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My book says:

For most of the small objects, both are same. But for mammoth ones, they are really different ones. And in a gravity-less environment, COG is absent; COM still exists.

Ok, what's the big deal when things are small and big? How these two : centre of mass & centre of gravity?

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/50107/2451 and links therein. $\endgroup$ – Qmechanic Dec 9 '14 at 16:46
  • $\begingroup$ @Qmechanic: Really I don't deem the linked one apt one to be marked as the same. Make a scrutiny! I have asked about how they differ in big bodies & he has asked something else. $\endgroup$ – user36790 Dec 9 '14 at 17:51
  • $\begingroup$ @Qmechanic: Sir, I have been unable to post any question here. A code is asked to coy and to paste then. But nothinng happens. Can you tell why?? Plz. $\endgroup$ – user36790 Dec 10 '14 at 18:16
  • $\begingroup$ Try first asking in chat and if no help then try meta. $\endgroup$ – Qmechanic Dec 10 '14 at 18:23
  • $\begingroup$ @Qmechanic: Hmmm... I have been able to post nothing due to the new CAPTCHA. If I could post a quo on meta, would there be any problem? No question can I post now! $\endgroup$ – user36790 Dec 11 '14 at 2:34
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Both values are computed as a position weighted average. For the center of mass we average the mass in this way, while for the center of gravity we average the effect of gravity on the body (i.e. the weight).

\begin{align} x_\text{com} &= \dfrac{\int x \, \rho(x) \,\mathrm{d}x}{\int \rho(x) \, \mathrm{d}x} \\ \\ x_\text{cog} &= \frac{\int x \, \rho(x)\, g(x) \,\mathrm{d}x}{\int \rho(x) \,g(x) \,\mathrm{d}x} \end{align}

Now, in the usual Physics 101 "near the surface of the Earth" convention $g(x)$ is constant so these two are equivalent. However, if the body is big enough that we need to account for either the changing strength or changing direction of gravity then they are no longer the same thing.

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    $\begingroup$ Hmm so CoM is just the centroid of the density field, while CoG is the centroid of the weight field? I somehow got through two physics degrees without ever realising that CoM and CoG aren't the same thing... $\endgroup$ – Mark K Cowan Dec 10 '14 at 21:03
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    $\begingroup$ Mark, well at least where I see people make a distinction this is the one they make. There are texts that do not distinguish at all. $\endgroup$ – dmckee Dec 10 '14 at 21:24
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    $\begingroup$ I suspect I'm a victim of thost latter texts... Seriously thanks for teaching me something potentially important that 4 years of university somehow managed to omit! $\endgroup$ – Mark K Cowan Dec 10 '14 at 22:40
  • $\begingroup$ So a spinning object free falling in a gravitational fiel in which is CoM and CoG will be spinning around an axis through its CoM, but will feel torque from the gravitational forces? @dmckee $\endgroup$ – Andrea Nov 15 '15 at 14:41
  • $\begingroup$ @AndreaDiBiagio Not sure I followed that comment. Answer to my best guess about what you meant: if the line between the CoM and the CoG is not co-linear with the net gravitational force than there will be a rotational tendency due to gravitation. $\endgroup$ – dmckee Nov 15 '15 at 17:16
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As a quick rehash of layman's terms definitions you've probably heard: The Center of Mass (CM) represents a single point where you could treat the object as a point particle, with the combined mass of the object. It's found by the average location of the mass of an object. The Center of Gravity (CG) is a point that represents the average pull of gravity on an object.

Close to the surface of the Earth, it may not be obvious to you why these are separate descriptions. You've probably learned by now that the weight, or pull of gravity, on an object is given by $F_g=mg$ where $F_g$ is the weight, or pull of gravity, of the object, $m$ is the mass of the object, and $g$ is the acceleration due to gravity at that location. You've also probably been told that $g=9.8 ~\mathrm{m/s^2}$ and that on that on a given planet, it's a constant...

...except it's not. The gravitational force between two objects depends on the distance between two objects, and is actually an inverse square relationship, which means that $F_g \propto \frac{1}{d^2}$. As you increase in altitude the acceleration due to gravity, $g$ decreases because you are further away from the center of the Earth. However, here on the planet Earth, changes in height on the order of magnitude of meters are very small when compared to the radius of the earth. However, for large objects, the size of the object itself is not negligible when compared to the external distance between it and the Earth.

Consider the Sears Tower. Its CG is about 1 millimeter below its CM. The reason why is because the base of the tower is closer to the center of the Earth than the top of the tower (by 442 m), and therefore receiving a slightly higher pull of gravity than the top of the tower. As a result, the CG is closer the the ground than the CM, because the part of the tower below the CM is being pulled by gravity (slightly) harder than the part of the tower above the CM.

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The centre of mass is the average point of the "mass" of the body, whereas centre of gravity is the average point of the "weight" that is mass times the local gravitational acceleration. For small objects both of them are almost same, but for large objects as the value of gravitational acceleration can change along the body (as the gravitational acceleration decreases the further an object is away from a planet), the centre of gravity can be different from the centre of mass!

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In our Moon, the center of gravity is not the same as the the center of mass. The result is that the Earth always sees the same side of the moon. This is becasue gravity pulls at the center of gravity, but the orbit is determined by the center of mass.

The center of mass determines the kinematics - how an object will rotate, spin, revolve, and orbit.

If the moon were symmetrical, these points would be the same.

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protected by Qmechanic Dec 11 '14 at 1:18

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