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## Main Question or Discussion Point

in my notes i have..

[itex]\Phi[/itex] = [itex]\int[/itex] [itex]\vec{g}\cdot[/itex]dA = g(4[itex]\pi[/itex]r

which yields [itex]\vec{g}[/itex]=-[itex]\frac{GM}{r^2}[/itex]

here's what i did independently..

[itex]\Phi[/itex] = [itex]\int[/itex] [itex]\vec{g}\cdot[/itex]dA = -[itex]\frac{GM}{r^2}[/itex](4[itex]\pi[/itex]r

but since [itex]\vec{g}[/itex]= -[itex]\vec{\nabla}[/itex][itex]\Phi[/itex]

ie...

[itex]\vec{g}[/itex]=[itex]\frac{d \Phi}{dr}[/itex]=[itex]\frac{4 \pi GM}{r}[/itex]

but this comes out to a different answer. is this still correct?

*to admins/moderators/mentors/etc: before i get in trouble again, i appologize if this is considered another hw problem. but i thought it wasn't, so i posted in this section

[itex]\Phi[/itex] = [itex]\int[/itex] [itex]\vec{g}\cdot[/itex]dA = g(4[itex]\pi[/itex]r

^{2}) = -[itex]\frac{GM}{r^2}[/itex](4[itex]\pi[/itex]r^{2})which yields [itex]\vec{g}[/itex]=-[itex]\frac{GM}{r^2}[/itex]

here's what i did independently..

[itex]\Phi[/itex] = [itex]\int[/itex] [itex]\vec{g}\cdot[/itex]dA = -[itex]\frac{GM}{r^2}[/itex](4[itex]\pi[/itex]r

^{2})but since [itex]\vec{g}[/itex]= -[itex]\vec{\nabla}[/itex][itex]\Phi[/itex]

ie...

[itex]\vec{g}[/itex]=[itex]\frac{d \Phi}{dr}[/itex]=[itex]\frac{4 \pi GM}{r}[/itex]

but this comes out to a different answer. is this still correct?

*to admins/moderators/mentors/etc: before i get in trouble again, i appologize if this is considered another hw problem. but i thought it wasn't, so i posted in this section

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