
#1
Oct2310, 03:01 PM

P: 90

1. The problem statement, all variables and given/known data
When the Earth, Moon, and Sun form a right triangle, with the Moon located at the right angle, as shown in the figure , the Moon is in its thirdquarter phase. (The Earth is viewed here from above its North Pole.) Find the magnitude of the net force exerted on the Sun. Find the direction of the net force exerted on the Sun. Give the direction relative to the line connecting the Sun and the Moon. 2. Relevant equations Gm_{1}m_{2}/R^{2} F_{net}=√F_{1}^{2}+F_{2}^{2} [tex]\theta[/tex]= inverse trig (whatever necessary) Mass sun=1.99E30 Mass earth= 5.98E24 mass moon= 7.35E22 radius sun to moon=1.50E11 radius moon to earth=3.85E8 3. The attempt at a solution F_{1}(SunMoon)=4.336E20 found by using gravity force eq. F_{2}(sun earth)= 3.53E22, however this is wrong because it should be 3.58E22, so that means the distance between the sun and the earth at this time would have to be 1.49E11, which doesn't make since because this should be the hypotenuse, right???. So when using the sencond value for F_{2}, and using hte pythagorean theory I get F_{net}=3.57E22, which is the right answer. NOW to find the direction, i thought you can use the cos inverse *F_{1}/F_{2}, but this turns out to be the wrong answer I have no idea what else to do. Please help! 



#2
Nov710, 12:28 PM

P: 1

I would've done cos^{1} also. Did you ever find out the correct answer? I had this exact problem, only I was tasked with finding F_{net} exerted on the moon. I found my angle by taking tan^{1} of F_{3}/F_{1}, where my value for F_{3} was 1.9814E20, the force of earthmoon. F_{1} is obviously sunmoon, only the value I calculated was 4.3438E20. I got the correct answer.




#3
Nov710, 04:23 PM

P: 90

yes finding the net force on the moon was easy, but still never found out for the sun



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