How Does Gravitational Force Influence Ocean Tides?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 3K views
Frank66
Messages
11
Reaction score
0
Hi,
it is well know that the force on a mass point outside to an homogeneous sphere is as the whole sphere's mass is concentred on its centre.
It seems that the reaction of this force (the total force of the point on the shepre) is applied on the centre of the sphere but, is it true?
If this is it true why tides?
Do you know a book where this is demonstrated?
thank you and excuse my english
 
Astronomy news on Phys.org
Frank66 said:
Hi,
it is well know that the force on a mass point outside to an homogeneous sphere is as the whole sphere's mass is concentred on its centre.
It seems that the reaction of this force (the total force of the point on the shepre) is applied on the centre of the sphere but, is it true?
If this is it true why tides?
Do you know a book where this is demonstrated?

No it isn't true. The attraction is greater for parts of the sphere that are closer to the point mass.
if the force of the moon was only applied to the center of the earth, and not to the oceans as well, you'd have MUCH bigger tides.
 
thank you,
Have you reference? Where can I found the calculus?
 
For a rigid body, it is as if the force were applied to the center of gravity (which is why it is called the center of gravity!) but for a body which can be thought of as made of independent particles, the force acts on the individual particles. We can (roughly) think of the Earth as rigid body but not water. That's why "tidal" effects on the Earth itself are small compared with ocean tides.