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- Thread starter wildkat7411
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bapowell

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No, I think all you need here is to apply conservation of energy,

[tex]\Delta KE = -\Delta PE[/tex],

where PE is the gravitational potential energy between the two objects.

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But since acceleration due to gravity increases as the distance between the two objects decreases, you cant use PE=massxgravityxheight since the acceleration is constantly changing for a given height. or does that not matter? I worked it out and i got that an object of mass one kilogram starting from rest at a distnce of 3.8x10^8 meters would be traveling at 1450 m/s at the point of impact, disregarding air resistance.

- #4

bapowell

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Right, you can't use [tex]PE = mgh[/tex]. Use Newton's Law of Gravitation.

- #5

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No you can't. you have to use

[tex]PE = -\frac{GMm}{r}[/tex]

Where r is the distance between the centers of m and M.

Just remember that [itex]\Delta PE[/itex] is the difference in PE between the start of the fall and the end of the fall.

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