Asteroid falling to earth, what will be minimum speed and KE on planet?

[overdrive7]

Homework Statement

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 = 9.8m/s^2, and at large distances, the force is zero. If a 20,000-kg asteroid falls to Earth from a very great distance away, what will be its minimum speed as it strikes the Earth's surface, and how much kinetic energy will it impart to our planet? You can ignore the effects of the Earth's atmosphere

Homework Equations

K=(1/2)(mv²)
g=~1/r^2, where r is the distance from the center of Earth to the asteroid
W=∫F_xdx

The Attempt at a Solution

I've seen a solution where the equation for potential energy (U) is involved, but it seems I'm supposed to solve this using calculus. I'm not quite sure where to start. I'm trying to use the integral of the work formula, but I'm not sure how to relate it to this problem where the gravity is involved. Thank you.

 

[Delphi51]

I'm glad to see you doing this, Overdrive. One should do it from basics until it gets boring, then use those more powerful formulas.

W=∫Fr*dr should do the trick. Your F will just be the big G formula for gravitational force. Show your work here if you want a check or help!

 

[overdrive7]

I'm glad to see you doing this, Overdrive. One should do it from basics until it gets boring, then use those more powerful formulas.

W=∫Fr*dr should do the trick. Your F will just be the big G formula for gravitational force. Show your work here if you want a check or help!

Thank you for the reply. By "big G formula", are you referring to the Gm1m2/r^2 formula? If so, is there a way to do it without the big G? I don't think I'm supposed to use the gravitational constant, because that is chapters away from where we are right now in our textbook (ch13 vs ch6).

 

[Delphi51]

Yes, that is the formula I meant.

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 = 9.8m/s^2

Could this be suggesting you derive that formula?
F proportional to 1/r² means F = k/r²
At the surface of the Earth this is mg = k/R² where R is the radius of Earth, so k = mgR²
and so F = k*r² becomes F = (mgR²)/r²
(Hopefully this mgR² has the same value as G*m*M.)
It does neatly avoid using G.

 

[overdrive7]

Thank you! I will use that and find the integral from infinity to the surface of the earth, since it says "at large distances, the force is 0", so the infinity would cancel out and leave the KE when it hits the surface.. I hope that's correct.

 

[Delphi51]

Sounds good!