anonymous
  • anonymous
When the Earth, Moon, and Sun form a right triangle with the Moon located at the right angle, as shown in the figure below, the Moon is approaching its third quarter. (The Earth is viewed here from above its north pole.) Find the magnitude and direction of the net force exerted on the Moon. Give the direction relative to the line connecting the Moon and the Sun. Figure: http://www.webassign.net/walker/12-21alt.gif answers in N for first question and ° counterclockwise from the Moon-to-Sun ray for second question. I have no idea how to do this question. Thank you!
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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experimentX
  • experimentX
find the force of gravitation attraction .. between two objects individually due to sun and earth on moon. Add the components of forces and get the resultant force
anonymous
  • anonymous
ok, I will try that. I'm a little confused on how to find the direction..
experimentX
  • experimentX
|dw:1333570007541:dw|

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experimentX
  • experimentX
in your case, since the force will be at right angles, one will act as x-component while other will act as y component ... and direction will be equal to arctan of ratio of y/x components
experimentX
  • experimentX
|dw:1333570155711:dw|
anonymous
  • anonymous
I'm still not getting it. I calculated the individual forces between sun and moon, and earth and moon. But then what? How do I find the net force? I know I don't just add them together, do I?
experimentX
  • experimentX
(Fs^2+Fe^2)^(1/2) magnitude direction arctan(Fs/Fe)
anonymous
  • anonymous
thank you! I think I get it now.
anonymous
  • anonymous
But I'm still not getting the right answer! I'm getting 4.8*10^26 J. I'm really not sure where I'm going wrong. Thought I calculated correctly.
experimentX
  • experimentX
I don't you will have J as your unit ... it should be N
anonymous
  • anonymous
That's what I meant, sorry. N
experimentX
  • experimentX
you know the formula right?? Gm1m2/r^2
anonymous
  • anonymous
yes
experimentX
  • experimentX
you know the values,,?? distances ... and masses
anonymous
  • anonymous
for sun to moon I got 4.35*10^26. And for earth to moon I got 1.98 *10^26
anonymous
  • anonymous
yes, I know all those values. That's why I'm confused.. So after I calculated my forces, I use the pythgaorean theorem to get the net force, right?
experimentX
  • experimentX
can you post the expression??
anonymous
  • anonymous
so I have (4.35*10^26)^2 + (1.98*10^26)^2= (net force)^2
anonymous
  • anonymous
I hope that makes sense. The masses are Earth=5.97*10^24 kg, Moon=7.35*10^22 kg, and sun=2.00*10^30 kg.
anonymous
  • anonymous
uh oh
anonymous
  • anonymous
I was supposed to convert to meters wasn't I?
anonymous
  • anonymous
damn it.
experimentX
  • experimentX
disance would be measured in meters
anonymous
  • anonymous
omg. I feel like such an idiot! I'll try again!
anonymous
  • anonymous
ok, I got net force correct! But now I'm getting direction wrong..
experimentX
  • experimentX
i am sure you will get it .. it's just tan inverser of ratio of one to other.
anonymous
  • anonymous
Got it. Thank you so much! I've spent like 2 hours an that problem..

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