Gravity and Velocity

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I recently saw the equation [tex] v= \sqrt{2 \int_{r_1}^{r_2} \frac{GM}{r}} [/tex] (G=gravitational constant, M=larger mass, r=radius). I know it has to do with the velocity of an object attracted by a larger object's gravity, which is pulling it a distance [tex] r_2-r_1 [/tex] taking into account the change in force of gravity. My question is: does this equation represent the instantaneous velocity at r1 or the average velocity? Also, how can this equation be changed to include the force of gravity of the smaller object on the larger one?
 

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Doc Al
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Edited: Please note my correction to the original equation!​

Firstly, that equation is not quite right. The integral (under the square root sign) should be:
[tex] \int_{r_1}^{r_2} \frac{GM}{r^2} dr = \left[ - \frac{GM}{r} \right]_{r_1}^{r_2}[/tex]

So your equation should read:
[tex] v= \sqrt{2 \int_{r_1}^{r_2} \frac{GM}{r^2} dr} [/tex]

My question is: does this equation represent the instantaneous velocity at r1 or the average velocity?
The corrected equation represents the instantaneous speed of the falling object at r1, assuming it started from rest at r2 (where r2 > r1).
Also, how can this equation be changed to include the force of gravity of the smaller object on the larger one?
The integral comes from calculating the increase in KE of the two mass system. If the smaller mass is a tiny fraction of the larger's mass, then all that energy becomes KE of the smaller mass--as in the given equation for v. (That's where that equation comes from.) But if you want to include the motion of the larger mass as well, you'll have to distribute that KE across both masses and apply conservation of momentum to find the relative speed of the masses.
 
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But if you want to include the motion of the larger mass as well, you'll have to distribute that KE across both masses and apply conservation of momentum to find the relative speed of the masses.
So, if I wanted to include the KE that the smaller mass' force of gravity produces, would the new equation be [tex] v= \sqrt{2 \int_{r_1}^{r_2} \frac{GMm}{r}} [/tex] where m is the smaller mass?
 
Doc Al
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Edited: Please note my corrections!​

No. Check and you'll see that the units don't match across that equation.

Before you worry about modifying the original equation, first understand how it was obtained; it starts with this statement of conservation of energy:
[tex] \frac{1}{2}mv^2 = \int_{r_1}^{r_2} \frac{GMm}{r^2} dr [/tex]
 
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Would the KE be distributed evenly?
 
Gib Z
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None of these problems would come up if you were thinking about it relativistically >.<. Equating (1/2)mv^2 as KE makes it obvious to me this is classical mechanics...As to your last post: It depends on your frame of reference :p

EDIT: MY BAD!! Classical Physics section >.< Sorry
 
Doc Al
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Oops... that equation as stated is incorrect

Note to Izzhov: I just realized (last night, after logging off) that there is an error in your original equation, which I have let propagate into my own. :redface: D'oh! So I will revise my answers accordingly. Stay tuned. (Same basic idea, though; perhaps you just miscopied the equation.)

Edited: Please note my corrections in posts 2 and 4!​
 
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Doc Al
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Would the KE be distributed evenly?
The distribution of KE depends on the relative size of the masses. If they were equal, then the KE would be distributed evenly. If they were wildly different, like a bowling ball or rocket compared to the earth, then the bowling ball or rocket would get just about all of the KE. (That's the assumption made in your orginal equation.)

The key is that both energy and momentum must be conserved. If the two masses start from rest, their initial--and final--momentum must be zero.
 
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So essentially what you're saying is that if [tex] \int_{r_1}^{r_2} \frac{GMm}{r^2} dr [/tex] is the KE, then [tex] \frac{M}{M+m} \ast \int_{r_1}^{r_2} \frac{GMm}{r^2} dr [/tex] is the amount of KE that the smaller mass gets, and [tex] \frac{m}{M+m} \ast \int_{r_1}^{r_2} \frac{GMm}{r^2} dr [/tex] is the larger mass's KE, and that this can be derived using conservation of momentum. Is this correct?
 
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Doc Al
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Exactly right!
 

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