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11-10-08LectPHY2048

# 11-10-08LectPHY2048 - Gravitation(continued Last time we...

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Gravitation (continued) Last time we discussed the force of gravity between spherical bodies and the superposition principle that let’s you calculate the net force of gravity on an object due to an array of other discrete masses (as the vector sum of the individual forces). Now we consider an example of the force of gravity between a spherical mass m 1 and an extended body. For the arrangement shown in the figure where the thin uniform rod has total mass m , what is the force of gravity between spherical mass m 1 and the rod of length L ?

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We will use the expression, 11 2 dm ˆ FF G m r r == GG Where dm is an infinitesimal mass element along the rod having infinitesimal length dr , as shown. Since each mass element dm lies at a different distance , r , from m 1 , our integration variable needs to reflect the variable that is changing (i.e. since it is r that is changing the integration needs to be over dr ) To make the conversion from dm we define the linear density of the rod, m L λ= (the mass per length)
mm d m LL d L Δ λ= = = Δ If we slice the rod into short lengths Δ L the mass of each such length is Δ m and their ratio still gives λ. Since this remains true in the limit of infinitesimals we have that, But there is no distinction between dL and dr allowing the replacement, which gives, dm dr But then, dm dr So that, 11 1 22 dm dr FG m G m rr λ == ∫∫ The conversion we needed.

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Removing constants from under the integral and integrating, dL 11 1 1 2 d d dr 1 1 1 FG m G m G m rr d L d + + ⎤⎛ ⎜⎟ ⎢⎥ + ⎦⎝ 1 d L d m G m dd L d ( d L )d ( d L ) + ⎡⎤ ++ + ⎣⎦ L m d(d L) + But, m L λ= m m = + So, 1 1 mm d(d L) = + Answer.
The problem here is not that each mass element dM of the ring is a different distance from m (they are all the same distance ) but rather that each element lies in a different direction and superposition requires us to sum the vector forces from the elements. Another example is that of a thin ring of matter of radius R and mass M that attracts a mass, m , which lies on the axis of the ring, some distance z from the ring center as shown. M z m dM m r φ The force on m from element dM lies along the direction toward dM with magnitude, dF G 2 mdM dF G r = We resolve this force into horizontal and vertical components .

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11-10-08LectPHY2048 - Gravitation(continued Last time we...

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