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The centres of the two identical spheres are $ 1.0m $ apart, if the gravitational force between the spheres be $ 1.0N $ , then what is the mass of each sphere? $ (G = 6.67 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}})$.

Answer
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Hint: In this question we need to find out the mass of each sphere. Therefore, to solve this question we will use Newton’s universal law of gravitation where the equation will represent the force $ (F) $ between the two masses $ (m) $ that are separated by a distance $ (r) $ and the equation is given by $ F = \dfrac{{Gmm}}{{{r^2}}} $ where $ G $ is the gravitational constant.

Complete step by step solution:
Let $ m $ be the mass of the each sphere,
According to the question, we are given that radius $ r = 1.0m $ and the gravitational force between the spheres is $ F = 1.0N $ ,
Now using the Newton’s law of gravitation the equation we have,
 $ F = \dfrac{{Gmm}}{{{r^2}}} $ ,
 $ (F) $ Is the gravitational force, $ (m) $ is the mass of the object, $ (r) $ is the distance separation between the objects and where $ G $ is the gravitational constant and the value of $ (G = 6.67 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}) $
 $ \Rightarrow m = \sqrt {\dfrac{{F{r^2}}}{G}} $
 $ \Rightarrow \sqrt {\dfrac{1}{{6.67 \times {{10}^{ - 11}}}}} = 0.378 \times {10^{11}}Kg $ .
Hence, the mass of each sphere will be $ 0.378 \times {10^{11}}Kg $ .

Note:
In this type of question where two objects are attracted with a gravitational force with some distance given, we will always go for the concept of Newton’s law of gravitation. We must note that in the formula of Newton’s law of gravitation $ F = \dfrac{{Gmm}}{{{r^2}}} $ , $ (F) $ is the gravitational force, $ (m) $ is the mass of the object, $ (r) $ is the distance separation between the objects and $ G $ is the proportionality symbol known as gravitational constant, also in this equation force is proportional to mass and distance which is related by $ G $ .We must remember the value of $ G $ which is $ (G = 6.67 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}) $ .