Why is there a dr in the second term of the gravitational force equation?

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SUMMARY

The discussion centers on the presence of the differential element "dr" in the gravitational force equation, specifically in the term involving the dot product of force and displacement vectors. The equation under scrutiny is F_g = (G*m_e*m)/r^2, where "dr" appears in the expression F⃗ g⋅dr⃗ = −(Gm_em/r^2)(drr_1⋅r_1 + r*dr*dθ*θ_1⋅r_1). Participants clarify that "dr" is included due to the nature of the vector calculus involved, despite being zero in the context of the dot product with perpendicular unit vectors. This confirms that the term does not contribute to the overall calculation.

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REVIANNA
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Homework Statement



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Homework Equations



I know that
##F_g=(G*m_e*m)/r^2)##
##dr⃗ =drr_1+rdθθ_1.##

##F⃗ g⋅dr⃗ =−(Gm_em/r^2)*r_1⋅(drr_1+rdθθ_1)##

##F⃗ g⋅dr⃗ =−(Gm_em/r^2)(drr_1⋅r_1+r*dr*dθ*θ_1⋅r_1)##

The Attempt at a Solution


[/B]
I don't understand why there is a dr in the 2nd term in the second bracket.
I have used r_1 and ##theta_1## for unit vectors.
I know how to proceed from this step (##r_1.r_1=1## whereas ##theta_1.r_1=0(perpendicular)##
 
Last edited:
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REVIANNA said:
I don't understand why there is a dr in the 2nd term in the second bracket.
There should not be a ##dr## there. It looks like a mistake.
In any case it doesn't matter, as the term is zero because the dot product is.
 
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andrewkirk said:
In any case it doesn't matter, as the term is zero because the dot product is.

yes as unit vector θ and unit vector r are perpendicular .
thanks.
 

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