# Meteor problem

by strugglin-physics
Tags: meteor
 P: 48 A meteor has a speed of 83.0 m/s when 811 km above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in 3.31 m. What is its speed just before striking the sand? So here is the formula that I have but it doesn't give me the right answer. v=square root of (Vi^2 + 2G(Mass of the Earth)/(Radius of the Earth*1-1/3.31) I get 9337.6 m/s. Does anyone see where my mistake is?
 P: 119 Use the following expression for potential energy $$V = -GMm/r$$ In the $$r$$ above, don't forget to inculde earth's radius too. Use energy conservation for the initial point and the point just before striking. You don't need to do anything with the sand spacetime www.geocities.com/physics_all/index.html
 P: 16 With the above equation, you need to use it twice! You must find the difference in PE between the two distances from the Earth's center. THe change in PE equals the gain in KE. It occurs to me. Wouldn't it be neat to see a meteor that ignores air resistance?
P: 48

## Meteor problem

So you mean use the formula
V=GMm/R+r - GMm/R and that will give me the velocity just before it hits? What about the initial velocity? Does that go in there somewhere?
Mentor
P: 40,262
 Quote by strugglin-physics So you mean use the formula V=GMm/R+r - GMm/R and that will give me the velocity just before it hits? What about the initial velocity? Does that go in there somewhere?
As Vector Sum and spacetime have explained, mechanical energy is conserved. This means that the decrease in gravitational PE will equal the increase in kinetic energy.
 P: 16 [QUOTE=strugglin-physics]So you mean use the formula V=GMm/R+r - GMm/R and that will give me the velocity just before it hits? QUOTE] That forumula is correct for increase in KE. Remember KE = 1/2 mv^2. This formula ought to be imprinted on the inside of your eyelids.

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