A Physics Problem

[yruc]

I am working on my homework and I have run into a problem and was wondering if someone out here can help me a bit. As with all homework help, please not just answers, but how to arrive at the solution.<br /><br />During an Apollo lunar landing mission, the command module continued to orbit the Moon at an altitude of 87.0 km. How long did it take to go around the Moon once? <br /><br />I figured out acceleration due to the gravity of the moon. <br /><br />g = G*Mass of moon / (radius of moon + distance from surface)^2 <br /><br />which is radial acceleration in which Ar = V^2/r <br /><br />Then using (G*mass of moon)/(radius of moon +87km) = (4pi*raidus of moon^2 ) / T^2(period of one revolution)<br /><br />Solve for T and I got 1.44 hours. This isn't correct.<br /><br />I looked on line and it appears that it took the command module about 2hours to orbit... but it didn't a distance from the surface of the moon. <br /><br />Thanks for any and all help!<br /><br />Yruc<br /><br /><br /><br />

 

[Leovinus]

2 hours sounds about right. You'd think that a orbit around a smaller body would take less time, but the gravity is less so it takes less speed to keep the orbit. Less speed = more time. <div class="Discussion_UserSignature"> </div>

 

[larper]

Did you mean "4pi*(radius_of_moon + 87)^2" ? <div class="Discussion_UserSignature"> <p><strong><font color="#ff0000">Vote </font><font color="#3366ff">Libertarian</font></strong></p> </div>

 

[Saiph]

You can also use keplars law: D^3=P^2....make sure to use the right units, and don't forget the propotion constant K = (M+m) divided by some other stuff IIRC.<br /><br />This isn't to far from what you have actually (bring your R's together and there you go).<br /><br /><br />Also note: You're using the radius of the moon + 87 km for the module's height. <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>

 

[yruc]

Thank you all for your help! I see where my problem was, I was using the radial accleration and finding that velocity instead of the tangental. Yes larper correct I did mean 4pi*(raidus of moon+87)^2 I did calculate it that way, just typed it incorrectly here. <br /><br />Again Thanks a bunch everyone for your help. The 1.95 hours is correct.<br /><br />Yruc

 

[rogerinnh]

OK, here's an additional interesting problem to solve. Given that you now know how to calculate the time it takes to orbit the moon, given a height above the surface, try to calculate the size of the equatorial orbit necessary for the orbit to be lunarstationary, that is, for the orbit to keep the spacecraft permanently positioned over a specific position on the lunar surface.

 

[larper]

L1 (or is it L2?) <div class="Discussion_UserSignature"> <p><strong><font color="#ff0000">Vote </font><font color="#3366ff">Libertarian</font></strong></p> </div>

 

[yruc]

The height above the surface of the moon would be<br /><br />d = orbit height above moon's surface<br />r = radius of the moon = 1 737.4 kilometers<br />G = gravitational constant = 6.67300 × 10^-11 m3 /kgs2<br />M = mass of the moon = 7.36 × 10^22 kilograms<br />P = Period. The Moon takes ~ 28 days to rotate once.<br />R^3 = (G * M * P^2 / (4*pi*pi))<br /><br />d = R - r<br /><br /><br />Don't have my calc with me but this the fomula or should be if I got it right.. I'll calculate it tonight after I get some sleep. (work nights goto school in mornings and sleep now :b)