phy,gravitation

piyopiyo

原帖由 x-Stephen-x 於 21-9-2008 23:39 發表
with air resistance,that means the satellite will decelerate(linearly) , so the satellite will go a lower potential level,which increase its linearly speed.
as T^2 = 4pi^2 r^3/ GM , r decrease=>t dec ...

I got ur meaning
how about B,angular momentum?-.-...is it mw??
if mw decreases...with m being constant...then w should decrease-,-...[[tk_30]]??

j32738

Are u using New way?
This one need to use Keplers LAW

you studied very fast...

piyopiyo

原帖由 j32738 於 22-9-2008 18:13 發表
Are u using New way?
This one need to use Keplers LAW

you studied very fast...

yes
however,i have a problem
according to wiki
angular momentum = rxp
where p is the linear momentum of the object and r is the radius
and x is the vector cross product[[tk_30]]

好複雜[[tk_30]]
但其實 r同p都係成90度, angular momentum即係都係 rp[[tk_42]]
r下跌,但p=mv,,v上升...
咁我都唔知r跌得勁定v升得勁
定係又要另外再prove[[tk_35]]

Zend

transform to one variable lor
and one more thing

r decreases, v increases, but v increases  and v = sqrt GM/r(it still follows a circular orbit, so the KE is due to circulation, and circulation=> centripetal force necessary for uniform circular motion


ne 個時候我應該仲係chapter 2

不過講樣事實比你聽, gravitation 好多年都無再出LQ
只得MC

x-Stephen-x

as mv^2/r = GMm/r^2 => v=(GM/r)^1/2 ,
angular momentum =mvr = m(GM)r^1/2 , as r decrease , angular momentum decrease

adamsteve

其實換個方法諗就係話
如果時間真係increase左 姐係佢速度慢左
速度慢左 根本就唔應該 或者應該話佢根本無能力再留係哥條orbit上面

咁佢都講得佢一定係哥條orbit度  所以速度一定唔會改變 所以時間都唔會變

x-Stephen-x

i mean 舊野係transverse deceleration but not radial...

j32738

原帖由 piyopiyo 於 22-9-2008 18:38 發表

yes
however,i have a problem
according to wiki
angular momentum = rxp
where p is the linear momentum of the object and r is the radius
and x is the vector cross product[[tk_30]]

好複雜[[tk_3 ...

I have never seen angular momentum
I think it is OUT C

also for moment of inertia