Whenever I ask this doubt that how can force perpendicular to objects velocity can change its direction but cant change its magnitude I get proof for only the magnitude will be constant that is: no component of force is parallel to velocity so it cant change the magnitude of velocity but I cant get proof for how can it change direction of motion. Every time they say that since force cant change magnitude it changes its direction. But I need proof of how orthogonal force change the direction of velocity.

edit-My problem is in my class my teacher said if acceleration is always perpendicular velocity then the magnitude of velocity will never change and proved it like this: $$\frac{d(v^2)}{dt} =\frac{d(v⃗.v⃗)}{dt}=a⃗ ⋅v⃗ =0$$ (since $a⃗$ and $v⃗$ are perpendicular to each other) So $v$ (speed=magnitude of velocity) is constant. After that he said since this acceleration cannot change magnitude of velocity so it will change only direction of velocity and said in general it is valid for any vector and said if a vector and its derivative are perpendicular to each other then only direction of vector will change and magnitude of that vector will remain constant. But I cannot understand how can perpendicular acceleration change the direction of velocity, I searched for the answer in the internet for hours and cant get a satisfactory answer I only get the same proof(that magnitude will remain constant) but in different different ways. Please if anyone can give a good explanation on how it changes direction of velocity(without changing its magnitude).

  • $\begingroup$ Hello Rahul Einstein and welcome to Physics SE. I believe that you should format your question a little bit better to make it clear. Additionally, you should provide some information as to what you have tried to tackle your problem and where you got stuck. I believe that what you need in order to solve your problem is to learn more about vectors. Vectors are "mathematical constructs" that have both magnitude and direction. In order to change the magnitude of a vector you add another one which as a component parallel to the first. Orthogonal vectors do not share a component that is parallel to $\endgroup$
    – ZaellixA
    Commented Oct 8, 2022 at 14:26
  • $\begingroup$ Please consider a particle moving on a circle -- or assume that the earth circles the sun on a circular orbit. Just think about this setup and ask yourself in which direction the velocity points, and into which direction the forces (=mass * acceleration = constant * a) points. $\endgroup$
    – Semoi
    Commented Oct 8, 2022 at 16:19
  • $\begingroup$ Hello ZaellixA thank you for the answer I have edited my question to make it more clear. $\endgroup$ Commented Oct 8, 2022 at 17:47
  • $\begingroup$ Please refer to chapter 1 of Kleppner and Kolenkow. In short the answer is that the change in velocity caused is a differential quantity thus can be ignored. Another way to look at it is that in uniform circular motion the change in velocity caused at any one point precisely cancels out by change in velocity at a point directly away from first one. That is if circle is centred at origin then changes in velocity due to points at x and - x cancel out since the force is constant. Thus to a very good aprrox the average velocity is constant since we are dealing with differential quantities. $\endgroup$ Commented Oct 8, 2022 at 19:08
  • $\begingroup$ The proof is not quite correct. You have to use the product rule:$$\frac{d}{dt}\langle\vec v,\vec v\rangle=\langle\frac{d}{dt}\vec v,\vec v\rangle+\langle\vec v,\frac{d}{dt}\vec v\rangle=\langle\vec a,\vec v\rangle+\langle\vec v,\vec a\rangle=0$$ $\endgroup$
    – Filippo
    Commented Oct 8, 2022 at 19:35

4 Answers 4


If a vector $\mathbf v$ is changing (so $\mathbf a = \frac{d}{dt} \mathbf v \neq 0$) but its magnitude is constant ($\frac{d}{dt}|\mathbf v|=0$), then the only thing that can be changing is its direction.

If you wish, you can write a 2-dimensional vector as $$\mathbf v = v \pmatrix{\cos(\theta)\\\sin(\theta)}$$ where $v\equiv |\mathbf v|$ and $\theta$ specifies its orientation with respect to the horizontal axis. From there, $$\frac{d}{dt}\mathbf v = \frac{dv}{dt} \pmatrix{\cos(\theta)\\\sin(\theta)} + v \frac{d\theta}{dt}\pmatrix{-\sin(\theta)\\\cos(\theta)} \equiv \mathbf a_\parallel + \mathbf a_\perp$$

It's easy to check that $\mathbf a_\parallel$ is parallel to $\mathbf v$, while $\mathbf a_\perp$ is perpendicular to $\mathbf v$. If the magnitude of the vector is constant - so $\frac{dv}{dt}=0$ - then only the second term survives, and is proportional to the rate of change of $\theta$. Therefore, $\mathbf a_\perp$ corresponds to a change in the direction of $\mathbf v$.

The extension to 3D is straightforward, and can be explicitly checked by writing $$\mathbf v= v\pmatrix{\sin(\theta)\cos(\phi)\\\sin(\theta)\sin(\phi)\\\cos(\theta)}$$ where now $\theta$ and $\phi$ are the polar and azimuthal angles, respectively.


As you stated, a force acting perpendicular to the velocity of a particle can't change its magnitude, as the component of the force acting along the direction of instantaneous velocity is zero.

The fact that such a force changes the direction of the particle's motion is simply because even though none of the force is increasing the particle's speed along its instantaneous direction of motion, ALL of the force is increasing the particle's speed along the direction of the force, i.e., its instantaneous acceleration is in the direction of the force.

Since this acceleration is not parallel to the particle's motion(it is gaining velocity in the direction of acceleration, that's what it means to accelerate), we can safely conclude that the particle must be changing its direction.

Similarly, for a force that's always perpendicular to a particle's velocity, it will, at all instants, only change the particle's direction of motion(towards the direction of the force), without changing its magnitude; an example being a central force, responsible for uniform circular motion.

@J.Murray has given an appropriate mathematical proof for the same.

  • 1
    $\begingroup$ Please correct me if I am wrong what I get after reading your answer is there is no other deeper reason just simply the reason that velocity has to change(since external force is acting) and because magnitude cant change(already proved) so we can conclude direction must me changing. And there also exist a mathematical proof to it. $\endgroup$ Commented Oct 9, 2022 at 6:39
  • $\begingroup$ Your conclusion is correct that if a non-zero external force is acting while the speed is constant, the direction must be changing, but that's an outcome of the imposed condition. What I wanted to convey was that if you breakdown the motion into its components, you can see that an external force acting perpendicular to the velocity $\vec v$is increasing its velocity along the direction perpendicular to its motion from $0$ to some $v_0$. The result is that the particle has the same magnitude of velocity $v$ along the initial direction, and a new velocity $v_0$ along the direction... $\endgroup$ Commented Oct 9, 2022 at 6:48
  • $\begingroup$ ... perpendicular to it. The new velocity being along the resultant of the initial velocity and the gain in velocity $$\vec v + \vec v_0 = \vec v_{net}$$ $\endgroup$ Commented Oct 9, 2022 at 6:54
  • $\begingroup$ But when the force is acting perpendicular to the velocity at ALL instants, like in circular motion, the force vector is constantly changing direction to be perpendicular to the motion of the particle. In this case the gain in velocity $v_0 = dv$ is very small, as the force acts on the particle in that particular direction for a very small time $dt$. $\endgroup$ Commented Oct 9, 2022 at 7:15
  • $\begingroup$ As the gain in velocity v0=dv is very small, as the force acts on the particle in that particular direction for a very small time dt so can be ignored why cant we ignore change in direction as it is also small $\endgroup$ Commented Oct 9, 2022 at 16:27

Acceleration, $\vec{a}=\frac{d\vec{v}}{dt}$. Let's assume $\vec{a} \perp \vec{v}$ at all times and $\vec{a}\neq \vec{0}$.

$\vec{a} \perp \vec{v} \Rightarrow \vec{v}\cdot \vec{a}=0$. i.e, $\vec{v}.\vec{a}=\vec{v}\cdot \frac{d\vec{v}}{dt}=0 \quad ...(\star)$

Consider: $\frac{dv^2}{dt}=\frac{d\vec{v}\cdot \vec{v}}{dt}=\vec{v}\cdot \left(\frac{d\vec{v}}{dt}\right)+\left(\frac{d\vec{v}}{dt}\right)\cdot\vec{v}=2\vec{v}\cdot\left(\frac{d\vec{v}}{dt}\right)$, using this in $(\star) $, we get:

$\quad \vec{v}.\vec{a}=\vec{v}\cdot \frac{d\vec{v}}{dt}=\frac{1}{2}\frac{dv^2}{dt}=0 \Rightarrow v^2=constant \quad ...(\star\star)$.

$\therefore$ whenever a non-zero acceleration acts perpendicular to the velocity at all times, magnitude of velocity stays constant. Now read the sequence of equations ($\star\star$) from right to left. It says: if the magnitude of velocity is a non-zero constant then dot product of acceleration and velocity is zero $\Rightarrow$ either $\vec{a}=\vec{0}$ or $\vec{a}\perp \vec{v}$ at all times.


Mathematically, acceleration is defined as, $$\vec{a}=\frac{d\vec{v}}{dt}$$ Hence, we can define change in velocity as $$\vec{v}(t+dt)-\vec{v}(t)=d\vec{v}=\vec{a}dt$$ $$\vec{v}(t+dt)=\vec{v}(t)+\vec{a}dt$$ Now, if initial velocity, $\vec{v}(t)\perp\vec{a}$, you would be able to show that final velocity, $\vec{v}(t+dt)$ is not parallel to the initial velocity by the simple vector addition in last equation. Hence, direction of velocity is changed by an acceleration perpendicular to it.

  • 1
    $\begingroup$ Yes I agree, we can prove in this way that the direction of the final velocity is slightly different from initial velocity (since adt is small) and this direction is continuously changing , but in the same way we can prove that there is a slight change in magnitude also which is continuously changing so magnitude of final velocity is slightly more than initial velocity so magnitude of velocity is not constant but there is a proof that magnitude of velocity will be constant. $\endgroup$ Commented Oct 9, 2022 at 6:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.