• Support PF! Buy your school textbooks, materials and every day products Here!

Gravitation Force Question | Tricky!

  • Thread starter Raza
  • Start date
  • #1
203
0
Hi, I am taking Physics Grade 12 at home, so I get teacher's help for 2 hours once a week. They basically give me booklets to do at home and just simply hand it in. But the negative side to this that it's only 2 hours of help and teachers don't know most of the questions (they're new). And also the books are COMPLETE crap; there is little or no explanation behind the physic's equation and leave you to think about equation yourself. There's more questions that I don't get but here's the 1st question from the booklet.

Homework Statement


The mass of the Moon is 7.35 x 1022kg. At some point between Earth and the Moon, the force of Earth's gravitational attraction on an object is cancelled by the Moon's force of gravitational attraction. If the distance between Earth and the Moon (center to center) is 3.84 x 105km, calculate where this will occur, relative to Earth.


Homework Equations


[tex]F_{g}=\frac{Gm_{1}m_{2}}{r^2}[/tex]
MMoon=7.35 x 1022kg
MEarth=5.98 x 1024kg
Gconstant=6.67 x 10-11N x m^2/kg^2


The Attempt at a Solution


I think it must be:
[tex]\frac{F_{(G)moon}}{F_{(G)earth}}=1[/tex]
and you are trying to figure out r.
 
Last edited:

Answers and Replies

  • #2
13
0
Using your solution, you've got a wrong answer? r^2 = (3.84E5)^5?
 
  • #3
203
0
Well, is my solution right?
 
  • #4
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
How about trying;

[tex]\frac{GM_e}{d^2}-\frac{GM_m}{(r-d)^2} = 0[/tex]

Where d is the distance from the earth to the equilibrium position and r is the distance from earth to the moon. Does that make sense?
 

Related Threads for: Gravitation Force Question | Tricky!

  • Last Post
Replies
6
Views
2K
Replies
3
Views
1K
Replies
5
Views
979
Replies
15
Views
8K
  • Last Post
Replies
0
Views
964
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
9
Views
2K
Top