Questions on Gravity

Here is an answer for number one:

While the Sun is pulling on you, it is also pulling on the Earth. Your equations need to reflect both facts. When they do, you see that the Sun's pull cancels out.

On the close side, for example, you are being pulled toward the Sun (and away from the Earth) by the Sun. By iteslf, this ought to make the Earth's gravity seem weaker on the close side.

Your equation shows this effect:
32 - 5 = 27


But the Earth is also being pulled toward the Sun (and toward you). By itself, this ought to make the Earth's gravity seem stronger on the close side. Your equation does not show this effect, but it should. When your equation is corrected the two effects cancel:

(32 + 5) - 5 = 32

The converse of this happens on the far side.

=====

BTW, you might want to brush up on your use of units. The surface gravity of the Earth is 32 ft/sec^2, not 32 ft/sec. In Physics the units you use with a number are AT LEAST as important as the number itself. We lost an expensive spaceship (the Mars Climate Orbiter) a few years ago because some experienced engineers goofed up on units, even though they had the numbers right.

And, the Sun's gravitational acceleration near the Earth's orbit is somewhat smaller than the value you used here. You have to be well inside Mercury's orbit to see the Sun's g field that strong. In our neck of the woods it is around one or two in/sec^2.

The Internet is a great resource for checking on stuff like this before you post. You still can (and will) make mistakes, but not nearly as many. I knew your number was high, but I didn't know what the right number was. So I did a Google search on "surface gravity sun" to find that it is about 28 g. That same site also gave me the radius of the Sun. Then I set up a spread sheet to calculate the strength of the Sun's g field at various distances. Piece-o-cake.

Regards,
LB

To LB, Is the explaination you posted the same as saying the Earth is in free fall into the solar gravity field? I have been wondering how to say this in the way you have if this is correct.

Yes.

Nice catch,
LB

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
And, the Sun's gravitational acceleration near the Earth's orbit is somewhat smaller than the value you used here. You have to be well inside Mercury's orbit to see the Sun's g field that strong. In our neck of the woods it is around one or two in/sec^2.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Hi Bert

I assumed the suns pull of around 5 ft/sec from below formula based on the inverse square relationship. If this is incorrect please let me know - I am not a mathmatician.

(425,000 suns radius / 93,000,000 miles from sun) * 896 ft/sec of suns pull

This gives around 5 ft/sec.

Also did you mean 18 miles / sec is wrong for our orbit speed? I got this from an encyclopedia.

And finally, I am sorry if my use of units irritates you but I am not trying to put a spaceship into space, I am merely trying to get answers.

Thanks Bert

David

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
[kingdavid]
I assumed the suns pull of around 5 ft/sec from below formula based on the inverse square relationship. If this is incorrect please let me know - I am not a mathmatician.

(425,000 suns radius / 93,000,000 miles from sun) * 896 ft/sec of suns pull
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Your prose description of the calculation is correct, but your actual equation does not have the ratio squared. If you don't square this ratio you will indeed get a number near 5. It is the wrong number.

Squaring it makes the calculated g field drop off much faster, matching the observed g field.


<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
[kingdavid]
Also did you mean 18 miles / sec is wrong for our orbit speed? I got this from an encyclopedia.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Nope, it looks about right.


<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
[kingdavid]
And finally, I am sorry if my use of units irritates you but I am not trying to put a spaceship into space, I am merely trying to get answers.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Thanks for your concern. Things like this don't irritate me - I'm just trying to help a fellow science enthusiast learn the ropes. I guess it's become sort of a reflex action. For the kind of stuff we are doing here, it's usually obvious what you really meant.

BTW, a rigorus units analysis of this problem would have helped you get the right answer above.

Regards,
LB

Here is an answer to questions 4, 5 and 6

These questions are related and need a substantial amount of explaining - more than is reasonable in a forum like this. You need to study celestial mechanics to really understand how these things work, but Dr. Van Flandern has written some excellent summary/introductory material to help get started. If these whet your appetite for more a text book and/or a college course would be the next step.

Go to the main page of this site and click the tabs Solar System/EPH. Along the left side of the EPH page is a list of articles. The one titled "The Original Solar System" will help you begin to see how things like this probably work.

His book "Dark Matter, Missing Planets and New Comets" goes into more detail, but still contains very little math. For the layman this is absolutely the best $26 you can ever spend on learning Physics (this is an opinion, of course). All of the issues you raise in these questions are covered explicitly in this book, with the possible exception of #6. You might have found some bad information, or you might have misinterpreted the information. Or I might be misinterpreting your description. Can you be more explicit?

Warning - in addition to a lot of really good basic science this book contains some controversial speculation. The article mentioned above also contains some controversial material.

Regards,
LB