Answer:
Explanation:
Be M1 and M2 the two massive objects and m the third one
The gravitational force over the third object at the beginning is zero (assuming that the position of the third object is x=0)
[tex]F_{m}=-G\frac{mM_{1}}{x_{1}^{2}}-G\frac{mM_{2}}{x_{2}^{2}}=0[/tex]
when the third object is displaced we have that the new position is x=Δx. Hence
[tex]F_{m}=-G\frac{mM_{1}}{(-x_{1}+\Delta x)^{2}}-G\frac{mM_{2}}{(x_{2}-\Delta x)^{2}}[/tex]
I attached a scheme of the system
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The gravitational force acting on the third object is given by Newton's Law
of universal gravitation.
Reasons:
The gravitational force, F, acting on the object is given by Newton's Law of
gravitation as follows;
[tex]\displaystyle F=G \cdot \frac{M_{1} \cdot m}{r^{2}} = G \cdot \frac{M_{2} \cdot m}{(x - r)^{2}}[/tex]
Where;
G = The universal gravitational constant
M₁, and M₂ = The mass of each of the two massive objects
m = The mass of the third object
r = The distance between the third object and the object of mass M₁
x = The distance between the two massive objects
Which gives;
[tex]\displaystyle \mathbf{\frac{M_{1} }{r^{2}}} = \frac{M_{2}}{(x - r)^{2}}[/tex]
Let the distance with which the object is displaced towards the mass M₁ = a, we have;
Force, F₁, on the third object from the object of mass M₁ after the displacement a is therefore;
[tex]\displaystyle F_1 = \mathbf{G \cdot \frac{M_{1} \cdot m }{(r-a)^{2}}}[/tex]
The force due to mass M₂ is; [tex]\displaystyle F_2 = G\cdot \frac{M_{2} \cdot m}{(x - (r-a))^{2}} = \mathbf{G \cdot \frac{M_{2} \cdot m}{(x - r+a)^{2}}}[/tex]
r > r - a, therefore;
F₁ > F
Similarly;
x - r < x + a - r, therefore;
F > F₂
Which gives;
F₁ > F₂
F₁ - F₂ acts in the direction of F₁
Therefore;
The total gravitational force, F = F₁ - F₂ acts in the direction of F₁, which is in the direction towards which the direction in which the third object is displaced.
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