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Using Gravitational constant to get the final velocity

by hyperkkt
Tags: force, gravitational, integral, position, speed
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hyperkkt
#1
Nov24-08, 05:27 PM
P: 2
1. The problem statement, all variables and given/known data
Evil Alien has put an Asteroid with
mass of 1,000,000kg to destroy mankind.
Distance from center of the Earth to Asteroid
(assume negligible center) is
10^8m and lets the gravitational force do the work.
Earth radius is 6.4 x 10^6m
and its mass is 5.98 x 10^24kg.
If the atmosphere extends out to 500km beyond
the surface and exrts an average friction force of 10^8N,
calculate the speed of the asteroid just before it hits the ground.
(Assume the asteroid rtains all of its mass as it travels through the atmosphere)



2. Relevant equation
Fgrav = Gm1m2/r^2


3. The attempt at a solution

I tried to use F=ma first
by doing it so I get
F=Gm1m2/r^2=m1a
a=Gm2/r^2

However, could not do anything more since there is no dt.

So, I tried taking an integral of it,
finding the work and set it equal to kinetic energy
(not sure whether indefinite/definite integral matters)
integral of Gm1m2/r^2 = -Gm1m2/r
-Gm2/r^2=v^2
but got this and it cannot happen
because v^2 cannot be - number...
v=i(imginary)

I am stuck at this point have no
further suggestion on what I should do.

I've been working on it for about an hour
and would appreciate any help.
Thanks in advance
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Dick
#2
Nov24-08, 06:12 PM
Sci Advisor
HW Helper
Thanks
P: 25,250
The potential energy PE(r)=-G*m1*m2/r. You find the change in potential energy by subtracting PE(r=10^8m)-PE(r=radius of the earth). The difference is positive, not negative.
hyperkkt
#3
Nov24-08, 06:27 PM
P: 2
Oh I see, but how do you get v after finding the change in PE?
And from there how do I apply it to the interval which frictino decreases the speed?

I may be understanding something wrong.
Thanks again for the reply

Dick
#4
Nov24-08, 09:53 PM
Sci Advisor
HW Helper
Thanks
P: 25,250
Using Gravitational constant to get the final velocity

Part of the change in potential energy goes to overcoming the friction. The friction work is just force times distance. The rest of the potential energy becomes kinetic energy. Equate that to 1/2*mv^2.


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