Two body escape velocity

In summary: For large masses, the barycentric escape velocity is not independent of the mass of the escaping object.
  • #1
dean barry
311
23
After calculating the gravitational PE using :
PE = ( G * m1 * m2 ) / d

Then i split the result into KE between the two bodies according to the ratio of the masses, then calculated the individual velocities from those (based on KE = ½ * m * v ²)

Any comments ?
 
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  • #2
How did you split the KE between the two bodies?
 
  • #3
According to my calculation, using your method, v1=v2 regardless of the mass ratio. But I may be doing something wrong.

Keep in mind that according to the conservation of momentum and the third law of motion:
m1v1 = m2v2
must be true.

nasu said:
How did you split the KE between the two bodies?

Assuming a ratio of 2 to 1, I did it like this: x + 2x = KE
So if KE=25, then KE1=8.333 and KE2=16.666.

I suck at math, so I have little confidence in my analogy. :)
 
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  • #4
[STRIKE]Your second part contradict your first part.
Why don't you use the conservation of momentum result?

If the mass ratio is 1:2, the KE ratio will be 1:4.[/STRIKE]
 
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  • #5
nasu said:
Your second part contradict your first part.
Why don't you use the conservation of momentum result?

If the mass ratio is 1:2, the KE ratio will be 1:4.
Yes, you're right. I'll try to fix that. Thanks.
 
  • #6
TurtleMeister said:
Yes, you're right. I'll try to fix that. Thanks.

No, I was wrong. Sorry for that. The square of the speeds are in the ratio 1:4 but not the KE. You were right the first time.
 
  • #7
nasu said:
No, I was wrong. Sorry for that. The square of the speeds are in the ratio 1:4 but not the KE. You were right the first time.

Ok, thanks. I'll go back and edit my post again. I've already reported this thread, so they may delete some of it.
 
  • #8
The split of the KE between the two bodies, is based on the mass ratio :
So :
KE (m1) = ( m2 / ( m1 + m2 ) ) * PE
KE (m2) = ( m1 / ( m1 + m2 ) ) * PE

Then the individual velocities from :
v ( m1 ) = sqrt ( ( KE ( m1 ) ) / ( ½ * m1 ) )
v ( m2 ) = sqrt ( ( KE ( m2 ) ) / ( ½ * m1 ) )

Ive ran this through as an example and the equal momentum is preserved.
Comments ?
 
  • #9
Basically, in asking if the total KE of both objects is equal to the original PE, this inquiry is based on the widely used statement that the mass of m2 is irrelavent in the caculation of escape velocity.
 
  • #10
Well, the gravitational potential energy is negative if it is zero at infinite separation, how do you split it into kinetic energy of two bodies?

ehild
 
  • #11
dean barry said:
Ive ran this through as an example and the equal momentum is preserved.
Comments ?

Yes, it is. I had the masses reversed in my original calculation.
 
  • #12
dean barry said:
The split of the KE between the two bodies, is based on the mass ratio :
So :
KE (m1) = ( m2 / ( m1 + m2 ) ) * PE
KE (m2) = ( m1 / ( m1 + m2 ) ) * PE

Then the individual velocities from :
v ( m1 ) = sqrt ( ( KE ( m1 ) ) / ( ½ * m1 ) )
v ( m2 ) = sqrt ( ( KE ( m2 ) ) / ( ½ * m1 ) )

Ive ran this through as an example and the equal momentum is preserved.
Comments ?
Apart from the sign error (PE is negative, KE is positive), it is correct.
You can also see that m1>>m2 means all the energy is used for m2 and v1 is negligible, in agreement with the usual definition of an escape velocity.
 
  • #13
Thanks for your input everyone, here's some more : I read on wikipedia that " the barycentric escape velocity is independent of the mass of the escaping object "
However, according to my previous thinking, the mass of the escaping object does alter the barycentric escape velocity, though at small values its virtually unnoticable, given a large enough value it does count.
Any comments ?
 
  • #14
dean barry said:
I read on wikipedia that " the barycentric escape velocity is independent of the mass of the escaping object "
That is an approximation (and an extremely good one!) for small masses.
 

1. What is the definition of "two body escape velocity"?

Two body escape velocity is the minimum speed required for an object to break free from the gravitational pull of another object and escape into an unbound orbit.

2. How is the two body escape velocity calculated?

The two body escape velocity is calculated using the equation: Ve = √(2GM/r), where Ve is the escape velocity, G is the gravitational constant, M is the mass of the object being escaped from, and r is the distance between the two objects.

3. Is the two body escape velocity the same for all objects?

No, the two body escape velocity varies depending on the mass and distance between the two objects. For example, the escape velocity for escaping the Earth's gravitational pull is different from escaping the Moon's gravitational pull.

4. How does the two body escape velocity differ from the three body escape velocity?

The two body escape velocity only takes into account the gravitational pull of two objects, while the three body escape velocity takes into account the gravitational pull of three objects. The calculation for the three body escape velocity is much more complex and depends on the mass and distance of all three objects.

5. Can the two body escape velocity be achieved by any object?

In theory, any object can achieve the two body escape velocity if it has enough speed. However, in practical terms, it may not be possible for all objects due to factors such as the object's mass, available propulsion, and other external forces acting on the object.

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