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I want an explanation for this:

"The centre of mass & centre of gravity of an extended body on the surface of earth is different for objects with sizes greater than 100m"

As a matter of fact I was under the impression that the centre of mass(CM) always coincides with the centre of gravity(CG).

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The centre of mass is always the same.

The centre of gravity depends on gravity's pull. If it pulls equally everywhere, then this pull as well averages out to be in the centre of mass.

But imagine having a very tall uniform object. A beam standing on the surface and reaching 300 km outwards i.e. The bottom is closer to earth where gravity is stronger, while there is less gravitational pull in the far end. The result is an averaged gravitational pull closer to the Earth surface.

The centre of mass is still in the very middle of the beam, but the centre of gravity is not anymore.

From the comment, let me add the note that the centre of mass regards one object and not necessarily a whole system. When Earth pulls in the beam in my example, then that gravitational force is an external force. The beam is the system in my example, not including Earth.

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  • $\begingroup$ If the object is extended to that height(~300Km) and obviously it should contain some mass. Then centre of mass of the system(object+earth) will not change ? I'm confused here. $\endgroup$
    – Mihir
    Commented Oct 13, 2017 at 7:47
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    $\begingroup$ @Maverick The centre of mass and centre of gravity are with respect to the object, not the entire system. The centre of mass is the point around which the mass is evenly balanced. The center of gravity is the point around which the gravitational force on each particle of the object is balanced. If this force of gravity is greater on one side (force gradient) then the center of gravity will be shifted in that direction—just as the centre of mass would be if the object had a higher mass density on one side (mass gradient). $\endgroup$
    – Kieran Moynihan
    Commented Oct 13, 2017 at 8:04
  • $\begingroup$ Thank you for that very clear explanation @KieranMoynihan. $\endgroup$
    – Steeven
    Commented Oct 13, 2017 at 8:32

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