anonymous
  • anonymous
The sun’s mass is about 2.7 x 107 times greater than the moon’s mass. The sun is about 400 times farther from Earth than the moon. How does the gravitational force exerted on Earth by the sun compare with the gravitational force exerted on Earth by the moon?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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BAdhi
  • BAdhi
for easiness take newtons gravitational equation as, \[F \propto \frac{M}{r^2}\] (since \(m\), \(G\) are common for both scenarios) for moon=> \[F_m\propto \frac{M_m}{r_m^2}\] for sun, \[F_s \propto \frac{M_s}{r_s^2} \] by dividing both, get the equation
anonymous
  • anonymous
im a little lost
BAdhi
  • BAdhi
well this is a bit of a easy manipulation method, but if its confusing i'll describe, The gravitational field equation is,\[F=\frac{GmM}{r^2}\] If the gravity force between the moon and earth is \(F_m\) and mass of moon is \(M_m\) and the distance between moon and earth is \(r_m\) and the mass of the earth is \(m\) the above equation becomes,\[F_m = \frac{GmM_m}{r_m^2}= (Gm) \frac{M_m}{r_m^2}\] For force between the sun and earth is \(F_s\) and mass of the sun is \(M_s\) and distance is \(r_s\) , equation becomes,\[F_s = \frac{GmM_s}{r_s^2} = (Gm) \frac{M_s}{r_s^2}\] Is it clear so far?

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anonymous
  • anonymous
yes
BAdhi
  • BAdhi
now what you need is to compare \(F_s\) with \(F_m\) or get a value for \(\displaystyle \frac{F_s}{F_m}\) For that what should you do?
anonymous
  • anonymous
divide F8/Fm
BAdhi
  • BAdhi
there you go.... any problems after that?
anonymous
  • anonymous
only curious what do you plug into it so you can divide
BAdhi
  • BAdhi
first divide and see what terms get eliminated by division... Then try to plugin the values given as data... Remember what is given is as ratios too... For example, they say sun is \(2.7\times 10^7 \) times larger than moon in mass. Simply what that mean is, \(M_s = 2.7\times 10^7 M_m\)
anonymous
  • anonymous
8/m?
BAdhi
  • BAdhi
whats that?
anonymous
  • anonymous
F8/Fm F cancels rights
BAdhi
  • BAdhi
you mean \(\displaystyle \frac{F_s}{F_m} = \frac{s}{m}\) ??
anonymous
  • anonymous
yah...is that wrong
BAdhi
  • BAdhi
oh dear... \(F_s\) and \(F_m\) are symbols as whole Just like \(A\) or \(B\) or any other symbols.. you jst cannot cut out parts of the symbol... simply what you are saying is like, \[\frac{V}{W} = \frac{1}{V}\] :P (by cutting out a apart of W with V since W is like two Vs)
anonymous
  • anonymous
my bad so what do i do than
BAdhi
  • BAdhi
If you are not comfortable with the notation, Jst replace \(F_m,F_s, M_m, M_s, r_m, r_s\) with normal english letters such as \(a,b,c,d,e,f\) etc. and try out.. And please try to show what you've tried out
anonymous
  • anonymous
no its not the symbols i just dont understand what im supposed to do to divide F8 by Fm
BAdhi
  • BAdhi
I have given you two equations for F_s and F_m. What you have to do is divide like, \(F_s = abc\), \(F_m = pqr\) \[\frac{F_s}{F_m} = \frac{abc}{pqr}\] Havent you done equation division before?
anonymous
  • anonymous
idk
BAdhi
  • BAdhi
ok here is a problem solve it for me.. find \(\frac{F}{T}\) \(F=ba^2\) \(T=bc\)
anonymous
  • anonymous
2b*c*a^2
BAdhi
  • BAdhi
I need the steps how you got there... not jst the answer..
anonymous
  • anonymous
idk how to explain it i just factored it out
anonymous
  • anonymous
i know if you divide a equation by an equation you actually combine and multiply the variables
anonymous
  • anonymous
there are 2 b(s) hence 2b theres 1 c hence times c and 1 a^2 hence times a^2
anonymous
  • anonymous
correct?
anonymous
  • anonymous
so f8/fm becomes 2f8m
anonymous
  • anonymous
\[2F _{8}m\]
BAdhi
  • BAdhi
well thats not how it works... I think you should work out more with your equations and symbols. before getting into this kinda applications.
anonymous
  • anonymous
its the last question i have than im done :(
BAdhi
  • BAdhi
\[\frac{F}{T} = \frac{ba^2}{bc}=\frac{a^2}{c}\] when the left hand side is divided same goes to the right hand side.. then if there are common symbols they jst cross out jst like what happen to the b here
BAdhi
  • BAdhi
Sorry I cannot jst give out the answer to this...
anonymous
  • anonymous
im not asking you to just help me understand what i need to do
anonymous
  • anonymous
ohhh ok i get it
BAdhi
  • BAdhi
So would you mind showing us how to get the answer..
anonymous
  • anonymous
yes but i just need to know what am i pluging in as the right hand side against F8/Fm
anonymous
  • anonymous
nvm i know
anonymous
  • anonymous
nvm i know what i need to plug in
anonymous
  • anonymous
\[\frac{ F _{8} }{ Fm }=\frac{ \frac{ (Gm)M _{m} }{ r _{m}^2} }{ \frac{ (Gm)M _{8} }{ r _{8}^2 } }=\frac{ \frac{ M _{m} }{ r _{m}^2 } }{ \frac{ M _{8} }{ r _{8}^2 } }\]
anonymous
  • anonymous
?
BAdhi
  • BAdhi
good
BAdhi
  • BAdhi
\[\frac{\left(\frac ab\right)}{\left( \frac c d\right)} = \frac a b \times\frac d c\] use this now
anonymous
  • anonymous
\[\frac{ M _{m} }{ r _{m}^2 }*\frac{ M _{8} }{ r _{8}^2 }=2M _{m8}r _{m8}^4\]
anonymous
  • anonymous
?
anonymous
  • anonymous
?
BAdhi
  • BAdhi
note that when the division is changed to multiplication, c/d is changed to d/c...so what you've written there is wrong also... as i have mentioned before, \(M_s\times M_m\) cannot be further simplified to \(2M_{ms}\) (u can, maybe with some other algebra :P). It should jst be \(M_sM_m\) and treat \(M_s\) as a whole which cannot be separated and same goes to \(M_m\)
anonymous
  • anonymous
ohh ok so how would I write it @Badhi
BAdhi
  • BAdhi
\[\frac{M_m}{r_m^2}\frac{r_s^2}{M_s}\] and then adjust it like this, \[\left(\frac{M_m}{M_s}\right)\left(\frac{r_s}{r_m}\right)^2\]

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