Falling object (gravity + kinematics)

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SUMMARY

The discussion focuses on calculating the speed of an object dropped from an altitude of one Earth radius above Earth's surface. The initial attempt used the gravitational force equation, F_g = GMm/(r^2), and the kinematic equation v^2 = v_0^2 + 2ah, leading to an incorrect conclusion. The correct approach involves recognizing that gravitational acceleration is not constant and applying energy conservation principles instead. The final correct speed formula is v = √(GM/R).

PREREQUISITES
  • Understanding of gravitational force and acceleration (F_g = GMm/r^2)
  • Familiarity with kinematic equations (v^2 = v_0^2 + 2ah)
  • Knowledge of energy conservation principles in physics
  • Basic concepts of potential energy (PE) in gravitational fields
NEXT STEPS
  • Study energy conservation in gravitational fields and its applications
  • Learn about variable acceleration and its implications in physics problems
  • Explore advanced kinematics and dynamics in non-constant acceleration scenarios
  • Review gravitational potential energy calculations beyond the Earth's surface
USEFUL FOR

Physics students, educators, and anyone interested in understanding gravitational dynamics and energy conservation principles in mechanics.

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



[tex]F_g=ma_g=\frac{GMm}{r^2}[/tex]

[tex]v^2=v_0^2+2ah[/tex]

The Attempt at a Solution



[tex]F_g=\frac{GMm}{(2R)^2}=ma_g[/tex]

[tex]a_g=\frac{GM}{4R^2}[/tex]

Now, plug that into the kintematics equation and get

[tex]v^2=0+2(\frac{GM}{4R^2})R[/tex]

[tex]v=\sqrt{\frac{GM}{2R}}[/tex]

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

The Attempt at a Solution



[tex]F_g=\frac{GMm}{(2R)^2}=ma_g[/tex]

[tex]a_g=\frac{GM}{4R^2}[/tex]
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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