Question

A 5.0-m-diameter solid aluminum sphere is launched into space. By how much does its diameter increase? Give your answer in mm.

Short answer

2.4$\mu m$

Step-by-step solution

Given Info

Aluminum sphere

Radius: 5m

Initial Pressure: 1atm

Final pressure: 0

New Radius (radius in space): ?

Formula Used

$B=\frac{∆P}{{\displaystyle \frac{∆V}{V}}}$

Where, B = Bulks Modulus

P = Pressure

V = Volume

Substituting Values

Let new radius of the sphere(in space) be x.

$$\begin{gathered} \Rightarrow 7\times {10}^{10}=\frac{101325}{{\displaystyle \frac{{\displaystyle \frac{4}{3}}{\mathrm{\pi x}}^3-{\displaystyle \frac{4}{3}}\mathrm{\pi }{5}^3}{{\displaystyle \frac{4}{3}}{\mathrm{\pi x}}^3}}} \\ \Rightarrow 7\times {10}^{10}=\frac{101325x^3}{x^3-125} \\ \Rightarrow (7x^3\times {10}^{10})-(875\times {10}^{10})=101325x^3 \\ \Rightarrow (7x^3\times {10}^{10})-101325x^3=875\times {10}^{10} \\ \Rightarrow x^3=\frac{875\times {10}^{10}}{(7\times {10}^{10})-101325} \\ \Rightarrow x^3=125.0001809 \\ \Rightarrow x=5.0000024119 \end{gathered}$$

And therefore change in radius is x - 5 = 0.0000024m or 2.4$\mu m$