Answer:
463.4 m/s
Explanation:
The escape velocity on the surface of a planet/asteroid is given by
[tex]v=\sqrt{\frac{2GM}{R}}[/tex] (1)
where
G is the gravitational constant
M is the mass of the planet/asteroid
R is the radius of the planet/asteroid
For the asteroid in this problem, we know
[tex]\rho=4.49\cdot 10^6 g/m^3[/tex] is the density
[tex]V=3.32\cdot 10^{12} m^3[/tex] is the volume
So we can find its mass:
[tex]M=\frac{\rho}{V}=(4.49\cdot 10^6 g/m^3)(3.32\cdot 10^{12}m^3)=1.49\cdot 10^{19} kg[/tex]
Also, the asteroid is approximately spherical, so its volume is given by
[tex]V=\frac{4}{3}\pi R^3[/tex]
where R is the radius. Solving the formula for R, we find its radius:
[tex]R=\sqrt[3]{\frac{3V}{4\pi}}=\sqrt[3]{\frac{3(3.32\cdot 10^{12}m^3)}{4\pi}}=9256 m[/tex]
So now we can use eq.(1) to find the escape velocity:
[tex]v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.49\cdot 10^{19}kg)}{9256 m}}=463.4 m/s[/tex]