The gravitational pull of the earth on an object is inverselyproportional to the

Question

The gravitational pull of the earth on an object is inverselyproportional to the square of the distance of the object from thecenter of the earth. At the earth's surface this force is equal tothe object's normal weight mg, where and at large distances, theforce is zero. If a 25,950-kg asteroidfalls to earth from a very great distance away, what will be itsminimum speed as it strikes the earth's surface, and how muchkinetic energy will it impart to our planet? You can ignore theeffects of the earth's atmosphere.
vmin = 1 m/s KE = 2 J vmin = 1 m/s KE = 2 J

Explanation / Answer

You just need to set your starting potential energy equal to thefinal kinetic energy. So, PE = GmM/r2 - GmM/r1    Here, r2 is going to be the radius of earth,and r1 is going to be infinity (so the second term is going toequal zero) So, G*25950*M/(Rearth) = 1/2 *25950 *V^2 Where the kinetic energy is the right side of the equation. and the V of the equation is the final velocity of theasteroid.

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