Question

a sphere of mass 40 kg is attracted by another sphere of mass 15 kg wit a force of 1/10 mg wt. find the value of gravity if centers of spheres are 20 cm apart

Answer 1

Answer on Question #55871-Physics-Astronomy-Astrophysics

A sphere of mass $40\mathrm{kg}$ is attracted by another sphere of mass $15\mathrm{kg}$ with a force of $1/10\mathrm{mg}$ wt. find the value of constant of gravity if centers of spheres are $20\mathrm{cm}$ apart

Solution

$$m _ {1} = 4 0 k g; m _ {2} = 1 5 k g; r = 0. 2 m.$$ $$F = \frac {1}{1 0} m g w t = \frac {1}{1 0} \cdot 1 0 ^ {- 3} g w t = \frac {1}{1 0} \cdot 1 0 ^ {- 6} k g w t = 1 0 ^ {- 7} k g w t = 9. 8 \cdot 1 0 ^ {- 7} N.$$

As

$$F = G \frac {m _ {1} m _ {2}}{r ^ {2}}$$ $$G = \frac {F r ^ {2}}{m _ {1} m _ {2}} = \frac {9 . 8 \cdot 1 0 ^ {- 7} (0 . 2) ^ {2}}{4 0 \cdot 1 5} \frac {N m ^ {2}}{k g ^ {2}} = 6. 5 3 3 \cdot 1 0 ^ {- 1 1} \frac {N m ^ {2}}{k g ^ {2}}$$

Answer: $6.533 \cdot 10^{-11} \frac{Nm^2}{kg^2}$ .

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