Using Gravitational constant to get the final velocityby hyperkkt Tags: force, gravitational, integral, position, speed 

#1
Nov2408, 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 



#2
Nov2408, 06:12 PM

Sci Advisor
HW Helper
Thanks
P: 25,170

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.




#3
Nov2408, 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 



#4
Nov2408, 09:53 PM

Sci Advisor
HW Helper
Thanks
P: 25,170

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.



Register to reply 
Related Discussions  
Final velocity of the mass  Introductory Physics Homework  3  
Final Velocity  Introductory Physics Homework  7  
Final velocity of an object accelerated across a given distance with constant power  Introductory Physics Homework  9  
Final velocity  Advanced Physics Homework  5  
final velocity of car  Introductory Physics Homework  5 