A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass, based on its size, of 5×109kg. It is approaching the Earth on a head-on course with a velocity of 860m/s relative to the Earth and is now 5.5×106km away. With what speed will it hit the Earth's surface, neglecting friction with the atmosphere?
I started this question by stating:
Kinetic Energy + Potential Energy = Kinetic Energy Final + Potential Energy Final since energy is conserved.
So: $$({\frac 1 2} mv^2)_{initial} + (mgh)_{initial} = ({\frac 1 2} mv^2)_{final} + (mgh)_{final}$$ but once it hits $h = 0$ we can cancel that out.
Next I had to solve for the current Potential energy which is where I got stuck. I'm using $Fg = \frac {GMm} {r^2}$ where r is the distance from the center of the earth to the asteroid. Plugging everything in I get:
$$\large \frac {(6.67*10^{-11})(5*10^9)(5.97*10^{24})} {(5.5*10^9+6371000)^2} = 65665.799 = Fg$$
Dividing by the mass of the asteroid gives me a value for g which I found to be $1.3*10^{-5}$ and I averaged that with 9.81 to get an average value for g which is 4.91.
I'm not sure if what I've done so far is on the right track or not but I'm starting to get confused and a bit frustrated, can someone point me in the right direction? I think i can solve for the PE now with my calculated g value but like I said I'm really unsure.
Best Answer
Remember that gravitational potential energy is not defined by GPE = mgh when not close to the Earth's surface. Neither do you use the formula
$$\frac{GMm}{r^2}$$
but rather another formula derived from it which is
$$GPE = \frac{-GMm}{r}$$
Please see this link for any further information regarding its derivation. Its hyperphyiscs. Sorry, if I had more time I would go into depth with it, but for now I can say is relates to the formula that energy equals the rate of change of force