A sphere of mass 40 kg is attracted by another sphere of mass 80kg by

1.	A sphere of mass 40 kg is attracted by another sphere of mass 80kg by a force of2.5 x 10 ^-6 N when the centres are 3x10^-2 m apart find the value of G
Answer» GIVEN: Mass of the first sphere (m₁) = 40 Kg. Mass of the SECOND sphere (m₂) = 80 Kg. Force (F) = 2.5 X 10⁻⁶ N. Distance between the spheres (d) = 3x10⁻² m. TO FIND: Value of G. SOLUTION: Use this formula: $\mapsto \sf{\green{F=G\dfrac{m_1 \times m_2}{d^2} }}$ Where: F is force. m₁ and m₂ are mass of the objects. d is distance. G is universal gravitation constant. Substitute. $\to \sf 2.5 \times 10^{-6} =G \times \dfrac{40 \times 80}{\left(3 \times 10^{-2}\right)^2}$ ▣ Use the radicle rule : (a×b)ⁿ=aⁿ×bⁿ to simplify the denominator. $\to \sf 2.5 \times 10^{-6} =G \times \dfrac{40 \times 80}{3^2 \times (10^{-2})^2}$ ▣ Use the radicle rule = (mᵃ)ᵇ=mᵃᵇ. Moreover, 3²=9 and 40×80 = 3200. $\to \sf 2.5 \times 10^{-6} =G \times \dfrac{3200}{9 \times (10^{-2 \times 2})}$ $\to \sf 2.5 \times 10^{-6} =G \times \dfrac{3200}{9 \times (10^{-4})}$ ▣ Now transpose. $\to \sf 2.5 \times 10^{-6} \times \dfrac{9 \times (10^{-4})}{3200} =G$ $\to \sf \dfrac{22.5 \times 10^{-6} \times (10^{-4})}{3200} =G$ $\to \sf \dfrac{22.5 \times 10^{-10} }{3200} =G$ $\to \sf \dfrac{225 \times 10^{-10} }{32\times 10^3} =G$ $\sf \to 7.03125 \times 10^{-13}N \cdot m^2/kg^2=G$ G is 7.03×10⁻¹³ N·m²/kg².