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Sagot :
Answer:
Following are the solution to the given question:
Explanation:
For crashing speed, we can use energy conservation:
kinetic energy [tex]= \frac{1}{2}\times m \times v^2[/tex]
potential energy [tex]= -\frac{GMm}{r}[/tex]
moon mass[tex]= 7.36\times 10^{22} \ kg[/tex]
Radius[tex]= 1738\ km[/tex]
[tex]\to (K + U) \ orbit = (K + U)\ crash\\\\\to \frac{1}{2}\times m \times v_o^2 - \frac{GMm}{(1738000 + 110000)} \\\\ \to \frac{1}{2}\times m \times vc^2 - \frac{GMm}{1738000}[/tex]
Calculating the mass drop for the leave:
[tex]\to \frac{vo^2}{2} - \frac{GM}{1848000}\\\\ \to \frac{vc^2}{2} - \frac{GM}{1738000}[/tex]
Solve the value for
[tex]vc = \sqrt{(vo^2 +2\times GM \times(\frac{1}{1738000} - \frac{1}{1848000}))}\\\\vc = \sqrt{(1600^2 +2\times 6.67\times 10^{-11} \times 7.36 \times 10^{22}\times (\frac{1}{1738000} - \frac{1}{1848000}))}\\\\vc = 1701 \ \frac{m}{s}\\\\[/tex]
The approach is correct but misrepresented in replacing 180 km instead of 110 km.
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