Falling object (gravity + kinematics)

AI Thread Summary
An object dropped from one Earth radius above the surface experiences varying gravitational acceleration as it falls. The initial attempt to calculate the final speed used constant acceleration, leading to an incorrect result. The correct approach involves using energy conservation principles instead of kinematics, as gravitational potential energy varies with distance from the Earth. The final speed just before impact is derived from the gravitational potential energy equation, yielding the correct result of v = √(GM/R). This highlights the importance of recognizing non-constant forces in physics problems.
mbrmbrg
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Homework Statement



An object is dropped from an altitude of one Earth radius above Earth's surface. If M is the mass of Earth and R is its radus, find the speed of the object just before it hits Earth.

Homework Equations



F_g=ma_g=\frac{GMm}{r^2}

v^2=v_0^2+2ah

The Attempt at a Solution



F_g=\frac{GMm}{(2R)^2}=ma_g

a_g=\frac{GM}{4R^2}

Now, plug that into the kintematics equation and get

v^2=0+2(\frac{GM}{4R^2})R

v=\sqrt{\frac{GM}{2R}}

But the correct answer is given as \sqrt{\frac{GM}{R}}, and I can't find my error.
 
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mbrmbrg said:
F_g=ma_g=\frac{GMm}{r^2}
Note that F_g and a_g are not constant, but are functions of r.
v^2=v_0^2+2ah
But this kinematic equation assumes constant acceleration.

The Attempt at a Solution



F_g=\frac{GMm}{(2R)^2}=ma_g

a_g=\frac{GM}{4R^2}
That's only the acceleration at the point r = 2R; as the object falls, the acceleration increases.

Your error is treating this as a constant acceleration problem. Instead of using kinematics, why not use energy conservation? (What's the general form for gravitational PE? Note that "mgh" is only valid near the Earth's surface--no good here.)
 
Thanks, got it now!
 
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