- #1

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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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- Thread starter dean barry
- Start date

- #1

- 311

- 23

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 ?

- #2

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How did you split the KE between the two bodies?

- #3

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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.

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. :)

Keep in mind that according to the conservation of momentum and the third law of motion:

m1v1 = m2v2

must be true.

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

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[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]

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

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Yes, you're right. I'll try to fix that. Thanks.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.

- #6

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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

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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

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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

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- #10

ehild

Homework Helper

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ehild

- #11

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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

mfb

Mentor

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Apart from the sign error (PE is negative, KE is positive), it is correct.

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 ?

You can also see that m

- #13

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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

mfb

Mentor

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That is an approximation (and an extremely good one!) for small masses.I read on wikipedia that " the barycentric escape velocity is independent of the mass of the escaping object "

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