Kinetic energy problem (masteringphysics)

[kottur]

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.

a) If a 30000kg 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?

Homework Equations

F=Gm1m2/r
sqrt(2K/m)

The Attempt at a Solution

I've tried to put the variables in equation F=Gmm/r and I'm using the Earth's radius from Wikipedia = 6371km. Should I be using m1=m2=30000kg? Plus I don't know how to get kinetic energy out of the force equation.
For part b I need to know the kinetic energy.

 

[danielakkerma]

Hello there, welcome to the board!
Firstly, a correction there, which is very essential.
In your equations, you should've used:
$\large U = -\frac{Gm_1m_2}{r}$
Which is the potential energy, not the force ;
Now you must consider this:
$\large -\Delta{U} = W = \frac{m_2v^2}{2}$
Recall that you're told that the comet/asteroid arrives from a great distance away, meaning that you can neglect its point of origin(somewhere in the infinity of space), and simply plug in the destination, literally, the surface of the earth.
M_1 in the above formulae is the mass of the earth(which you can also find on wikipedia), The universal gravitational constant(G) is also there, and don't forget to convert R to meters before employing it.
Have fun,
Daniel

 

[kottur]

Thank you very much for the help!

From the equation U=-Gm1m2/r I got a minus value but when I plugged that into masteringphysics it told me to check my signs. Why wasn't the negative outcome correct but the positive one was? Is it because U=K2-K1 where K2 is zero?

 

[danielakkerma]

You should've taken note of this:

$\large -\Delta{U} = W = \frac{m_2v^2}{2}$

It's the negative change in the potential energy of a conservative force(such as gravity), that equals Work, and not just U.
Daniel

 

[kottur]

Thank you!