Three uniform spheres are located at the corners of an equilateral triangle. Each side of the triangle has a length of 1.17 m. Two of the spheres have a mass of 3.97 kg each. The third sphere (mass unknown) is released from rest. Considering only the gravitational forces that the spheres exert on each other, what is the magnitude of the initial acceleration of the third sphere?

Resultant acceleration on the third sphere is

(3.352 × 10⁻¹⁰) m/s²

Explanation:

Newton's law states that the gravitational attractive force between two masses

m and M at some distance, d, apart.

F = (GMm/d²)

The diagram of the three masses placed at the edges of an equilateral triangle is presented in the attached image.

The resultant force, Fᵣ = F₁₃ cos 30° + F₂₃ cos 30°

This is only a sum of vertical components of the forces, as the horizontal components of the forces cancel out.

Note that since m₁ = m₂ = m = 3.97 kg,

F₁₃ = F₂₃ = F

Let the mass of the third sphere = M

F = (GMm/d²)

Recall that

The resultant force, Fᵣ = F₁₃ cos 30° + F₂₃ cos 30°

Fᵣ = F cos 30° + F cos 30° = 2F cos 30° = 1.732 F

Net force on the third sphere, Fᵣ = Ma

Fᵣ = 1.732 F = Ma

Recall that F = (GMm/d²)

Ma = 1.732 F = (1.732GMm/d²)

Ma = (1.732GMm/d²)

a = (1.732Gm/d²)

G = (6.674 × 10⁻¹¹) Nm²/kg²

m = 3.97 kg

d = 1.17 m

a = (1.732 × 6.674 × 10⁻¹¹ × 3.97)/(1.17²)

a = (3.352 × 10⁻¹⁰) m/s²

Hope this Helps!!!

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.

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