In the traditional sense, acceleration means the rate at which velocity is changing. Visually, we tend to associate this with the magnitude of velocity.

If the centripetal acceleration of the moon is $2.8 \times 10^{-3} ~\mathrm {m/s^2}$, where the magnitude of velocity remains constant, what does this value mean? How can this value be interpreted intuitively, just as "regular" acceleration is so intuitive?

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    $\begingroup$ The magnitude of the velocity in each direction in changing. The centripetal acceleration is the same acceleration you're used to, it's just always directed towards the center of orbit. $\endgroup$ Jul 1, 2016 at 9:31
  • $\begingroup$ @immibis where does it say that the velocity is constant in the question? $\endgroup$
    – JimmyJames
    Jul 1, 2016 at 14:27
  • $\begingroup$ " Visually, we tend to associate this with the magnitude of velocity." That's something that you should get over. Velocity is a vector and changing it's direction is changing the quantity just as much as changing it's magnitude. $\endgroup$ Jul 1, 2016 at 16:32

3 Answers 3


what does this value mean?

This value means for maintaining of moon on the circular path, we need a force by amount of $\textrm{mass of the moon}\times\textrm{that value}$

How can this value be interpreted intuitively?

According to first law of Newton, a moving body will move on a straight line uniformly unless a force is exerted on it. enter image description here

As you can see in the figure, a force must exist so that pulls the moon towards the circle (otherwise, the moon won't move on the circle). The centripetal acceleration is this force divided by the mass of the moon.


Acceleration is defined as the time rate of change of velocity - be it in its direction or magnitude or both.

Velocity $\mathbf v$ in polar coordinates is given by:

$$\mathbf v= \dot {\mathbf r} = \dot r\mathbf{e_r} + r~\dot \theta\mathbf{e_{\theta}}\;.\tag 1$$

And then acceleration $\bf a$:

$$\mathbf a = \dot{ \mathbf v} = \left[\ddot r - r\dot \theta ^2\right]\mathbf{e_r} + \left[r\ddot \theta + 2~\dot r\dot \theta\right]\mathbf{e_\theta}\;.\tag 2$$

This is the crux of the kinematics involved in terms of polar coordinates.

Now, in circular motion, $\dot r= 0$ which leads $(2)$ to

$$\mathbf a = \dot{ \mathbf v} = - r\dot \theta ^2\mathbf{e_r} +r\ddot \theta \mathbf{e_\theta}\;.\tag {2-a}$$

The radial acceleration is the centripetal acceleration.

More precisely, in circular motion

$$\mathbf v= r\dot \theta \mathbf{e_\theta}$$

Therefore, the acceleration, is given by

$$\mathbf{a} = \frac{\mathrm d\mathbf v}{\mathrm dt}= r\frac{\mathrm d^2 \theta}{\mathrm dt^2}\mathbf{e_\theta} + \underbrace{r\frac{\mathrm d\theta}{\mathrm dt} \left(\frac{\mathrm d\mathbf {e_\theta}}{\mathrm dt}\right)}_\textrm{radial acceleration}\;.$$


\begin{align}\mathbf{a_r} \equiv \textrm{centripetal acceleration} &= r\frac{\mathrm d\theta}{\mathrm dt} \left(\frac{\mathrm d\mathbf {e_\theta}}{\mathrm dt}\right)\\ &= v~(-\omega \mathbf{e_r})\;.\end{align}

Centripetal acceleration measures the time rate change of direction of the velocity vector i.e., $\mathbf{e_\theta}\;.$

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    $\begingroup$ Can someone explain to me how this qualifies as way to a 'intuitively' understand this? $\endgroup$
    – JimmyJames
    Jul 1, 2016 at 16:42
  • $\begingroup$ I really don't know why it got downvoted. I am not sure how can anything be clearer from that. $\endgroup$
    – user36790
    Jul 14, 2016 at 9:10
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    $\begingroup$ @JimmyJames Perhaps not intuitive to someone without vector calculus. However, once you master that, then these are indeed intuitive concepts. If you've not done this, then you'll benefit from looking at these kinds of problems afresh when you have. $\endgroup$ Jul 14, 2016 at 11:21

Take a rope with a small weight on one end and spin around on a point such that the rope is taut. You will feel force on your arm 'pulling' on it towards the weight. That force exists because that's what you are exerting on the object to prevent constantly change it's direction. The value you refer to is the acceleration portion of that force from $F=ma$. If you attached a small spring scale to the rope, you could measure that force. One of my professors gave me a hard time (rightly, I suppose) about thinking about this from the perspective of the accelerated frame but it is much more intuitive to me.

  • $\begingroup$ Thinking from an accelerated frame is quite OK and sound, and indeed in some rigid body problems it's almost indispensable. However, I think your prof was giving you a hard time because he /she probably wanted you to get proficient with using Newton's laws from inertial frames before shifting on to accelerated ones. $\endgroup$ Jul 14, 2016 at 11:20

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