Question

Suppose the moon were held in its orbit not by gravity but by a massless cable attached to the center of the earth. What would be the tension in the cable? Use the table of astronomical data inside the back cover of the book

Short answer

The tension in the cable is$F_c=2.01\times {10}^{20}\mathrm{N}$.

Step-by-step solution

Given Information

From the table on the back cover of book Astronomical Data we can find that the value of the moon we needed.

$m_{\text{moon}}=7.36\times {10}^{22}\mathrm{kg}$$d_{\text{from earth}}=r=3.84\times {10}^8\mathrm{km}$$\text{period}=t=27.3\text{days}=2,358,720\mathrm{s}$

Explanation

$F_c=\frac{m{v}^2}{r}$$F_c=\frac{m(r\omega {)}^2}{r}$$F_c=mr{\omega }^2$$F_c=mr{\left(\frac{2\pi }{t}\right)}^2$$F_c=\frac{4{\pi }^{2}mr}{t^2}$

$F_c=\frac{4{\pi }^2\left(7.36\times {10}^{22}\mathrm{kg}\right)\left(3.84\times {10}^8\mathrm{km}\right)}{(2,358,720\mathrm{s}{)}^2}$

${\mathit{F}}_c=\mathbf{2}\mathbf{.}\mathbf{01}\mathbf{\times }\mathbf{10}\mathrm{N}$

Final Answer

The tension in the cable is${\mathit{F}}_c=\mathbf{2}\mathbf{.}\mathbf{01}\mathbf{\times }\mathbf{10}\mathrm{N}$.