Gravitational field from two point masses: why is there an r factor?

[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,