# Does a positive cosmological constant increase or decrease gravity?

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In de Sitter space with a positive cosmological constant $\Lambda$, it is usually said that gravity is smaller than the value given by $1/r^2$, because $\Lambda$ *reduces* the attraction of masses at large distances.

However, many people (Verlinde, Milgrom, etc.) argue that a positive $\Lambda$ *increases* gravitational attraction at large distances. For example, MOND (with its new constant $a_0$) conjectures a higher gravitational attraction than $1/r^2$ (at large distances).

How can this apparent contrast be resolved?

asked Jun 22
recategorized Jun 22

Mond doesn't provide a valid illustration, a0 is not the gravitational constant.

Verlinde and Milgrom likely said that given some observational data and working on fitting, if you raise the cosmological constant, you would need to raise also the 'gravitational attraction'. Could you provide references to their exact claims?

## 1 Answer

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MOND is not a replacement for dark energy but for dark matter, so of course that leads to a higher attraction. A cosmological constant has to do with dark energy and leads to an extra repulsive term so instead of ȑ=-GM/r² we have ȑ=-GM/r²+Λc²r/3

answered Jul 22 by (10 points)
edited Jul 23 by Yukterez

This is exactly what was written in the question. This is not an answer.

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