I've read again that a space telescope is to orbit a point of equal gravity on a line connecting Earth and Sun. How is this possible, is there curvature causing attraction throwards this point?

Thrusters place it there. It was a target specifically shot at by engineers.

"At Lagrange points, the gravitational pull of two large masses precisely equals the centripetal force required for a small object to move with them. These points in space can be used by spacecraft to reduce fuel consumption needed to remain in position."

https://solarsystem.nasa.gov/resources/754/what-is-a-lagrange-point/

 

To me it looks like at the Lagrange points space would not be curved at all (for small enough distances). The space grid would be just cubes of equal size.

12 minutes ago, Willem F Esterhuyse said:

To me it looks like at the Lagrange points space would not be curved at all (for small enough distances). The space grid would be just cubes of equal size.

It's just Earth. Not enough gravity to worry about relativity. Newtonian gravity is close enough.

There has to be some curvature though, because the point orbits the Sun. I believe that plot represents pseudopotentials in Earth's orbital reference frame (non-inertial in the Newtonian model), so they include the effect of a centrifugal force that would compensate for the Sun's attraction (the relativistic curvature as seen by a distant observer).

Edited January 30, 20232 yr by Lorentz Jr

The picture shows L4 as having repulsive gravity and L1, L2 and L3 to have attractive gravity from one direction and repulsive gravity for another direction. How does the gravity of Earth add together with the gravity of the sun to cause L3 to have neutral gravity? It isn't a subtractive operation. I don't believe L2 and L3 have neutral gravity.

2 minutes ago, Willem F Esterhuyse said:

I don't believe L2 and L3 have neutral gravity.

I don’t care what you believe 

 Neither does nature 

The law states: F = GM_Sm/r² + GM_Em/(r+d)² and this must equal zero for a neutral gravity point. It doesn't.

17 minutes ago, Willem F Esterhuyse said:

How does the gravity of Earth add together with the gravity of the sun to cause L3 to have neutral gravity? It isn't a subtractive operation.

The simple answer is that L3 is pretty much the same as the orbit of Earth itself. You can't really tell from the diagram, but L3 has to be infinitesimally farther from the Sun than Earth is, because Earth makes a tiny increase in the gravity at that point.

17 minutes ago, Willem F Esterhuyse said:

I don't believe L2 and L3 have neutral gravity.

They don't. They're the points where the orbital period is the same as Earth's. Farther away from L2, the period is longer. Closer to Earth, it's shorter.

8 minutes ago, Willem F Esterhuyse said:

The law states: F = GM_Sm/r² + GM_Em/(r+d)² and this must equal zero for a neutral gravity point. It doesn't.

They're pseudopotentials. They include the effects of centrifugal force in Earth's orbital reference frame. If the gravitational field were literally zero, the objects could stay in place while Earth orbited away from them, and that's not what we want. 🙂

Edited January 30, 20232 yr by Lorentz Jr

12 minutes ago, Lorentz Jr said:

They don't.

Then they must not state that the telescope orbits this point.

13 minutes ago, Willem F Esterhuyse said:

Then they must not state that the telescope orbits this point.

Excuse me, I guess I was wrong about that term. You said "The law states: F = GM_Sm/r² + GM_Em/(r+d)² and this must equal zero for a neutral gravity point." That's zero gravity, not neutral gravity. Neutral gravity is where an object's gravitational orbit appears to be stationary relative to another orbiting body, in that body's orbital reference frame, not where it's stationary in an inertial frame.

Nothing literally orbits the L points. They all orbit the Sun, and their defining characteristic is that their orbital period is the same as Earth's.

Edited January 30, 20232 yr by Lorentz Jr