Physics
posted by James on .
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.

The sum of potential and kinetic energy is zero. Potential energy per mass (if zero at infinite distance) is GM/R. Use the fact that the earth's radius is 6378 km. M is the mass of the Earth.
For meteor (1),
GM/(250,000*10^3)+ (2100)^/2 =
GM/6378*10^3)+ Vimpact^2/2
For meteor 2,
GM/(250,000*10^3)+ (2100)^/2 =
GM/8500*10^3)+ Vmax^2/2
Look up GM and solve.
If Vmax exceeds the escape velocity, it will not return, if this is considered a twobody problem. Because the sun plays a larger role than Earth at large meteor distances, the meteor could return to the same distance from the sun and the ecliptic plane of the Earth's orbit, but the chances are very small the Earth will be there at the time.