1. First of all, you'll have to excuse my lack of understanding of gravity. What I'd like to discuss, and hopefully come to comprehend, is Einstein's description of gravity regarding space time.

I have often seen diagrams depicting planets and stars resting on a surface (the grid that represents space time), and the 'indentation' if you will that these objects make on said surface. From what I know of the theory of gravity (and I use the term theory very loosely), is that this 'indentation' in space time is essentially gravity. When another object falls in to this indentation, it continues to orbit the other object in a somewhat circular motion, due to the distortion of space time.

*I fear that my understanding of gravity may be severely incorrect, thus I am seeking your help in clarifying it.*

What I'd like to ask is that if I understand correctly, and this 'indentation' or distortion of space time is the cause of gravity, would there not be a prior need for gravity in order for object B to fall in to this gravitational field of object A? Would the lack of gravity not allow object B to pass over the area created by object A unaffected? Or do objects literally travel 'on top of' space time, similar to a ball rolling on a table, rather than floating per se?

I'm sorry if this question is hard to follow. Any input is greatly appreciated!

David.

2.

3. That is the problem with this "rubber sheet" analogy to describe gravity, it does make you think that it needs gravity to make gravity work. The important thing is that it is just an analogy, a visual metaphor, for how GR describes gravity. And not a very good one.

Another way of looking at it is that mass curves spacetime "towards" itself and that causes objects to move in that direction.

I'm afraid I don't really know a better intuitive model.

4. Originally Posted by Strange
Another way of looking at it is that mass curves spacetime "towards" itself and that causes objects to move in that direction.
This is a reasonably straight forward way of describing it, and certainly helps with my understanding. Thanks for your input! The only problem I see with this, with my current understanding of gravity, is why this curvature wouldn't cause planets to continue to move inwards toward their stars, and eventually be enveloped? Logic would have me believe that this curvature wouldn't cause a planet to orbit a star, rather it would increase it's acceleration due to the declination on which it is travelling, until the point it is destroyed by the star.

I believe what I'm picturing (a downward slope toward the star) may again be a flaw of the "rubber sheet" analogy, and that I'm applying laws of physics which are not relevant in the case of gravity. These are the problems I'm hoping to weed out with your help!

5. Originally Posted by DavidT
Originally Posted by Strange
Another way of looking at it is that mass curves spacetime "towards" itself and that causes objects to move in that direction.
This is a reasonably straight forward way of describing it, and certainly helps with my understanding. Thanks for your input! The only problem I see with this, with my current understanding of gravity, is why this curvature wouldn't cause planets to continue to move inwards toward their stars, and eventually be enveloped? Logic would have me believe that this curvature wouldn't cause a planet to orbit a star, rather it would increase it's acceleration due to the declination on which it is travelling, until the point it is destroyed by the star.

I believe what I'm picturing (a downward slope toward the star) may again be a flaw of the "rubber sheet" analogy, and that I'm applying laws of physics which are not relevant in the case of gravity. These are the problems I'm hoping to weed out with your help!

If you imagine the rubber sheet to have zero friction, then you could take a marble, with zero friction, and spin it around like on a roulette table. Since there is no friction, it will "orbit" the mass in the middle for ever.

6. Originally Posted by DavidT
This is a reasonably straight forward way of describing it, and certainly helps with my understanding. Thanks for your input! The only problem I see with this, with my current understanding of gravity, is why this curvature wouldn't cause planets to continue to move inwards toward their stars, and eventually be enveloped? Logic would have me believe that this curvature wouldn't cause a planet to orbit a star, rather it would increase it's acceleration due to the declination on which it is travelling, until the point it is destroyed by the star.
This is exactly the same, whether you consider gravity as a Newtonian force or GR's curved spacetime. For this type of problem, they will give almost identical answers. Objects in orbit are continually falling. It is just that their sideways motion exactly matches the rate at which the planet "falls away" beneath them.

Consider throwing a ball: it will travel a few metres before gravity pulls it down to the ground. No shoot a canon ball: it will travel faster and hence further before its path curves down to the ground. Shoot a canon ball at escape velocity and it will fall to the ground at the same rate the horizon curves away from it: orbit!

7. Originally Posted by KALSTER
If you imagine the rubber sheet to have zero friction, then you could take a marble, with zero friction, and spin it around like on a roulette table. Since there is no friction, it will "orbit" the mass in the middle for ever.
Thank you for your response Kalster. I think you may have finally cleared this up for me. Am I right in saying that friction would slow the marble down, allowing it to succumb to the gravity and fall inwards. Where as the marble unaffected by friction will be travelling at a constant speed without losing acceleration, thus, as the marble is still essentially travelling straight, and it is the straight line itself that is curved, it will forever travel in this 'straight line' until another force acts upon it?

That is the best way I can describe my understanding of how gravity works after the information you've offered. I hope you can understand it, as I had a lot of trouble wording it!

8. Originally Posted by Strange
Consider throwing a ball: it will travel a few metres before gravity pulls it down to the ground. No shoot a canon ball: it will travel faster and hence further before its path curves down to the ground. Shoot a canon ball at escape velocity and it will fall to the ground at the same rate the horizon curves away from it: orbit!
Thank you Strange! This is the exact description I was hoping for! So I wasn't wrong, per se, I simply didn't understand all that was going on. I can see where I went wrong, and I think I now understand how gravity works.

Thank you so much for all your help.

9. Originally Posted by Strange
Originally Posted by DavidT
This is a reasonably straight forward way of describing it, and certainly helps with my understanding. Thanks for your input! The only problem I see with this, with my current understanding of gravity, is why this curvature wouldn't cause planets to continue to move inwards toward their stars, and eventually be enveloped? Logic would have me believe that this curvature wouldn't cause a planet to orbit a star, rather it would increase it's acceleration due to the declination on which it is travelling, until the point it is destroyed by the star.
This is exactly the same, whether you consider gravity as a Newtonian force or GR's curved spacetime. For this type of problem, they will give almost identical answers. Objects in orbit are continually falling. It is just that their sideways motion exactly matches the rate at which the planet "falls away" beneath them.

Consider throwing a ball: it will travel a few metres before gravity pulls it down to the ground. No shoot a canon ball: it will travel faster and hence further before its path curves down to the ground. Shoot a canon ball at escape velocity and it will fall to the ground at the same rate the horizon curves away from it: orbit!
Uh, not quite, shoot the cannon ball at orbital velocity (~7 km/s for earth) and it falls to the ground at the same rate as the ground falls away beneath it; shoot it at escape velocity (11.2 km/s) and it, well, escapes

10. Originally Posted by MeteorWayne
Uh, not quite, shoot the cannon ball at orbital velocity (~7 km/s for earth) and it falls to the ground at the same rate as the ground falls away beneath it; shoot it at escape velocity (11.2 km/s) and it, well, escapes
Thank you.

11. [/QUOTE]If you imagine the rubber sheet to have zero friction, then you could take a marble, with zero friction, and spin it around like on a roulette table. Since there is no friction, it will "orbit" the mass in the middle for ever.[/QUOTE]

There are no static bodies in nature. Hence, the assumed mass in the middle and associated rubber sheet are moving bodies. Then what happens to the orbital path of the marble?

12. If you imagine the rubber sheet to have zero friction, then you could take a marble, with zero friction, and spin it around like on a roulette table. Since there is no friction, it will "orbit" the mass in the middle for ever.
There are no static bodies in nature. Hence, the assumed mass in the middle and associated rubber sheet are moving bodies. Then what happens to the orbital path of the marble?
Since all motion is relative, your question makes no sense. Moving relative to what? You have to pick a reference frame to define that motion, and since any reference frame you choose is completely arbitrary, you are perfectly justified in choosing one in which the central mass is at rest ( it really makes no sense to even talk about the sheet as having a motion). This choice of reference frame is just as good as any other, and the choice of reference frame has no effect on the orbit of the marble relative to the central mass. It makes no difference if you consider the central mass as moving or not.

13. Originally Posted by MeteorWayne
Uh, not quite, shoot the cannon ball at orbital velocity (~7 km/s for earth) and it falls to the ground at the same rate as the ground falls away beneath it; shoot it at escape velocity (11.2 km/s) and it, well, escapes
So am I right in thinking that geostationary orbital velocity = escape velocity?

14. Originally Posted by Strange
Originally Posted by MeteorWayne
Uh, not quite, shoot the cannon ball at orbital velocity (~7 km/s for earth) and it falls to the ground at the same rate as the ground falls away beneath it; shoot it at escape velocity (11.2 km/s) and it, well, escapes
So am I right in thinking that geostationary orbital velocity = escape velocity?
No, it is just the orbital velocity and distance at the equator that matches the earth's rotation speed around it's axis. That way the satellite would appear to be stationary in the sky, while in fact it is still orbiting normally.

15. Originally Posted by KALSTER
No, it is just the orbital velocity and distance at the equator that matches the earth's rotation speed around it's axis. That way the satellite would appear to be stationary in the sky, while in fact it is still orbiting normally.
Of course. What was I thinking .... (maybe I wasn't)

16. Originally Posted by Strange
Originally Posted by MeteorWayne
Uh, not quite, shoot the cannon ball at orbital velocity (~7 km/s for earth) and it falls to the ground at the same rate as the ground falls away beneath it; shoot it at escape velocity (11.2 km/s) and it, well, escapes
So am I right in thinking that geostationary orbital velocity = escape velocity?
No, again, if it's at escape velocity, it escapes, it is no longer in orbit. All geostationary orbit means is that the earth's surface beneath spins at the same speed as the satellite orbits.

Orbital velocity of the ISS (392 km) = 7.68 km/s
Orbital velocity at geostationary orbit (35,786 km) = 6.88 km/s
Orbital velocity at the distance of the moon (384,400 km) = 2.26 km/s

Escape velocity = 11.2 km/s

Edit: Sorry, I see KALSTER got there first, but I did add some numbers

17. Originally Posted by Strange
Originally Posted by KALSTER
No, it is just the orbital velocity and distance at the equator that matches the earth's rotation speed around it's axis. That way the satellite would appear to be stationary in the sky, while in fact it is still orbiting normally.
Of course. What was I thinking .... (maybe I wasn't)
I know the feeling.

18.

19. That's good one Speed; thanx!

20. Originally Posted by MeteorWayne
Originally Posted by Strange
Originally Posted by MeteorWayne
Uh, not quite, shoot the cannon ball at orbital velocity (~7 km/s for earth) and it falls to the ground at the same rate as the ground falls away beneath it; shoot it at escape velocity (11.2 km/s) and it, well, escapes
So am I right in thinking that geostationary orbital velocity = escape velocity?
No, again, if it's at escape velocity, it escapes, it is no longer in orbit. All geostationary orbit means is that the earth's surface beneath spins at the same speed as the satellite orbits.

Orbital velocity of the ISS (392 km) = 7.68 km/s
Orbital velocity at geostationary orbit (35,786 km) = 6.88 km/s
Orbital velocity at the distance of the moon (384,400 km) = 2.26 km/s

Escape velocity = 11.2 km/s

Edit: Sorry, I see KALSTER got there first, but I did add some numbers
Your first number is correct, however orbital speed for geostationary is 3.34 km/sec and for the Moon is 1.02 km/sec

Also I should point out that the escape velocity you give is from the surface of the Earth, at the distance of the different orbits given, it would be sqrt(2) times larger than the orbital velocity. (10.86 km/s, 4.72 km/s and 1.44 km/s respectively)

21. I'll recheck my calculations!

22. Originally Posted by SpeedFreek
Actually, one way to deal with the "pulled down by what?" question is to imagine the rubber sheet as accelerating normal to the plane of the sheet. This has the bonus of illustrating an equivalence between inertial and gravitational mass.

23. Originally Posted by Janus
it would be sqrt(2) times larger than the orbital velocity
Aha, that's what I was thinking of (or failing to think of).

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