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Two body escape velocity

by dean barry
Tags: body, escape, velocity
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dean barry
#1
Jul4-14, 07:42 AM
P: 67
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 = * m * v )

Any comments ?
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nasu
#2
Jul4-14, 09:34 AM
P: 1,988
How did you split the KE between the two bodies?
TurtleMeister
#3
Jul4-14, 09:58 AM
TurtleMeister's Avatar
P: 759
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.

Quote Quote by nasu View Post
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. :)

nasu
#4
Jul4-14, 10:08 AM
P: 1,988
Two body escape velocity

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.
TurtleMeister
#5
Jul4-14, 10:15 AM
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P: 759
Quote Quote by nasu View Post
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.
nasu
#6
Jul4-14, 11:33 AM
P: 1,988
Quote Quote by TurtleMeister View Post
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.
TurtleMeister
#7
Jul4-14, 11:47 AM
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P: 759
Quote Quote by nasu View Post
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.
dean barry
#8
Jul5-14, 06:25 AM
P: 67
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 ) ) / ( * m1 ) )
v ( m2 ) = sqrt ( ( KE ( m2 ) ) / ( * m1 ) )

Ive ran this through as an example and the equal momentum is preserved.
Comments ?
dean barry
#9
Jul5-14, 06:53 AM
P: 67
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.
ehild
#10
Jul5-14, 07:00 AM
HW Helper
Thanks
P: 10,769
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
TurtleMeister
#11
Jul5-14, 09:09 AM
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P: 759
Quote Quote by dean barry View Post
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.
mfb
#12
Jul5-14, 10:03 AM
Mentor
P: 12,081
Quote Quote by dean barry View Post
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 ) ) / ( * m1 ) )
v ( m2 ) = sqrt ( ( KE ( m2 ) ) / ( * 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.
dean barry
#13
Jul7-14, 04:51 AM
P: 67
Thanks for your input everyone, heres 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 ?
mfb
#14
Jul8-14, 01:22 PM
Mentor
P: 12,081
Quote Quote by dean barry View Post
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.


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