Jump to content


Photo

Does The Moon Disprove The Reason Behind The Earth's Wobble?


  • This topic is locked This topic is locked
21 replies to this topic

#21 ikester7579

ikester7579

    Member

  • Member
  • PipPipPipPipPip
  • 12,500 posts
  • Gender:Male
  • Location:Florida
  • Interests:God, creation, etc...
  • Age: 48
  • Christian
  • Young Earth Creationist
  • I'm non-denominational

Posted 03 March 2009 - 09:14 PM

The moon's gravity does not differentiate between water and Earth mass.  The moon would still pull on the Earth the same whether it was dry or covered in water.  The oceans don't weaken the grip gravity has on the Earth.


I disagree because distance also has an effect on how strong gravity is. 7 miles difference would be less pull. And the water just might affect that as well. Has there been any tests done at the bottom of the ocean to confirm the moon's gravity pull is the same as on the surface?

Or would it continue to rotate.  It's been a long time since I took a basic mechanics class.

View Post


Unless the actual speed of orbit of the moon actually helps the wobble keep going?

#22 Guest_tharock220_*

Guest_tharock220_*
  • Guests

Posted 03 March 2009 - 10:45 PM

I disagree because distance also has an effect on how strong gravity is. 7 miles difference would be less pull. And the water just might affect that as well. Has there been any tests done at the bottom of the ocean to confirm the moon's gravity pull is the same as on the surface?


Obviously the moon's gravitational force would be weaker at the bottom of the ocean than at the surface since you would be dividing by a larger number. The point was that the gravitational force between the Earth and moon would the largly the same with or without the water, ie a change of about 1.5%. This assumes your using water in calculating Earth's mass, but as you no doubt know, water moves independently of the Earth.

What is your reasoning that water would have any effect on the gravitational force anyway?? Newton's law of universal gravitation says Fgrav = ((G)(m1)(m2)/r/^2. No need for surface area, density, materials, etc.




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users