Question

a sphere of mass 40 kg is attracted by another sphere of mass 80 kg by a force of 2.5 multiplyed 10 raise to power minus 6 and when their centres are 30 mm apart. find the value of g

Answer 1

We know that gravitational force between two bodies is given by -

F = G m1m2/r²
where m1,m2 are masses of the bodies
G is Universal gravitation constant.
r is the distance between two objects from their centers.

So here
F = 2.5 × 10^-6
r = 30mm or 30 × 10^-3 m
m1 = 40kg
m2 = 80kg
we have to find the value of G

so plug the values in the formula

we get
2.5×10^-6 = G × 80×40/3×10-²

so we get G = 0.0023.. × 10^-8

Answer 2

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) = 30mm = 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.