth's gravitational force is not the cause of why it moves in a circle, it only provides the centripetal force necessary and centripetal force causes no whatsoever displacement to the moon.

In case you are interested in a more mathematical approach, this can actually be proven entirely using geometry. To proof this you need to to know a few things about vectors.

If $\vec v=(v_x,v_y,v_z)$ is the velocity vector of the moon then $v_x$, $v_y$ and $v_z$ represent the components of the speed in the x, y and z direction.

You can calculate the speed of the moon by calculation the length of the velocity vector $$v=|v|=\sqrt{v_x^2+v_y^2+v_z^2}$$ If you square the length you get $v^2=v_x^2+v_y^2+v_z^2$, which I will use later. Lastly the dot product between two vectors is defined to be $$\vec a\cdot\vec b=a_xb_x+a_yb_y+a_zb_z=|a||b|\cos \alpha$$ $\alpha$ is the angle between the two vectors. This means that the dot product is zero if the two vectors are perpendicular

Proof

If no work is done on the moon the kinetic energy must be constant. So the derivative of the kinetic energy must be zero. So we take the derivative by applying our $v^2$ substitution and taking the constants out of the derivative. $$\frac{dKE}{dt}=\frac{d}{dt}(\tfrac{1}{2}mv^2)=\tfrac{1}{2}m\cdot\frac{d}{dt}(v_x^2+v_y^2+v_z^2)$$ Then apply the chain rule to each of the components. $$\frac{d}{dt}(v_x^2+v_y^2+v_z^2)=2v_x\frac{dv_x}{dt}+2v_y\frac{dv_y}{dt}+2v_z\frac{dv_z}{dt}=2v_xa_x+2v_ya_y+2v_za_z$$ In which we recognize the dot product: $$\frac{d}{dt}v^2=2\vec v\cdot\vec a$$ So the derivative of the kinetic energy becomes $$\frac{dKE}{dt}=m\vec v\cdot\vec a$$

If the orbit is circular, the velocity is always perpendicular to the acceleration (and the force). So the kinetec energy doesn't change and no work is done on the moon.