Can an Astronaut Escape an Asteroid with a 1 Meter Jump?

AI Thread Summary
The discussion revolves around calculating the maximum radius of an asteroid from which an astronaut can escape with a 1-meter jump, assuming the asteroid has a density similar to Earth. Participants express confusion about the relationship between the jump height and escape velocity, noting that a 1-meter jump cannot equate to escape velocity, which allows for infinite height. The conversation highlights the need for clarity in the question's wording and the conditions under which the jump is analyzed. To solve the problem, it is suggested to derive the asteroid's radius in terms of Earth's radius and density, using the escape velocity formula. Overall, the thread emphasizes the complexities of applying energy conservation principles in this context.
gagge78
Messages
2
Reaction score
0

Homework Statement


We assume an asteroid with density similar to earth. An astronaut performs a spot jump of 1 meter. Which is the maximum radius of the asteroid in order to escape the astronaut from the asteroid?

Homework Equations


Vesc=√2GM/R=R√8ΠGρ/3


The Attempt at a Solution


Vesc,earth/Vesc,ast=Rearth/Rast→ Rast=Rearth Vesc,ast/Vesc,earth
So the values of Rearth and Vesc,earth are known i need one more equation in order to calculate Vesc,ast which will be the minimum velocity that is necessary to reach 1 meter above the ground. So i am stuck!
 
Physics news on Phys.org
I am having difficulty understanding how a person doing a jump of only 1 m. from the asteroid can possibly escape. Can you give us the exact wording of the question?

AM
 
Also for me is difficult. But as i can understand speaking for an imaginary asteroid we try to find the necessary radius for which the escape velocity is equal with the begging velocity that we need for a jump of 1 meter (according to the energy conservation: v0=√2gh).
 
I suspect the wording of the question is different than what is posted here. Are you translating from a language other than English?

A velocity that results in a jump of 1 meter height can never be equal to the escape velocity, which produces a jump of infinite height. This is a simple logical fact. Perhaps you mean the velocity that would produce a 1 m jump height under different conditions -- but we need to know what those conditions are.

p.s. Welcome to Physics Forums.
 
gagge78 said:
Also for me is difficult. But as i can understand speaking for an imaginary asteroid we try to find the necessary radius for which the escape velocity is equal with the begging velocity that we need for a jump of 1 meter (according to the energy conservation: v0=√2gh).
There appears to be language/translation problem here.

Is the 1m. the height of the jump of the person on the Earth surface? Do you want to know the radius of the asteroid such that the energy used in such a jump on the Earth would provide the escape energy from the asteroid?

If that is the question, write out the expression for r, the asteriod radius, in terms of the radius, R, and density, ρ, of the Earth (ρ = Mearth/4∏R^3/3; Masteroid = ρ4∏r^3/3). Then use that in the equation for escape velocity.

AM
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top