Two concentric uniform thin spherical shells of masses 74 kg and 14 kg and radii 32 m and 89 m, respectively. what is the gravitational force on a particle of mass 130 kg located (a) a distance 11 m from the center of the shells: $f1= pn (b) a distance 68 m from the center of the shells: $f2= pn (c) a distance 103 m: $f3= pn

a) F= 0 N, b)  F = 138.8 pN , c)    F = 68.66 pN

Explanation:

The universal gravitation force is

         F = G m₁ m₂ / r²

When this equation is used, the mass of a spherical body can be considered at its center, in general when using Gaussian surfaces the force is produced by the mass inside the surface. This means that if the mass this force of the point of analysis does not produce force

With these arguments we will apply them to our case

a) r = 11  m

This distance is less than the radius of the two spherical shells, so the net effect of these distributions is zero

      F = 0 N

b) r = 68 m

The point is between the two shells, therefore the gravitational force is

      F = G m₁ m₂ / r²

      F = 6.67 10⁻¹¹ 130 74/68²

      F = 13.88 10⁻¹¹ N  

     F = 138.8 pN

c) r = 103 m

The point is outside the two spheres, so the two exert gravitational attraction to the body as if the entire mass were in its center

       F = G m₁m₂ / r² + G m₁ m₃ / r²

       F = G m₁ (m₂ + m₃) / r²

       F = 6.67 10⁻¹¹ 130 (14 * 74) / 103²

       F = 6.67 19⁻¹¹ 130 84/103²

       F = 6.8655 10⁻¹¹ N

       F = 68.66 pN

c because the action was the roller coaster reaches bottom of hill (action) reaction would be blood tends to drain away from head

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