Gravitational field

  • #1
1,197
1
the question is http://home.earthlink.net/~urban-xrisis/phy.jpg [Broken]

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

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

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

what did I do wrong?
 
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Answers and Replies

  • #2
Doc Al
Mentor
45,136
1,433
UrbanXrisis said:
[tex]g=\frac{GM}{r^2+a^2}[/tex]
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...
[tex]g=2 \frac{GM}{r^2+a^2}[/tex]
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.
 
  • #3
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1
horizontally:
[tex]g=\frac{GM}{r^2}+\frac{GM}{r^2}[/tex]
vertically:
[tex]g=\frac{GM}{a^2}-\frac{GM}{a^2}=0[/tex]

so then g would be: [tex]g=2\frac{GM}{r^2}[/tex]

not quite sure what to do
 
  • #4
VietDao29
Homework Helper
1,424
3
That's WRONG.
Vertically:
[tex]\frac{GM\sin{\alpha}}{a^2 + r^2} - \frac{GM\sin{\alpha}}{a^2 + r^2} = 0[/tex]
Horrizontally:
[tex]\frac{GM\cos{\alpha}}{a^2 + r^2} + \frac{GM\cos{\alpha}}{a^2 + r^2} = \frac{2GM\cos{\alpha}}{a^2 + r^2}[/tex]
Find [itex]\cos{\alpha}[/itex]. Can you handle it from here?
Viet Dao,
 
  • #5
1,197
1
[tex]\cos{\alpha}=\frac{r}{a^2+r^2}[/tex]

[tex]g=\frac{2GM\frac{r}{a^2+r^2}}{a^2 + r^2}[/tex]

???
 
  • #6
VietDao29
Homework Helper
1,424
3
That's not correct. Recheck your [itex]cos{\alpha}[/itex].
Viet Dao,
 

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