Answer:
[tex]v=516526.9m/s[/tex]
Explanation:
The force to which the object of mass m is attracted to a star of mass M while being at a distance r is:
[tex]F=\frac{GMm}{r^2}[/tex]
Where [tex]G=6.67\times10^{-11}Nm^2/Kg^2[/tex] is the gravitational constant.
Also, Newton's 2nd Law tells us that this object subject by that force will experiment an acceleration given by F=ma.
We have then:
[tex]ma=\frac{GMm}{r^2}[/tex]
Which means:
[tex]a=\frac{GM}{r^2}[/tex]
The object departs from rest ([tex]v_0=0m/s[/tex]) and travels a distance d, under an acceleration a, we can calculate its final velocity with the formula [tex]v^2=v_0^2+2ad[/tex], which for our case will be:
[tex]v^2=2ad=\frac{2GMd}{r^2}[/tex]
[tex]v=\sqrt{\frac{2GMd}{r^2}}[/tex]
We assume a constant on the vecinity of the surface because d=0.025m is nothing compared with [tex]r=5\times10^3m[/tex]. With our values then we have:
[tex]v=\sqrt{\frac{2GMd}{r^2}}=\sqrt{\frac{2(6.67\times10^{-11}Nm^2/Kg^2)(2\times10^{30}Kg)(0.025m)}{(5\times10^3m)^2}}=516526.9m/s[/tex]
