Falling object (gravity + kinematics)

In summary, the question asks for the speed of an object dropped from an altitude of one Earth radius above the surface of Earth. Using the equations F_g=ma_g=\frac{GMm}{r^2} and v^2=v_0^2+2ah, the attempt at a solution finds the correct answer to be \sqrt{\frac{GM}{R}}, not \sqrt{\frac{GM}{2R}} as originally thought due to assuming constant acceleration. Instead, energy conservation can be used to find the correct answer.
  • #1
mbrmbrg
496
2

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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  • #2
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.)
 
  • #3
Thanks, got it now!
 

1. What is the formula for calculating the velocity of a falling object?

The formula for calculating the velocity of a falling object is v = gt, where v is the velocity (in meters per second), g is the acceleration due to gravity (9.8 meters per second squared), and t is the time (in seconds).

2. Does the mass of an object affect its acceleration when falling?

No, the mass of an object does not affect its acceleration when falling. All objects, regardless of their mass, accelerate at the same rate due to gravity (9.8 meters per second squared).

3. How does air resistance affect the motion of a falling object?

Air resistance, also known as drag, can slow down the motion of a falling object. As an object falls, it pushes air molecules out of the way, creating a force in the opposite direction of the object's motion. This force increases as the object's velocity increases, eventually reaching a point where it is equal to the force of gravity, resulting in no further acceleration.

4. What is the difference between free fall and terminal velocity?

Free fall is the motion of an object when gravity is the only force acting upon it. Terminal velocity, on the other hand, is the maximum velocity an object can reach when falling, taking into account the force of air resistance. Once an object reaches terminal velocity, its velocity remains constant because the forces acting upon it are balanced.

5. How do you calculate the distance traveled by a falling object?

The formula for calculating the distance traveled by a falling object is d = 1/2gt^2, where d is the distance (in meters), g is the acceleration due to gravity (9.8 meters per second squared), and t is the time (in seconds). This formula assumes that the object starts at rest and is not affected by air resistance.

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