# Deriving the Roche limit(d) for body of mass M and satellite m

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• PhysicsEnjoyer31415
PhysicsEnjoyer31415
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So i recently studied some basic gravitation and tried deriving the roche limit(d) for body of mass M and satellite m but when i used the custom formula to get the roche limit of earth and moon , i was off by 195 km , is that normal or is my formula incorrect or is it because newtonian gravity becomes inaccurate at planetary scale?

1. It's definitely not a problem with Newtonian gravity. In the planetary regime it's completely kosher.
2. Are you comparing your result with the value that used to be on Wikipedia (as this page does: https://www.astronomicalreturns.com/2021/06/roche-limit-radius-of-disintegration.html)? It's not there any longer, as isn't their derivation, and I don't feel like dredging the edit history as to why. But I it was divergent from what you can find elsewhere (e.g. Zelik & Gregory, Astronomy and Astrophysics, similar derivation here: https://www.astro.umd.edu/~hamilton/ASTR630/handouts/RocheLimit.pdf) - it had a 2 under a cube root in one place, instead of 3, giving too low a value. Possibly because it ignored orbital motion, but it's just a guess. You ignore that too, but it's not the main issue.
3. Even using the ex-Wiki value, you're not off by 195 km, you're off by 195km-6370km. Your d is distance above Earth's surface, not the distance from the centre that the ~9500 km indicates.
4. If I see correctly what you did there, the mistake in the derivation is that you've set up the initial equation for where the gravity between the two bodies is equal. This is not the condition for disintegration, as the satellite can very well be accelerated in its entirety by such gravitational potential and not suffer for it, as long as it's done uniformly. You need to set up the equation by comparing the tidal forces on the edge of the satellite to its self-gravity. Check the references above for example derivations.

PhysicsEnjoyer31415
Bandersnatch said:
1. It's definitely not a problem with Newtonian gravity. In the planetary regime it's completely kosher.
2. Are you comparing your result with the value that used to be on Wikipedia (as this page does: https://www.astronomicalreturns.com/2021/06/roche-limit-radius-of-disintegration.html)? It's not there any longer, as isn't their derivation, and I don't feel like dredging the edit history as to why. But I it was divergent from what you can find elsewhere (e.g. Zelik & Gregory, Astronomy and Astrophysics, similar derivation here: https://www.astro.umd.edu/~hamilton/ASTR630/handouts/RocheLimit.pdf) - it had a 2 under a cube root in one place, instead of 3, giving too low a value. Possibly because it ignored orbital motion, but it's just a guess. You ignore that too, but it's not the main issue.
3. Even using the ex-Wiki value, you're not off by 195 km, you're off by 195km-6370km. Your d is distance above Earth's surface, not the distance from the centre that the ~9500 km indicates.
4. If I see correctly what you did there, the mistake in the derivation is that you've set up the initial equation for where the gravity between the two bodies is equal. This is not the condition for disintegration, as the satellite can very well be accelerated in its entirety by such gravitational potential and not suffer for it, as long as it's done uniformly. You need to set up the equation by comparing the tidal forces on the edge of the satellite to its self-gravity. Check the references above for example derivations.
Hmm it seems i need to study more , i am in high school still so i am not aware of calculations and formulae regarding tidal forces but ill try to understand that. Also wasnt aware of wiki article , i just spontaneously decided to do it without any previous references .

Tom.G
My comment about the Wiki was just me trying to track down where you got the ~9500km number from, that you were comparing your own result to. I should have just asked you instead of playing detective.

BTW, you don't need to learn anything more than what you used in your attempt to follow what's being done (check that second link). It's pretty much just a matter of setting up the forces properly, followed by some algebra.

PhysicsEnjoyer31415
Oh i got that 9500 number from google and some pdf by nasa if i remember it was like 9496 km or something.I saw the Rochelimit pdf and it helped a lot i was not using it correctly enough sorry

Tom.G

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