# Kinetic Energy and Velocity

1. Feb 19, 2008

### clope023

[SOLVED] Kinetic Energy and Velocity

1. The problem statement, all variables and given/known data

The gravitational pull of the earth on an object is inversely proportional to the square of the distance of the object from the center of the earth. At the earth's surface this force is equal to the object's normal weight mg, where g=9.8, and at large distances, the force is zero.

a) If a 50000kg asteroid falls to earth from a very great distance away, how much kinetic energy will it impart to our planet? You can ignore the effects of the earth's atmosphere.

b) What will be its minimum speed as it strikes the earth's surface?

2. Relevant equations

Wtot = 1/2m(v2)^2-1/2m(v1)^2

Wgrav = mg

3. The attempt at a solution

a) W = mg = (50000kg)(9.8m/s^2) = 4.9 x 10^5 J (wrong)

b) W = 1/2mv^2, v = $$\sqrt{2W/m}$$ = 4.4 (definetly wrong)

as you can see I tried using the work energy theorem with the given data, but nothing was right, any help is appreciated.

2. Feb 19, 2008

### Staff: Mentor

W = mg is a formula for the weight of something near the earth's surface (not work).

Hint: Look up the gravitational PE between two bodies as a function of distance.

3. Feb 19, 2008

### clope023

but no distance was given, I know Ugrav = mgd, d being distance...

4. Feb 19, 2008

### Staff: Mentor

Ug = mgd only applies near the earth's surface, which is not the case here. Look for another version of a gravitational PE formula.

5. Feb 19, 2008

### clope023

so I guess you're talking about the gravitational PE formula where the force approaches zero:

U = -GMm/r

but no distance is given so I can't find a radius, I'm assuming you're using:

Ugrav + Wtot = K

but I don't understand how I can get it without a given distance so I can find a radius for the potential energy, do I use the radius of the earth?

edit: so apparently -GMm/r = -1/2mv^2, but again I'm stumped about the radius.

Last edited: Feb 19, 2008
6. Feb 19, 2008

### clope023

tried it with the earth's radius and got the answer for part a)

GMM/r_earth = K = 3.1x10^12 J w00t!

now for part b)

would it be $$\sqrt{2K/m}$$?

7. Feb 19, 2008

### clope023

lol, I'm answering all my own questions.

it is what I just posted above, thanks to Doc Al for pointing me in the right direction with the proper formula.

8. Feb 19, 2008

### Staff: Mentor

Good work!

9. Feb 25, 2008

### rsala

can anyone summarize how to solve this problem, i do not understand how to use this formula GMm/r

what is G,? 9.8?, does it give you potential energy of the object? doesnt the question ask for kinetic?

thanks for the help

10. Feb 26, 2008

### Staff: Mentor

G is the universal gravitational constant, which appears in Newton's law of gravity. Since mechanical energy is conserved, knowing the change in gravitational potential energy allows you to calculate the change in kinetic energy. Read: Gravitational Potential Energy

11. Feb 26, 2008

### qspeechc

Use conservation of energy. At very large distances, the asteroid has zero potential energy. We can see this is true because U = k/r (k is a constant), so as r tends to infinity, U tends to zero, so very large r we can neglect U (gravitational energy). Then it becomes trivial to solve using conservation of energy. It starts with at least zero kinetic energy.