Mass M is distributed uniformly over a rod of length L. Find the gravitational field at P at a distance a on perpendicular bisector.
GM2a(L2+4a2)
2GMa(L2+a2)
none
2GMa(L2+4a2)1/2
A
2GMa(L2+4a2)1/2
B
2GMa(L2+a2)
C
GM2a(L2+4a2)
D
none
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Solution
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dm=MLdx
dE=GMLdx(x2+a2)dEnet=dEcosθ=GMadxL(x2+a2)3/2
E=∫dEcosθ=GMaL∫L/2−L/2dx(x2+a2)3/2
Put x=atanθ then dx=asec2θdθ
E=GMaL∫asec2θdθa3sec3θ=GMLa∫cosθdθ
=GMLasinθ=GMxLa√x2+a2∫L/2−L/2
=GM(2L)aL(L2+4a2)1/2=2GMa(L2+4a2)1/2
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