Question

Three very small spheres of mass 2.50 kg, 5.00 kg, and 8.00 kg are located on a straight line in space away from everything else. The first one is at a point between the other two, 8.0 cm to the right of the second and 20.0 cm to the left of the third. Compute the net gravitational force it exerts.

Answer 1

Answer:

The net gravitational force it exerts is $F_{net}=9.66*10^{-8}N$

Explanation:

Newton's Law of Gravitation can be written as

$F=\frac{Gm_{1}m_{2}}{r^{2} }$

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

$F_{21} =\frac{Gm_{1}m_{2}}{x_{21}^{2} }$ is the force that 2 feels because of 1, and

$F_{31} =\frac{Gm_{1}m_{3}}{x_{31}^{2} }$ is the force that 3 feels because of 1.

If we replace the data in those previous equations, we have that

$F_{21} =\frac{G(2.5)(5) }{(0.08)^{2} }=1.3*10^{-7}N$

$F_{31} =\frac{G(2.5)(8) }{(0.2)^{2} }=-3.34*10^{-8}N$

Finally, adding both results, the net force the sphere 1 exerts is

$F_{net}=9.66*10^{-8}N$