Two masses and spring oscillation

  • #76
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Well.. it is quite confusing that both post#69 and post#71 are correct ,value of ##v## is coming out same ,yet the ##v## in the two posts are different .

Does ##v## in post#71 represent the linear speed of the planet in circular orbit around the CM ?
 
  • #77
ehild
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In post #69 you got the speed of the reduced-mass hypothetical planet:

[tex]v_r^2=G\frac {m+M}{D}[/tex]
In post #71, you derived the speed of the real planet:

[tex]v^2=G\frac {M^2}{(M+m)D}[/tex]

They are not the same.
 
  • #78
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It is better to get the angular speed. It does not change with the reduced mass approach. So ω^2=G(M+m)/D3. The radii of the orbits both of the planet and the Sun is obtained from the condition that the CM is in rest in the applied frame of reference.
$$ \frac{G(M+m)\mu}{D^2}=\mu ω^2D $$

$$ ω^2 = \frac{G(M+m)}{D^3} $$

Now , ## v = ωr_2 ## where ## r_2 = \frac{MD}{m+M}##

$$ v^2 = \frac{G(M+m)}{D^3} (\frac{MD}{m+M})^2 $$

$$ v^2 = \frac{GM^2}{(m+M)D} $$

This result matches the one in post#71 .

Is this how you wanted me to approach ?
 
  • #79
ehild
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$$ \frac{G(M+m)\mu}{D^2}=\mu ω^2D $$

$$ ω^2 = \frac{G(M+m)}{D^3} $$

Now , ## v = ωr_2 ## where ## r_2 = \frac{MD}{m+M}##

$$ v^2 = \frac{G(M+m)}{D^3} (\frac{MD}{m+M})^2 $$

$$ v^2 = \frac{GM^2}{(m+M)D} $$

This result matches the one in post#71 .

Is this how you wanted me to approach ?
Yes, it was. :smile: That is the "real" speed of the planet. And the speed of the Sun centre is ωr1

ehild
 
  • #80
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Thank you very much :)

But I do not understand the importance of this reduced mass approach . We could have calculated the speed as in post#71 .

What could have been asked in the question where the conversion of two body problem in one would have been a superior approach ?
 
  • #81
ehild
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How would you solve the problem of double stars in general ( not assuming circular orbit)? When you need the period and the data of the orbits?

ehild
 
  • #82
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OK.

Please have a look at the attachment .I have attached portion of a chapter in the book.

The motion of a planet is analogous to that of an electron revolving the nucleus just like in Bohr Model of the hydrogen atom .

If the nucleus is considered stationary then $$ \frac{1}{4\piε_0}\frac{e^2}{r^2}
= \frac{mv^2}{r} $$
But if take into account the motion of nucleus around the CM ,then do we simply replace ##m## by ##\mu ## where ##\mu =\frac{mM}{m+M}## ,where m is the mass of electron and M is the mass of proton(nucleus).

The LHS in both the cases remain same whereas the RHS in the stationary case has ##m## and the moving nucleus case has ##\mu## .

$$ \frac{1}{4 \piε_0}\frac{e^2}{r^2} = \frac{\mu v^2}{r} $$

But this is not correct because the 'v' in this case is that of the hypothetical reduced mass ,not the actual speed of the electron.

What is your opinion ? Is the book getting it wrong or am I interpreting it incorrectly ?
 

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  • #83
ehild
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Here the force is proportional to the product of charges instead of the product of the masses.

ehild
 
  • #84
vela
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Well, it is quite confusing that both post#69 and post#71 are correct, value of ##v## is coming out same, yet the ##v## in the two posts are different .

Does ##v## in post#71 represent the linear speed of the planet in circular orbit around the CM ?
Try calculating ##v_1## and ##v_2##, the speed of each body relative to the CM, using your approach in post 71. The bodies move in opposite directions, so their relative speed is given by ##v_1+v_2##. Calculate that sum and compare it to the result you got in post 69. Don't make any approximations based on the assumption that ##M \gg m##.
 
  • #85
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Here the force is proportional to the product of charges instead of the product of the masses.

ehild
Right . But according to book ,if we need to account for the motion of nucleus we simply need to replace ##m## by ##\mu## to get the speed .(Or am I understanding it incorrectly)

$$ \frac{1}{4 \piε_0}\frac{e^2}{r^2} = \frac{\mu v^2}{r} $$

This gives $$v^2 = \frac{1}{4 \piε_0}\frac{e^2(m+M)}{rmM}$$

But this is incorrect value of speed .

The correct result can be obtained on similar lines to post#71 .

$$ \frac{1}{4 \piε_0}\frac{e^2}{r^2} = \frac{mv^2}{r_2} $$ ,where ##r_2 = \frac{Mr}{m+M}## is the distance of the electron from the CM .

$$ \frac{1}{4 \piε_0}\frac{e^2}{r^2} = \frac{mv^2}{r_2} $$

$$v^2 = \frac{1}{4 \piε_0}\frac{e^2M}{rm(m+M)}$$ . This is the correct speed .

How is book interpretation correct ?
 
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  • #86
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How is book interpretation correct ?
You're comparing apples to oranges. Reread the first part of post 72 and do the calculation I suggested in post 84. Hopefully, that will clear up your confusion.
 
  • #87
ehild
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The Bohr model predicts the spectrum of hydrogen atom. It postulates that the angular momentum is integer multiple of (h/2pi) and determines the possible energy levels. The angular momentum and the energy are those of the whole Hydrogen atom. The angular momentum and energy of the whole atom can be considered as those of a single particle with mass equal to the reduced mass of the electron-nucleus system. https://www.google.com/url?sa=t&rct...mG2jGkozqYJcq5A&bvm=bv.68445247,d.bGQ&cad=rjt

The reduced mass correction explains the isotope shift of the spectral lines of Deuterium. In case of stationary nuclei, the spectra of both the Hydrogen and the Deuterium atoms would be the same.

ehild
 

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