phy,gravitation

piyopiyo

Due to air resistance,changes may occur to a satellite orbiting the earth(assume nearly circular orbit)
Which of the changes below is incorrect?
A. The total mechanical energy of the satellite will decrease
B.The angular momentum of the satellite about the earth's center will decrease
C.The linear speed of the satellite will increase
D.The time needed for the satellite to complete a revolution will increase.

why D[[tk_02]]

Zend

Kepler's third law. T^2 = kr^3
ad the question ask which one is incorrect

piyopiyo

原帖由 Zend 於 21-9-2008 20:33 發表
Kepler's third law. T^2 = kr^3
ad the question ask which one is incorrect

That question is placed at the part of "escape speed",which is before Kepler's law[[tk_02]]

Zend

有無satellite 先?
有都未叫過[[bao_34]]

piyopiyo

原帖由 Zend 於 21-9-2008 20:51 發表
有無satellite 先?
有都未叫過[]

Escape speed is before satellite[[tk_02]]

Zend

mv^2/r = GMm/r^2

mrw^2 = GMm/r^2
w^2 = GM/r^3
(2pi/T)^2 = GM/r^3
4pi^2 / T^2 = GM/r^3
T^2 = 4pi^2 r^3/ GM

then ok ?
it is derived form centripetal force ja ma?

piyopiyo

however,with air resistance,will the centripetal force still be equal to the gravitational force [[tk_24]]

Zend

原帖由 piyopiyo 於 21-9-2008 09:50 PM 發表
however,with air resistance,will the centripetal force still be equal to the gravitational force [[tk_24]]

when a object circulate around another object, they ware supposed to be the same.

air resistance affects the total energy, and then KE and PE ..

the KE comes from... orbiting around the Earth...

x-Stephen-x

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 decrease , so D is wrong and C is true

Zend

decelerate linearly?