Understanding Mercury's Precession

  • Thread starter Thread starter OnceKnown
  • Start date Start date
  • Tags Tags
    Precession
Click For Summary
The discussion revolves around a lab assignment focused on understanding Mercury's precession, where the user is struggling with the equations provided. They derived an equation related to the effective potential but found the resulting radius to be unexpectedly low and with incorrect units. Clarification was sought regarding the necessity of including the gravitational constant (G) in the calculations, which is essential for both Newtonian gravity and General Relativity. Additionally, the user encountered issues with a negative value in a radical while working through Einstein's equation for Mercury's perihelion, indicating potential calculation errors or formula inaccuracies. Accurate unit matching and the inclusion of G are crucial for resolving these issues.
OnceKnown
Messages
20
Reaction score
0
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/Physics%20197/labs/Mercury%20Precession.pdf

My problem relies to Query #1 of the lab assignment, with the equation
\frac{V(r)}{m} = - \frac{M}{r} + \frac{(\frac{L}{m})^{2}}{r^{2}}

I derived it to be equal to:
\frac{d}{dr} \frac{V(r)}{m} = \frac{M}{r^{2}} - \frac{2(\frac{L}{m})^{2}}{r^{3}}

So I was told that M = mass of the Sun = 1.99 x 10^{30} kg
m = mass of Mercury = 3.28 x 10^{23} kg
L = angular momentum of Mercury = 9.11 x 10 ^{38} \frac{kg m^{2}}{s^{2}}

I equated the derived equation to 0 and solved for " r " to get:
r_{o} = \frac{2(\frac{L}{m})^{2}}{M}

which r_{o} = 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?
 
Last edited:
Astronomy news on Phys.org
M/r should be GM/r with the gravitational constant G.
 
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!
 
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).
 
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:
38d9454e-7706-4f90-b903-0246a5047d03_zpsdf163876.jpg


2nd Page:
image-1_zps5276ba49.jpe


Let me know if you can read the images, sorry for the small writing.
 
The units in the last equation do not match. Just check them in your formulas, and you will find the error.
 
UC Berkely, December 16, 2025 https://news.berkeley.edu/2025/12/16/whats-powering-these-mysterious-bright-blue-cosmic-flashes-astronomers-find-a-clue/ AT 2024wpp, a luminous fast blue optical transient, or LFBOT, is the bright blue spot at the upper right edge of its host galaxy, which is 1.1 billion light-years from Earth in (or near) a galaxy far, far away. Such objects are very bright (obiously) and very energetic. The article indicates that AT 2024wpp had a peak luminosity of 2-4 x...

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
680
  • · Replies 9 ·
Replies
9
Views
779
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
6
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
2K