Answer:
F = 303,615 10⁻¹¹ N
Explanation:
Let's analyze this problem a little, problem we are asked to find the attractive force of the large sphere and a small sphere, we can find separately the attractive force between the full large sphere and the small sphere, Let's call this force F1 and on the other hand the force is between a sphere representing the gap and the small sphere, let's call this outside F2, the net bone force of the large sphere with gap is the subtraction of these two forces.
F = F₁ -F₂
Let's start by finding the force of the sphere of complete
F₁ = G M₁ m / r²
the masses of the sphere is M = 388 kg and the distance is r = d = 14m
F₁ = 6.67 10⁻¹¹ 380 27/14²
F₁ = 356.50 10⁻¹¹ N
Now let's calculate the mass of the gap if large sphere
let's use the concept of density
ρ = m / V
the volume of the gap is
V = 4/3 π r³
For the radius of the hole they tell us that it touches one side and the center of the sphere, therefore its diameter is the radius of the large sphere
d = 2.3 m
r_hole = 1.15 m
V_hole = 4/3 π r³
V_hole = 4/3 π 1.15³
V_hole = 6.37 m³
let's look at the density of the large sphere
ρ = m / V
V = 4/3 π r³
V = 4/3 π 2.3³
V = 50,965 m³
ρ = 388 / 50,965
ρ = 7.613 kg / m³
this is the density of the sphere without the gap
the mass of the gap is
m_hole = ρ V
m_hole = 7,613 6.37
m_hole = 48.49 kg
the distance from the hole to the small building
r₂ = d - r_hueco
r₂ = 14 - 1.15
r₂ = 12.85 m
the strength of this is
F₂ = 6.67 10⁻¹¹ 48.49 27 / 12.85²
F₂ = 52,885 10⁻¹¹ N
The strength of the sphere with the gap is
F = F₁ -F₂
F = (356.50 - 52,885) 10⁻¹¹
F = 303,615 10⁻¹¹ N
