Answer to Question #248449 in Molecular Physics | Thermodynamics for Kelani

Question

Now the pair decide to solve a homework problem to check their understanding.

The densest known element is osmium with a density of 𝜌 = 22.4 g/cm³. Suppose two small spheres of osmium with radii 2.60 cm are in contact with each other. Find the gravitational force of attraction between them, given that G = 6.67 ✕ 10⁻¹¹ N · m²/kg².

Recall that 𝜌 = mass/volume

 and V_sphere = 4/3𝜋r³.

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Answer 1

We use the gravitation law and the definition for the mass of the two galium spheres (m₁ and m₂) given the volume and density, and we also substituted the distance between the spheres as 2R or two times the radius of each sphere since they are in contact to each other, this gives the following:

$F=G\cfrac{m_1m_2}{r^2}=G\cfrac{(\rho\cdot \frac{4}{3}\pi R^3) (\rho\cdot \frac{4}{3}\pi R^3)}{(2R)^2} \\ \text{after we rearrange, we find} \\F =\frac{4}{9}G \pi^2 \rho^2R^4=G \Big( \frac{2\pi \rho R^2 }{3} \Big)^2$

Now we can proceed to substitute and use the unit factors that are necessary to find the attraction force in Newtons:

$F=(6.67\times 10^{-11}\,N · m^2/kg^2) \Bigg( \cfrac{2\pi (22.4\frac{g}{cm^3}) (2.60\,cm)^2\cdot \frac{100\,cm}{1\,m} }{3(1000\,g/kg)} \Bigg)^2 \\ \therefore F=6.7086\times10^{-11}\,N$

In conclusion, the gravitational force of attraction between the two gallium spheres is 6.71 X 10^-11 N.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.