Answer:
The net gravitational force it exerts is [tex]F_{net}=9.66*10^{-8}N[/tex]
Explanation:
Newton's Law of Gravitation can be written as
[tex]F=\frac{Gm_{1}m_{2}}{r^{2} }[/tex]
where G is the Gravitational Constant, m1 and m2 are the masses of two objects, and r is the distance between them. In this case, the spheres are loacted in straight line, so instead of a vector r, we have a distance x in meters. The distances and masses are given in the problem, and the smaller sphere is between the other two spheres. This means the sphere 1 is in the middle, the sphere 2 is on the left of 1, and the sphere 3 is on the right of 1, so
[tex]F_{21} =\frac{Gm_{1}m_{2}}{x_{21}^{2} }[/tex] is the force that 2 feels because of 1, and
[tex]F_{31} =\frac{Gm_{1}m_{3}}{x_{31}^{2} }[/tex] is the force that 3 feels because of 1.
If we replace the data in those previous equations, we have that
[tex]F_{21} =\frac{G(2.5)(5) }{(0.08)^{2} }=1.3*10^{-7}N[/tex]
[tex]F_{31} =\frac{G(2.5)(8) }{(0.2)^{2} }=-3.34*10^{-8}N[/tex]
Finally, adding both results, the net force the sphere 1 exerts is
[tex]F_{net}=9.66*10^{-8}N[/tex]
