- #1
iScience
- 466
- 5
in my notes i have..
\Phi = \int \vec{g}\cdotdA = g(4\pir2) = -\frac{GM}{r^2}(4\pir2)
which yields \vec{g}=-\frac{GM}{r^2}
here's what i did independently..
\Phi = \int \vec{g}\cdotdA = -\frac{GM}{r^2}(4\pir2)
but since \vec{g}= -\vec{\nabla}\Phi
ie...
\vec{g}=\frac{d \Phi}{dr}=\frac{4 \pi GM}{r}
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
\Phi = \int \vec{g}\cdotdA = g(4\pir2) = -\frac{GM}{r^2}(4\pir2)
which yields \vec{g}=-\frac{GM}{r^2}
here's what i did independently..
\Phi = \int \vec{g}\cdotdA = -\frac{GM}{r^2}(4\pir2)
but since \vec{g}= -\vec{\nabla}\Phi
ie...
\vec{g}=\frac{d \Phi}{dr}=\frac{4 \pi GM}{r}
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
Last edited:
