a) [tex]1.96\cdot 10^{-16} m/s^2[/tex]
The gravitational field strength near the surface of the asteroid is given by:
[tex]g=\frac{GM}{(R+h)^2}[/tex]
where
G is the gravitational constant
M is the mass of the asteroid
R the radius of the asteroid
h is the distance from the surface
Substituting the data of the asteroid:
[tex]M=3.2\cdot 10^3 kg[/tex] is the mass
[tex]R=30 km = 30000 m[/tex] is the radius of the asteroid
[tex]h=3 km = 3000 m[/tex] is the distance from the surface
We find
[tex]g=\frac{(6.67\cdot 10^{-11})(3.2\cdot 10^3)}{(30000+3000)^2}=1.96\cdot 10^{-16} m/s^2[/tex]
b) i) [tex]5.53\cdot 10^9 s[/tex]
The acceleration of the astronaut popped out at 3 km from the surface is exactly that calculated at part a):
[tex]g=1.96\cdot 10^{-16} m/s^2[/tex]
So, since its motion is at constant acceleration, we can find the time he takes to reach the surface using suvat equations:
[tex]s=ut+\frac{1}{2}gt^2[/tex]
where
s = 3 km = 3000 m is his displacement to reach the surface
u = 0 is his initial velocity
t is the time
Solving for t,
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(3000)}{1.96\cdot 10^{-16} m/s^2}}=5.53\cdot 10^9 s[/tex]
b) ii) [tex]1.08\cdot 10^{-6} m/s[/tex]
Again, we can use another suvat equation:
[tex]v=u+gt[/tex]
where
v is the final velocity
u is the initial velocity
g is the acceleration of gravity
t is the time
Since we have
u = 0
[tex]t=5.53\cdot 10^9 s[/tex]
[tex]g=1.96\cdot 10^{-16} m/s^2[/tex]
The velocity of the astronaut at the surface will be
[tex]v=0+(1.96\cdot 10^{-16} m/s^2)(5.53\cdot 10^9)=1.08\cdot 10^{-6} m/s[/tex]
b) iii) 175 years
The duration of one year here is
[tex]T=3.16\cdot 10^7 s[/tex]
And the time it takes for the astronaut to reach the surface of the asteroid is
[tex]t=5.53\cdot 10^9 s[/tex]
Therefore, to find the number of years, we just need to divide the total time by the duration of one year:
[tex]n=\frac{t}{T}=\frac{5.53\cdot 10^9 s}{3.16\cdot 10^7}=175[/tex]
So, the astronaut will take 175 years to reach the surface.
