Escape velocity from asteroid

In summary: 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.Can you please clarify what the question is?
  • #1
gagge78
2
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!
 
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  • #2
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
 
  • #3
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).
 
  • #4
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.
 
  • #5
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
 

1. What is escape velocity from an asteroid?

Escape velocity from an asteroid is the minimum speed required for an object to escape the gravitational pull of the asteroid and not fall back to its surface. It is a measure of the asteroid's gravitational strength and depends on its mass and size.

2. How is escape velocity from an asteroid calculated?

The escape velocity from an asteroid is calculated using the formula:
Ve = √(2GM/R)
where Ve is the escape velocity, G is the gravitational constant, M is the mass of the asteroid, and R is the distance from the center of the asteroid to the object.

3. Is escape velocity from an asteroid the same for all objects?

No, the escape velocity from an asteroid can vary depending on the object's mass and distance from the asteroid. It also depends on the object's velocity and direction of motion.

4. Can an object achieve escape velocity from an asteroid?

Yes, if an object is able to reach or exceed the escape velocity from an asteroid, it can escape the asteroid's gravitational pull and continue traveling in space.

5. Why is escape velocity from an asteroid important?

Escape velocity from an asteroid is important for understanding how objects behave in space and for planning space missions. It also helps determine the potential for resources and mining on asteroids.

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