Equation Help: Gravitationally Bound Rubble Pile Asteroid Orbit

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In summary, the conversation discusses a gravitationally bound rubble pile asteroid and the question of what minimum initial velocity would be required for the surface material to achieve a circular orbit. It also touches on the concept of critical spin and escape velocity, and the relevance of angular momentum in this scenario.
  • #1
higginsdj
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Hi all,

I have a gravitationally bound rubble pile asteroid. This asteroid reaches and exceeds 'critical' spin where centripetal force exceeds gravitational force and surface material is launched off the surface.

At what minimum initial velocity would the surface material need to be to achieve a circular orbit? Just looking for the equation or a reference site.

Cheers

David
 
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  • #2
Hi David! :smile:

As soon as it exceeds critical spin, any loose rock (on the top of the highest "mountain", say) will go into circular orbit (hopefully just missing the mountain on its return!).

(In practice, it'll collide will other rocks, and either crash, or get thrown out into some random elliptical orbit, where it will continue to collide with other rocks and eventually they'll all settle down to some regular pattern.)
 
  • #3
I though this to, but then I might have confused myself. ie for my object, at Critical spin, the surface of the primary is traveling at 2.28 m/s. BUT escape velocity is 3.28 m/s. At the surface Fg = 0.00706 m/s^2. So when I spin fast enough for Fc = 0.00706, The surface object is traveling at 2.28 m/s and will effectively just be weightless on the surface where it is, higher velocities will actually launch it.

BUT, if the object is now disconnected from the parent, Fg will still be acting on it, the more distant the object gets from the parent the slower it's orbital velocity and thus like a projectile it should return to the surface so I am assuming this all has something to do with Angular Momentum BUT I am having trouble connecting all the dots.

Cheers

David
 
  • #4
Hi David! :smile:
higginsdj said:
I though this to, but then I might have confused myself. ie for my object, at Critical spin, the surface of the primary is traveling at 2.28 m/s. BUT escape velocity is 3.28 m/s.

Escape velocity is irrelevant … it's the velocity for reaching infinity, not for orbiting.
BUT, if the object is now disconnected from the parent, Fg will still be acting on it, the more distant the object gets from the parent the slower it's orbital velocity and thus like a projectile it should return to the surface so I am assuming this all has something to do with Angular Momentum BUT I am having trouble connecting all the dots.

A projectile doesn't have to return to the surface, it only has to keep falling.

As Isaac Newton pointed out, if you throw an apple hard enough horizontally, it will go into orbit … it keeps falling, but the Earth curves away beneath it fast enough to counter the falling.
 
  • #5



Hi David,

To determine the minimum initial velocity required for the surface material of a gravitationally bound rubble pile asteroid to achieve a circular orbit, we can use the equation for orbital velocity, which is given by v = √(GM/r), where G is the gravitational constant, M is the mass of the asteroid, and r is the radius of the orbit.

In this case, we can rearrange the equation to solve for the minimum initial velocity (v) by setting it equal to the critical spin velocity, which is when the centripetal force equals the gravitational force. This can be represented as v = √(GM/r) = √(2GM/r), where the factor of 2 accounts for the critical spin condition.

To find a more specific value, we would need to know the specific mass and radius of the asteroid in question. However, this equation can be used as a reference site for calculating the minimum initial velocity for any gravitationally bound rubble pile asteroid.

I hope this helps! Let me know if you have any further questions.

Best,
 

FAQ: Equation Help: Gravitationally Bound Rubble Pile Asteroid Orbit

What is a gravitationally bound rubble pile asteroid orbit?

A gravitationally bound rubble pile asteroid orbit refers to the orbit of a small, irregularly shaped asteroid that is held together by its own gravity rather than being a solid object. These types of asteroids can be found in the asteroid belt between Mars and Jupiter.

How is the orbit of a gravitationally bound rubble pile asteroid different from a regular asteroid?

A regular asteroid has a more defined, elliptical orbit due to its solid and compact structure. A gravitationally bound rubble pile asteroid, on the other hand, has a more irregular and unpredictable orbit due to its loose and fragmented composition.

What factors influence the orbit of a gravitationally bound rubble pile asteroid?

The orbit of a gravitationally bound rubble pile asteroid is primarily influenced by the gravitational pull of nearby planets, as well as the asteroid's own shape and composition. Other factors such as collisions with other objects and solar radiation can also affect its orbit.

Why is studying the orbit of a gravitationally bound rubble pile asteroid important?

Studying the orbit of a gravitationally bound rubble pile asteroid can provide valuable insights into the formation and evolution of small bodies in our solar system. It can also help us better understand the potential risks of these asteroids colliding with Earth.

How can equations help us understand the orbit of a gravitationally bound rubble pile asteroid?

Equations, such as the Newton's Law of Gravitation and Kepler's Laws of Planetary Motion, can help us calculate the trajectory and behavior of a gravitationally bound rubble pile asteroid in its orbit. They can also assist in predicting future movements and potential collisions with other objects in the solar system.

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