Calculating Impact Speed of Asteroid on Earth - Neglecting Atmospheric Friction

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To calculate the impact speed of an asteroid on Earth, one must consider the varying gravitational force, as it changes with distance from the Earth. The initial approach using the formula v^2 = u^2 + 2as is flawed because it assumes a constant acceleration, which is not applicable in this scenario. Instead, a more appropriate method involves using energy conservation principles or integrating the gravitational force over the distance. The discussion highlights the importance of recognizing the non-constant nature of gravitational acceleration when calculating impact velocities. Ultimately, a correct approach will yield the accurate speed at which the asteroid will hit the Earth's surface.
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Q:
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 4.85×10^9 kg. It is approaching the Earth on a head-on course with a velocity of 593 m/s relative to the Earth and is now 4.85×10^6 km away. With what speed will it hit the Earth's surface, neglecting friction with the atmosphere?

I found g using (mass of earth)(mass of sat)/(r^2)
of course since x = 4.85×10^6 km = 4.85×10^9 m
so g=
(5.98e24)/(4.85e9) <--since one of the r's cancels out with the mass of sat
then i used v^2 = vo^2 + 2ax
using g as a and everything else is pretty explanitory.

Where'd I go wrong?
 
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.. i forgot the G in the equation... but yeah it's still wrong..
 
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You can use the equation v^2=u^2+2as, if "a" is constant. When we usually use this forumla (ex: projectile motion),we assume the value of g to be fairly constant.

In this case, the gravitational force is not constant, and is varying with distance. So you cannot use the forumla v^2=u^2 + 2as.

Ill give you a hint. There is another way (a much easier one) you should approach the problem with. What do you usually use when you want to find final velocities?
(Example, an object is dropped at a height h.What is its velocity when it reaches the ground?)
 
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