orbital calculations

[ianke]

How long would it take for a ping pong ball to orbit a bowling ball at 1m distance in space with no gravitational or frictional influence from outside the model.<br /><br />the bowling ball= 7kg, the ping pong ball= 1g<br /><br />I see them on rubber sheets all the time. lol <br />I am not a student. It is not my homework. I just think it is interesting to know. Thanks in advance ! <div class="Discussion_UserSignature"> </div>

 

[5billionyearslater]

It is a redundant question, to be honest!<br /><br />If a bowling ball and ping pong ball were put into space, they would both (probably) be forced into the orbits of the nearest strongest gravitation field.<br /><br />They, themselves would never generate their own gavitation tugs and there is no way a ping pong ball can 'orbit' a bowling ball.<br /><br />However, I think what you are trying to do is make a hypothetical question about a tiny star and a tiny orbiting body, which cannot exist in the scale you are wanting.

 

[nexium]

5 is correct, a 7 kilogram object has almost no gravity at a distance of one meter, but we have found asteroids that orbit each other. If you think 7 million tons, 100 meters from the mass center, I think the calculation is possible for any small mass orbiting 100 meter from the mass center of the 7 million ton asteroid.<br />A 7 million ton asteroid has one million cubic meters, if the average density is 7.<br />Since the pair is in free fall, larger mass and stronger gravity fields nearby have little effect on the way the pair orbit each other, but out gasing of either body, solar wind, light pressure, changing gravity fields and tidal effects can cause the lighter object to collide with the larger or be ejected from it's orbit around the 7 million ton asteroid. Sorry I do not know how to calculate specifics. Neil

 

[vogon13]

I don't have my calculator handy, but just working out the density and separation in my head, I would expect the orbital period to be several hours. <br /><br />Accurate initial placement of the ping pong ball will be difficult, I would not expect a circular orbit. This experiment will need to be conducted in darkness, and outside of the Roche limit of any other mass.<br /><br />An interesting perturbation of this experiment would be to release a small cloud of titanium dioxide powder and then to watch rings form around the bowling ball. Rings might take years to form, and the bowling ball will need to be slightly oblate.<br /><br /> <div class="Discussion_UserSignature"> <p><font color="#ff0000"><strong>TPTB went to Dallas and all I got was Plucked !!</strong></font></p><p><font color="#339966"><strong>So many people, so few recipes !!</strong></font></p><p><font color="#0000ff"><strong>Let's clean up this stinkhole !!</strong></font> </p> </div>

 

[ianke]

Thanks to all on this! while I realize it could not happen in reality, it still seems cool to ponder the question and responses to it. <br /><br />Vogon13: Cool add on th the original problem. I LMAO! <div class="Discussion_UserSignature"> </div>

 

[Saiph]

it's not a redundant question, it's a hypothetical. The part of the question that helps is "no gravitational force outside the model".<br /><br />I.e. we are dealing only with the gravity from the bowling and ping pong balls.<br /><br />And, they themselves are able to generate gravity, and can orbit eachother. However the escape velocity is so small that nearly any perturbation will ruin the orbit. <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>

 

[tony873004]

Using the same formula as tigerbitten, I get 80.75 hours. Here's a link to an online calculator where you can play with different numbers. <br /><br /><br />http://orbitsimulator.com/gravity/articles/period1.html<br /><br /><br />Plug in 7.001 kg for mass, 1 m for a, and change the output units to hours.<br /><br />Writing the Hill Sphere formula to solve for a:<br /><br />a = 1/((m/(3M))^(1/3))<br /><br />and plugging in the numbers for this problem:<br />a = 1/((7.001/(3*5.97e24))^1/3 = 136766113 m, or 136,766 km<br /><br />shows that for the bowling ball's Hill Sphere with respect to Earth to be 1 meter, the bowling ball must be 136,773 km from Earth. This is about a third of the way to the Moon.<br /><br />So the ping pong ball could orbit the bowling ball from a distance of 1 meter provided it was at least this far from Earth. However, for a long-term stable orbit it should be 2-3 times the Hill Sphere radius.

 

[ianke]

Thanks for the link<br />Much appreciated! <div class="Discussion_UserSignature"> </div>