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Understanding Mercury's Precession 
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#1
Apr1113, 08:56 PM

P: 20

I was recently given a lab to work at home but I am having trouble understanding the formulas that they have used.
The link to this lab Assignment: http://classroom.sdmesa.edu/ssiegel/...Precession.pdf My problem relies to Query #1 of the lab assignment, with the equation [itex]\frac{V(r)}{m}[/itex] =  [itex]\frac{M}{r}[/itex] + [itex]\frac{(\frac{L}{m})^{2}}{r^{2}}[/itex] I derived it to be equal to: [itex]\frac{d}{dr}[/itex] [itex]\frac{V(r)}{m}[/itex] = [itex]\frac{M}{r^{2}}[/itex]  [itex]\frac{2(\frac{L}{m})^{2}}{r^{3}}[/itex] So I was told that M = mass of the Sun = 1.99 x 10[itex]^{30}[/itex] kg m = mass of Mercury = 3.28 x 10[itex]^{23}[/itex] kg L = angular momentum of Mercury = 9.11 x 10 [itex]^{38}[/itex] [itex]\frac{kg m^{2}}{s^{2}}[/itex] I equated the derived equation to 0 and solved for " r " to get: r[itex]_{o}[/itex] = [itex]\frac{2(\frac{L}{m})^{2}}{M}[/itex] which r[itex]_{o}[/itex] = 7.729, which should be the radius at the effective potential minimum But that doesn't make sense at all since the number is so low and the units doesn't seem right. Can someone help with this? 


#2
Apr1213, 04:07 AM

Mentor
P: 12,037

M/r should be GM/r with the gravitational constant G.



#3
Apr1213, 11:43 AM

P: 20

Hi Mfb,
The equations were based off Newton's attempt to study Mercury's precession. And the point of the lab is to see why he was off in his calculations. I'm wondering if the GM/r with with the gravitational constant G was used for the first attempt to calculate Mercury's precession by Newtonian methods? Thank you for your help! 


#4
Apr1313, 07:50 AM

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P: 12,037

Understanding Mercury's Precession
Of course. You need the gravitational constant both in Newtonian gravity and General Relativity, otherwise the equations do not make sense (and the units to not match either).



#5
Apr1313, 04:52 PM

P: 20

I have one last question on the lab assignment regarding to Query #5.
I'm getting an incorrect expression when working with Einstein's equation to determine the perihelion of Mercury. At the end of the equation I'm getting a negative number within the radical of my 2nd pic near the end. I don't know if I did a calculation incorrectly or the formula was given to me incorrectly. I also added in the gravitational constant G to the formula. 1st Page: 2nd Page: Let me know if you can read the images, sorry for the small writing. 


#6
Apr1313, 05:26 PM

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P: 12,037

The units in the last equation do not match. Just check them in your formulas, and you will find the error.



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