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Escape velocity of an iron asteroid

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Homework Statement
Jack can jump upwards a distance 1.4 meters when he is on the surface of the Earth. What is his initial velocity when he jumps?
Most of the time, however, Jack is in space. He is an asteroid miner: he looks for asteroids made out of useful metals. His job involves landing on asteroids and taking samples for analysis.

Jack knows that when he is on an asteroid, the local gravitational force can be very small. He must be careful -- if he stands up too fast or steps too hard, he could launch himself into space and never return.

One day, Joe lands on a small spherical asteroid of iron which has a radius of R=280 meters. What is the escape velocity from this body?

"Nope," says Joe, "it would be too dangerous to walk around on that asteroid. I might accidentally push myself away and never return."

What is the minimum radius a spherical asteroid of pure iron could have and still be "safe" for Jack? That is, how big must an iron asteroid be in order to prevent Jack from leaping off it by accident?
Homework Equations
F= -G(m1)(m2)/(radius)^2
G=(6.67*10^-11)
Homework Statement: Jack can jump upwards a distance 1.4 meters when he is on the surface of the Earth. What is his initial velocity when he jumps?
Most of the time, however, Jack is in space. He is an asteroid miner: he looks for asteroids made out of useful metals. His job involves landing on asteroids and taking samples for analysis.

Jack knows that when he is on an asteroid, the local gravitational force can be very small. He must be careful -- if he stands up too fast or steps too hard, he could launch himself into space and never return.

One day, Joe lands on a small spherical asteroid of iron which has a radius of R=280 meters. What is the escape velocity from this body?

"Nope," says Joe, "it would be too dangerous to walk around on that asteroid. I might accidentally push myself away and never return."

What is the minimum radius a spherical asteroid of pure iron could have and still be "safe" for Jack? That is, how big must an iron asteroid be in order to prevent Jack from leaping off it by accident?
Homework Equations: F= -G(m1)(m2)/(radius)^2
G=(6.67*10^-11)

0^2 = initial velocity^2 + 2(-9.8m/s^2)(1.4m)
initial velocity =5.23m/s

Since ∫F= -G(m1)(m2)/(radius)^2 = G(m1)(m2)/radius
KE=(1/2)(m1)(velocity^2) = G(m1)(m2)/radius
m2=717.228*10^9 kg since it is a iron sphere

thus: ( (1/2) (velocity^2) = (6.67*10^-11) (717.228*10^9 kg) /250m
escape velocity=0.584m/s

For part 3, do I take 5.23m/s to substitute for velocity and solve for radius: (1/2)(m1)(velocity^2)=(m1)(m2)/radius
 
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For part 3, do I take 5.23m/s to substitute for velocity and solve for radius: (1/2)(m1)(velocity^2)=(m1)(m2)/radius
Yes. Of course the mass depends on radius as well.
 
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since (1/2)(m1)(velocity^2)=G(m1)(m2)/radius
m2=(4/3)(pi)(radius^3)(7.874 g/cm^3)
thus: (1/2)(5.23m/s)^2=(6.67*10^-11)( (4/3)(pi)(radius^3)(7.874 g/cm^3) ) /(radius)
radius= 32.2 *10^3
Is this correct?
 
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correction: radius=78.8 * 10^3
Is the units correct in terms of meter?
 

Filip Larsen

Gold Member
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Can you think of a way to check that your result is correct? Earlier you calculated the escape velocity for a small sphere of known radius, so perhaps you can do the same for this new sphere radius and compare that value to something?

By the way, for the first part of the problem you write 250m in your work, but the description mentions 280m.
 
1,901
201
since (1/2)(m1)(velocity^2)=G(m1)(m2)/radius
m2=(4/3)(pi)(radius^3)(7.874 g/cm^3)
thus: (1/2)(5.23m/s)^2=(6.67*10^-11)( (4/3)(pi)(radius^3)(7.874 g/cm^3) ) /(radius)
radius= 32.2 *10^3
Is this correct?
No. The calculation should give the right result if you express the density of iron in kg/m^3. You must have mad e some other error as well. The larger number you give in post #4 is incorrect as well.
The calculation is easier if you compute a formula of how the escape speed depends on the radius first.
 

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