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I come across with a question, Asked in JEE MAINS 2014

Question stated below

From a sphere of mass M and radius R, a smaller sphere of radius R/2 is carved out such that the cavity made in the original sphere is between its centre and the periphery. (See figure). For the configuration in the figure where the distance between the centre of the original sphere and the removed sphere is 3R, the gravitational force between the two spheres is :

enter image description here

I think about centre of mass concept, as outside of any irregular shaped object, Gravitational Intensity(Field) is same as if we assume Centre of mass of irregular shaped object as point mass and later work upon the calculation

So, I solved it by assuming ,we have two point masses one M/8 and other 7M/8 separated at (3R + R/14) units apart, the R/14 term which I get is centre of mass of remaining sphere from centre of original sphere(one can figure this out)

So, Gravitational Force between them is (G (7M/8) (M/8)) / (43R/14)^2 = 343 GM^2 / 29584 R^2

but answer to this questions is 41GM^2/3600R^2

Please Enlighted us

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    $\begingroup$ "Gravitational Intensity(Field) is same as if we assume Centre of mass of irregular shaped object as point mass and later work upon the calculation" - no it isn't, see physics.stackexchange.com/q/92951 - "You're in the milky way galaxy, whose center of mass is nowhere near Earth - so why are you feeling a force towards Earth?" $\endgroup$
    – sqek
    Jun 7 at 12:34
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    $\begingroup$ Consider the case of two large round identical masses connected by a thin wire. The center of mass is at the center of the wire. You are next to the center of the wire. What force do you feel? $\endgroup$
    – Jon Custer
    Jun 7 at 13:21

2 Answers 2

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I think about centre of mass concept, as outside of any irregular shaped object, Gravitational Intensity(Field) is same as if we assume Centre of mass of irregular shaped object as point mass and later work upon the calculation

This is false. It only holds exactly for objects with spherical symmetry, and it holds approximately when you are very far away from irregular objects. (Specifically, the distance from the source must be much greater than the size of the source.) Neither of those conditions applies here.

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  • $\begingroup$ thanks for clarifying $\endgroup$
    – 5 Dots
    Jun 8 at 4:09
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As I recall, in finding the force of gravity between two symmetrical spheres, you can treat each as a point mass located at its center. (The proof would require finding the net force on a spherical shell in a radial field.) Given this, I would start by finding the force between two solid spheres, and then subtract the force from the mass which is not in the hole.

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  • $\begingroup$ ohh thanks sir. $\endgroup$
    – 5 Dots
    Jun 8 at 4:08

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