What is the Distance from Earth to the Moon where Gravitational Pull is Equal?

[UrbanXrisis]

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?)

 

[whozum]

A point between the two? Just use the same method you used in the other post. It will work for this example.

 

[UrbanXrisis]

I tried, I got a nonreal number

http://home.earthlink.net/~urban-xrisis/clip_image002.jpg

 

[whozum]

384,400,000meters between Earth and moon

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

F_e = F_m

F = G\frac{Mm}{r^2}

\frac{5.98x10^{24}}{x^2} = \frac{7.35x10^{22}}{y^2}

x + y = 3.84 x 10^8

You don't 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.

 

[Werg22]

Does this distance takes account of the radius of both?

 

[jdavel]

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?

 

[UrbanXrisis]

I got x to equal something between 3 and 4 when I graphed the function...

http://home.earthlink.net/~urban-xrisis/clip_image002.jpg

 

[whozum]

\frac{5.98x10^{24}}{x^2} = \frac{7.35x10^{22}}{y^2}

M_ey^2 = M_mx^2

x+y = r

M_e(r-x)^2 = M_mx^2

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

\sqrt{\frac{M_m}{M_e}}x = r-x

\sqrt{\frac{M_m}{M_e}}x+x = r

(\sqrt{\frac{M_m}{M_e}}+1)x = r

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

r = 3.84 x 10^8m

 

[whozum]

A much easier way would be to realize that:

a = \frac{GM}{r^2}

and plot a for Earth and a for the moon, and find the intersect point.