Solving Gravitational Field: What Did I Do Wrong?

[UrbanXrisis]

the question is http://home.earthlink.net/~urban-xrisis/phy.jpg

g=\frac{GM}{x^2}
x=\sqrt{r^2+a^2}
g=\frac{GM}{(\sqrt{r^2+a^2})^2}
g=\frac{GM}{r^2+a^2}

since there are 2 masses...
g=2 \frac{GM}{r^2+a^2}

my book's answer is:
g=\frac{2MGr}{(r^2+a^2)^{3/2}}

what did I do wrong?

 

[Doc Al]

g=\frac{GM}{r^2+a^2}

OK, this is the field due to one of the masses. Note that it's a vector. What is its direction?

since there are 2 masses...
g=2 \frac{GM}{r^2+a^2}

Since the field contributions are vectors, they must be added as such. (You can double it only if the vectors were in the same direction.) Hint: The vertical components will cancel.

 

[UrbanXrisis]

horizontally:
g=\frac{GM}{r^2}+\frac{GM}{r^2}
vertically:
g=\frac{GM}{a^2}-\frac{GM}{a^2}=0

so then g would be: g=2\frac{GM}{r^2}

not quite sure what to do

 

[VietDao29]

That's WRONG.
Vertically:
\frac{GM\sin{\alpha}}{a^2 + r^2} - \frac{GM\sin{\alpha}}{a^2 + r^2} = 0
Horrizontally:
\frac{GM\cos{\alpha}}{a^2 + r^2} + \frac{GM\cos{\alpha}}{a^2 + r^2} = \frac{2GM\cos{\alpha}}{a^2 + r^2}
Find \cos{\alpha}. Can you handle it from here?
Viet Dao,

 

[UrbanXrisis]

\cos{\alpha}=\frac{r}{a^2+r^2}

g=\frac{2GM\frac{r}{a^2+r^2}}{a^2 + r^2}

?

 

[VietDao29]

That's not correct. Recheck your cos{\alpha}.
Viet Dao,