Answer:
The radius minimum that not exceed the 6.83 times free-fall accelerations
r=152.403m
Explanation:
The centripetal acceleration is given by.
[tex]a_{c}=\frac{V^{2} }{r}[/tex]
Where
[tex]V=101 \frac{m}{s}\\a_{c}=6.83*g\\g=9.8\frac{m}{s^{2}} \\[/tex]
[tex]a_{c}=66.934 \frac{m}{s^{2}}[/tex]
Obtaining the expression for the radius.
[tex]a_{c}=\frac{V^{2} }{r}\\r=\frac{V^{2} }{a_{c}}[/tex]
[tex]r=\frac{(101\frac{m}{s})^{2}}{66.934\frac{m}{s^{2}}}[/tex]
[tex]r=\frac{10201 \frac{m^{2} }{s^{2}}}{66.934\frac{m}{s^{2}}}[/tex]
[tex]r=152.403m[/tex]
This question involves the concepts of centripetal acceleration and tangential speed.
The minimum radius of the planet's path should be "152.25 m".
Centripetal acceleration is the acceleration of a body due to its circular motion.The centripetal acceleration of the planet can be given by the following formula:
[tex]a_c=\frac{v^2}{r}\\\\[/tex]
where,
Therefore,
[tex]67\ m/s^2=\frac{(101\ m/s)^2}{r}\\\\r=\frac{(101\ m/s)^2}{67\ m/s^2}\\\\[/tex]
r = 152.25 m
Learn more about centripetal acceleration here:
https://brainly.com/question/17689540
