1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equal gravitational pull

  1. Apr 22, 2005 #1
    anyone know the distance from earth (to the moon) where the gravitational pull of the moon is equal to the gravitational pull of the earth? (or any way I could calculate it?) :confused:
     
    Last edited: Apr 22, 2005
  2. jcsd
  3. Apr 22, 2005 #2
    A point between the two? Just use the same method you used in the other post. It will work for this example.
     
  4. Apr 22, 2005 #3
    I tried, I got a nonreal number

    chk here
     
    Last edited: Apr 22, 2005
  5. Apr 22, 2005 #4
    384,400,000meters between earth and moon

    earth mass is 5.98 x 10^24
    moon mass is 7.35 x 10^22

    [tex] F_e = F_m [/tex]

    [tex] F = G\frac{Mm}{r^2} [/tex]

    [tex] \frac{5.98x10^{24}}{x^2} = \frac{7.35x10^{22}}{y^2} [/tex]

    [tex] x + y = 3.84 x 10^8 [/tex]

    You dont need to add the radius of the earth and moon on the fractions on the topright. You can regard them as point masses. Solving the system above should get you the correct answer.
     
  6. Apr 22, 2005 #5
    Does this distance takes account of the radius of both?
     
  7. Apr 22, 2005 #6
    UrbanXrisis,

    First of all, NEVER, EVER, EVER start plugging in numbers until you have solved the equation for the variable you need. It wastes time, and it creates countless chances for errors.

    Second, it looks to me as though you're using the radius of the earth and moon in the denominators of your law of gravity equations. Why? How is the distance r in the denominator defined?
     
  8. Apr 23, 2005 #7
    I got x to equal something between 3 and 4 when I graphed the function...

    chk here
     
  9. Apr 23, 2005 #8
    [tex] \frac{5.98x10^{24}}{x^2} = \frac{7.35x10^{22}}{y^2} [/tex]

    [tex] M_ey^2 = M_mx^2 [/tex]

    [tex] x+y = r [/tex]

    [tex] M_e(r-x)^2 = M_mx^2 [/tex]

    [tex] \frac{M_m}{M_e} = \left(\frac{r-x}{x}\right)^2 [/tex]

    [tex] \sqrt{\frac{M_m}{M_e}}x = r-x [/tex]

    [tex] \sqrt{\frac{M_m}{M_e}}x+x = r[/tex]

    [tex] (\sqrt{\frac{M_m}{M_e}}+1)x = r [/tex]

    [tex] \frac{r}{\left(\sqrt{\frac{M_m}{M_e}}+1\right)} = x [/tex]

    [tex]r = 3.84 x 10^8m[/tex]
     
  10. Apr 23, 2005 #9
    A much easier way would be to realize that:

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

    and plot a for earth and a for the moon, and find the intersect point.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Equal gravitational pull
Loading...