The figure shows a spherical hollow inside a lead sphere of radius R = 2.3 m; the surface of the hollow passes through the center of the sphere and “touches” the right side of the sphere. The mass of the sphere before hollowing was M = 388 kg. With what gravitational force does the hollowed-out lead sphere attract a small sphere of mass m = 27 kg that lies at a distance d = 14 m from the center of the lead sphere, on the straight line connecting the centers of the spheres and of the hollow?

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