Posted by James on Tuesday, March 17, 2009 at 10:47pm.
Two meteoroids 250,000 km from Earth are moving at 2.1 km/s. Meteor one is headed in a straight path for Earth, while meteor two is on a path that will come within 8500 km from the Earth's center. (A) What is the speed of the first meteroid when it strikes Earth. (B) What is the speed of the second meteoroid at its closest approach to Earth. (c) Will the second meteroid ever return to Earth's vicinity?
I do not get this do you use something like v=sqr(GM/r)? I know part C is no, but how do you get part A and B. Any help would be greatly appreciated.

physics  Damon, Wednesday, March 18, 2009 at 9:24am
Yes, sort of
(1/2)m v^2 = original (1/2) m v^2 + loss of potential energy as the mass nears earth
That loss is
(G M m/r far away  G M m /r near)
for r far away use 2.5*10^8
for r near use radius of earth for the one that hits
for r near use radius of earth + 8.5*10^6 for the one that comes close

reverse sign physics  Damon, Wednesday, March 18, 2009 at 9:26am
(G M m/r far away  G M m /r near)
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