* Gravitational Field

  • Thread starter dekoi
  • Start date
  • #1
Compue the magnitude and direction of the gravitational field at a point P on the perpendicular bisector of the line joining two objects of equal mass seperated by a distance of 2a.
So:
[MASS] ---------------- 2a ------------------[MASS]
.......|
.......r
.......|
.......|
.......P
To calculate the resultant field, i assumed that the x direction of the resultant field is 0. Since the masses are equal, and therefore exert equal but opposite forces. Hence Resulant(X) = 0 N.
The y direction, according to my calculations, has a resultant field of [tex]\frac{2GM}{r^2}[/tex]. Since mass 1 exerts a force of [tex]\frac{GM}{r^2}[/tex] and so does mass 2.
Shouldn't the resultant gravitational field be the field in the y direction i calculated and mentioned above?
The correct answer is [tex]\frac{2GMr}{(a^2+r^2)^{3/2}}[/tex] towards the center of mass. HOW?
 

Answers and Replies

  • #2
lightgrav
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What were you going to use for "r" in your formula?
there are three interesting distances in this diagram:
1) a
2) p
3) sqrt(a^2 + p^2)
 
  • #3
For the 'y' resultant gravitational field, r is simply r as drawn in the diagram.
 
  • #4
lightgrav
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each source mass can be treated as if it exerted an independent Force on the "test mass" (at the field point). The "r" on the bottom of the g = GM/r^2 formula is the distance from the Source Mass to the field point (the place of interest).
The gravitational field contribution decreases with distance from the source, each source pulls independently.
(They're not smart enough to know where the other source masses are)
[You should NEVER use a special variable (like "r") as a label in a diagram]
I presume you've added Force vectors before, that were not co-linear?
here you have two more to add. Yes, the x-components cancel.
the "p/sqrt(a^2 + p^2)" is opp/hyp
 
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  • #5
I'm confused.
Say the diagram is like this:
[MASS] ---------------- 2a ------------------[MASS]
.......|
.......|
.......|
.......|
.......P

Where r is the hypotenuse.


So the horizontal forces cancel. And each vertical force should be equal to [tex]\frac{GMr}{r^2 - a^2}[/tex]. Shouldn't the resultant force be two times that? ([tex]\frac{2GMr}{r^2 - a^2}[/tex]
 
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  • #6
lightgrav
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each contribution to the gravity field is GM/r^2 .
Its "vertical" component is g sin(theta) = g p/r ,
with p^2 = (r^2 - a^2) being the offset from x-axis.

How did you get (r^2 - a^2) on the bottom?

An alternative form of contribution formula for a source mass is
g = GM (r_vector)/|r^3| , where r_vector is the vector (components) from the field point to the source mass.
 
Last edited:

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