Respuesta :
Answer: 117.626m/s
Explanation:
The escape velocity [tex]V_{esc}[/tex] is given by the following equation:
[tex]V_{esc}=\sqrt{\frac{2GM}{R}}[/tex] (1)
Where:
[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
[tex]M[/tex] is the mass of the asteroid
[tex]R[/tex] is the radius of the asteroid
On the other hand, we know the density of the asteroid is [tex]\rho=3.84(10)^{8}g/m^{3}[/tex] and its volume is [tex]V=2.17(10)^{12}m^{3}[/tex].
The density of a body is given by:
[tex]\rho=\frac{M}{V}[/tex] (2)
Finding [tex]M[/tex]:
[tex]M=\rhoV=(3.84(10)^{8} g/m^{3})(2.17(10)^{12}m^{3})[/tex] (3)
[tex]M=8.33(10)^{20}g=8.33(10)^{17}kg[/tex] (4) This is the mass of the spherical asteroid
In addition, we know the volume of a sphere is given by the following formula:
[tex]V=\frac{4}{3}\piR^{3}[/tex] (5)
Finding [tex]R[/tex]:
[tex]R=\sqrt[3]{\frac{3V}{4\pi}}[/tex] (6)
[tex]R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}[/tex] (7)
[tex]R=8031.38m[/tex] (8) This is the radius of the asteroid
Now we have all the necessary elements to calculate the escape velocity from (1):
[tex]V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}[/tex] (9)
Finally:
[tex]V_{esc}=117.626m/s[/tex] This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.