- #1
iScience
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- 5
in my notes i have..
[itex]\Phi[/itex] = [itex]\int[/itex] [itex]\vec{g}\cdot[/itex]dA = g(4[itex]\pi[/itex]r2) = -[itex]\frac{GM}{r^2}[/itex](4[itex]\pi[/itex]r2)
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]r2)
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 homework 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]r2) = -[itex]\frac{GM}{r^2}[/itex](4[itex]\pi[/itex]r2)
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]r2)
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 homework problem. but i thought it wasn't, so i posted in this section
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