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Universal Graviation 
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#1
Dec1912, 12:53 PM

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1. The problem statement, all variables and given/known data
Let (S) be a satellite between the earth and the moon such that (S) is at a distance dm from the moon. All forces on the the satellite are null (equal zero). Find the distance dm . Given: Distance between moon and earth is 38*10^4km and the mass of earth is 81 times the mass of the moon. 3. The attempt at a solution Well it's been a long time since I've done any universal gravitation. It would be nice if someone can just give me an idea on how to start and I'll go from there. 


#2
Dec1912, 01:01 PM

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If the distance between the moon & the satellite is d_{m} , then what is the distance between the satellite & the earth ? 


#3
Dec1912, 01:04 PM

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#4
Dec1912, 01:31 PM

P: 584

Universal Graviation
Let's name the distance from earth to the satellite by D. So D=ddm.
And by using g=(GM)/D^2 we get that the distance from the earth is 6.36*10^3 km. So that means the distance from the satellite to the moon is dD=3.8*10^56.36*10^3=3.74*10^5 km. Is that correct?? 


#5
Dec1912, 01:36 PM

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Please fill in and explain some steps. My approach would be set the magnitudes of the following two forces equal to each other. The force exerted on the satellite by the moon. 


#6
Dec1912, 01:49 PM

P: 584

By setting the forces equal to one another i got in the end:
Me/Mm=dm^2/de^2 and we know that the mass of the earth is 81 times the mass of the moon so we get 81Mm/Mm=(dm^2)/(de^2) and then we cancel with the mass of the moon and square root both sides and we are left with. √81=dm/de and we know that the distance from the satellite and the earth is ddm. and by doing some simple algebra we find that the distance to the moon is d/10 the total distance between the moon and earth so that gives us 3.8*10^4 km. 


#7
Dec1912, 02:13 PM

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[itex]\displaystyle G \frac{M_m\, m}{{d_m}^2}=G \frac{M_e\, m}{(dd_m)^2}\,,\ [/itex] where M_{m} is the moon's mass, M_{e} is the earth's mass, m is the satellite's mass, d_{m} is the satellite's distance from the moon, and d is the monn's distance from earth. M_{e} = 81 M_{m}. Plugging this in & doing some manipulation gives: [itex]\displaystyle \frac{M_m}{{d_m}^2}=\frac{81M_m}{(dd_m)^2}\ [/itex] [itex]\displaystyle\quad\to\quad dd_m=d_m\sqrt{81}=9d_m\ [/itex] [itex]\displaystyle\quad\to\quad d_m=\frac{d}{10}\ [/itex] Well, yes, your answer is correct. 


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