# Marion and thornton classical dynamics

Hello,

Does anyone have marion and thornton's classical dynamics book? I have a possible error that I wanted to point out. Actually, goldstein's classical mechanics book would work also since I found the same "error" in there as well.

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siddharth
Homework Helper
Gold Member
I have the second edition of goldstein's book. Why don't you just ask your question here, and those who have the book/s will be able to help.

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George Jones
Staff Emeritus
Gold Member
Hello,

Does anyone have marion and thornton's classical dynamics book? I have a possible error that I wanted to point out. Actually, goldstein's classical mechanics book would work also since I found the same "error" in there as well.

I have both books, but not in their latest editions. I might not be able to look at them until tomorrow.

What error do these books have in common?

OK. I am pretty sure this is not an not an error now that I found it in both books.

equation 4.93 in goldstein:

$$m\frac{d^2x}{dt^2}=-2m\omega v_z \sin \theta$$

They show that

$$v_z = -gt$$

So the first equation becomes $$m\frac{d^2x}{dt^2}= 2m\omega gt \sin \theta$$

What I DO NOT understand is how they integrate that twice to obtain

$$x = \omega g t^3 \sin \theta /3$$

To do that, they need the initial condition $\frac{dx}{dt}(t=0) = 0$. I do not understand how they get that initial condition!

Also, in a footnote of Goldstein, they say "It is easy to show, using Eq (4.93), that a particle projected upward will fall back to the ground westward of its original launching spot." That is extremely confusing to me.

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Hootenanny
Staff Emeritus
Gold Member
OK. I am pretty sure this is not an not an error now that I found it in both books.

equation 4.93 in goldstein:

$$m\frac{d^2x}{dt^2}=-2m\omega v_z \sin \theta$$

They show that

$$v_z = -gt$$

So the first equation becomes $$m\frac{d^2x}{dt^2}= 2m\omega gt \sin \theta$$

What I DO NOT understand is how they integrate that twice to obtain

$$x = \omega g t^3 \sin \theta /3$$

To do that, they need the initial condition $\frac{dx}{dt}(t=0) = 0$. I do not understand how they get that initial condition!

Also, in a footnote of Goldstein, they say "It is easy to show, using Eq (4.93), that a particle projected upward will fall back to the ground westward of its original launching spot." That is extremely confusing to me.
Could you be more specific? In what context is the above equation derived?

x is the eastward coriolis deflection of a free-falling object in the Northern Hemisphere

Hootenanny
Staff Emeritus
Gold Member
x is the eastward coriolis deflection of a free-falling object in the Northern Hemisphere
What is the initial velocity of the free-falling body?

zero

Cleonis
Gold Member
Also, in a footnote of Goldstein, they say "It is easy to show, using Eq (4.93), that a particle projected upward will fall back to the ground westward of its original launching spot." That is extremely confusing to me.

Let me first discuss what is happening when an object is dropped from height.
(In all of the following discussion I'm neglecting air friction.)

The trajectory of an object that is released from altitude is a ballistic trajectory. It has an initial tangential velocity, from co-rotating with the Earth. You can think of the starting point of the trajectory as the apogee of an eccentric orbit.

On its way down from apogee the angular velocity of the object in orbit increases. (Compare the extremely eccentric orbit of Halley's comet; falling towards the Sun it speeds up.)

That is why an object released from altitude will not fall perpendicular to the ground, it will reach the surface eastward of its launch coordinates.

Now the case of an object fired upward.
That trajectory is like that of an orbiting object moving upward to its apogee.
In climbing up the object loses angular velocity. So relative to the Earth the object will "fall behind". Later, during the descent, it will get that angular velocity back, but in the meantime the Earth has turned underneath all that.
Therefore an object fired straight up will land westward of the coordinates that it has been launched from.

There's a Java applet on my website designed for exploring precisely that kind of scenario: http://www.cleonis.nl/physics/ejs/ballistics_and_orbits_simulation.php" [Broken]

Cleonis
http://www.cleonis.nl

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