Answer:
[tex]1.98\cdot 10^{20} N[/tex]
Explanation:
The tension in the cable would be then equal to the gravitational force that keeps the Moon in circular orbit around the Earth. So, we just need to calculate the magnitude of this force, which is given by:
[tex]F=G\frac{mM}{r^2}[/tex]
where:
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}[/tex] is the gravitational constant
[tex]m=7.35\cdot 10^{22} kg[/tex] is the mass of the Moon
[tex]M=5.97\cdot 10^{24} kg[/tex] is the mass of the Earth
[tex]r=3.84\cdot 10^8 m[/tex] is the distance between the Earth and the Moon
Substituting these numbers into the formula, we find:
[tex]F=(6.67\cdot 10^{-11} ) \frac{(7.35 \cdot 10^{22})(5.97\cdot 10^{24})}{(3.84\cdot 10^8)^2}=1.98\cdot 10^{20} N[/tex]
