Calculating Minimum Asteroid Size for Astronaut Safety

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To determine the minimum asteroid size for astronaut safety, the average density of the asteroid is 5000 kg/m^3, and the astronaut's mass is 91 kg. The astronaut can jump 0.6 m on Earth, which provides a basis for calculating the necessary escape velocity on the asteroid. The escape velocity formula, v = sqrt(2GM/R), can be used to relate the astronaut's jump to the asteroid's size. The key is to find the asteroid's mass and radius to ensure the astronaut can safely jump without floating away into space. Understanding these calculations will help establish the minimum diameter required for safety.
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Sorry guys, one more question for now. It is the only other one I couldn't really get on this homework assignment.

Homework Statement


An astronaut is about to explore an asteroid of average density \rho=5000 kg/m^3. He is worried that he may accidentally jump from its surface and float off into space. How big (what diameter) must it be before he can neglect this possibility? The astronaut knows that his mass, m, including space suit, is 91 kg. He also knows that on the Earth he can raise his center of gravity 0.6 m by jumping with his space suit on. It seems reasonable to assume that the maximum energy output of his legs will be the same on the asteroid as on the earth. For simplicity assume M>>m, such that M does not move significantly during the action, and then check this assumption [How much does the asteroid move relative to the astronaut?].


Homework Equations


\rho = m/V

V = 4/3\pir^3


The Attempt at a Solution


I figure if I can get the mass of the asteroid I can calculate the radius, and hence, the volume. I am not sure what clue the 0.6 m jumping on Earth is supposed to give me. Also, why is m of the astronaut given if we can neglect it, since M>>m?
 
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oops, I meant to say "calculate the radius, and hence the diameter". Also, I am not sure now why I included V, as it doesn't say anything about the asteroid being spherical.
 
Regarding the 0.6 m jump on the Earth: What does this translate to in terms of velocity? Do you know what escape velocity is?
 
I'm not sure...I know escape velocity is \sqrt{2GM}/R
 
The astronaut changes his velocity from 0 to some value to enable him to jump 0.6 m into the air on the Earth. He will get this same "delta-V" on the asteroid. (That's one of the givens.) So relate that to escape velocity.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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