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

  • Thread starter Thread starter UrbanXrisis
  • Start date Start date
  • Tags Tags
    Gravitational Pull
Click For Summary

Homework Help Overview

The discussion revolves around determining the distance from Earth to the Moon where the gravitational pull from both bodies is equal. Participants are exploring gravitational forces and their calculations in the context of celestial mechanics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss methods for calculating the distance, including setting gravitational forces equal and using point mass approximations. Questions arise about the inclusion of the radii of the Earth and Moon in calculations, as well as the implications of nonreal numbers in solutions.

Discussion Status

The discussion is active with various approaches being explored. Some participants have provided guidance on methods to solve the problem, while others are questioning assumptions and the setup of equations. There is no explicit consensus on a single method or solution yet.

Contextual Notes

Participants are navigating constraints such as the need for accurate definitions of distance and gravitational force, as well as the potential impact of including or excluding the radii of the celestial bodies in their calculations.

UrbanXrisis
Messages
1,192
Reaction score
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:
Physics news on Phys.org
A point between the two? Just use the same method you used in the other post. It will work for this example.
 
I tried, I got a nonreal number

http://home.earthlink.net/~urban-xrisis/clip_image002.jpg
 
Last edited by a moderator:
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 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.
 
Does this distance takes account of the radius of both?
 
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?
 
I got x to equal something between 3 and 4 when I graphed the function...

http://home.earthlink.net/~urban-xrisis/clip_image002.jpg
 
Last edited by a moderator:
[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]
 
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.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
3K
Replies
21
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
2
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 28 ·
Replies
28
Views
2K
Replies
1
Views
2K
Replies
5
Views
2K