Observe the given figure and find the the gravitational force between m1 and m2.

Answer:
The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N
Explanation:
The details of the given masses having gravitational attractive force between them are;
m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m
The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;
[tex]F =G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}[/tex]
Where;
F = The gravitational force between m₁ and m₂
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
r₂ = 0.1 m + 0.15 m = 0.25 m
Therefore, we have;
[tex]F = 6.67430 \times 10^{-11} \ N \cdot m^2/kg \times \dfrac{20 \ kg\times 50 \ kg}{(0.1 \ m+ 0.15 \ m)^{2}} \approx 1.06789 \times 10^{-6} \ N[/tex]
The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N