Author Topic: Keplers (Read 1808 times) Tweet Share

naved_s9994 · « **on:** June 05, 2009, 10:07:10 pm »

on which types of questions do we use Keplers law...

and do we only use it, when do objects are rotating, in STABLE orbit??

please... thanks

bigtick · « **Reply #1 on:** June 05, 2009, 11:14:01 pm »

Use it when you climb a spiral stairs or ride a merry-go-round.

naved_s9994 · « **Reply #2 on:** June 05, 2009, 11:24:58 pm »

funny?

Quote from: bigtick on June 05, 2009, 11:14:01 pm

Use it when you climb a spiral stairs or ride a merry-go-round.

kaanonball · « **Reply #3 on:** June 05, 2009, 11:41:10 pm »

most often you use it when, you have a satellite orbiting a central body.

where $\frac{r^3}{T^2}$ is a constant for all satellites orbiting the body.

/0 · « **Reply #4 on:** June 05, 2009, 11:48:55 pm »

The satellite must be in CIRCULAR orbit. This can be seen from its derivation (I think it's worth learning this):

$\frac{mv^2}{R} = G\frac{Mm}{R^2}$ (here mv^2/r implies it MUST be circular motion)

$\Rightarrow v^2 = G\frac{M}{R}$

But $v = \frac{2\pi R}{T}$ (again, 2piR/T implies circular motion)

So $\left(\frac{2\pi R}{T} \right)^2 = G\frac{M}{R}$

Simplify gives:

$\Rightarrow T^2 = \frac{4\pi^2}{GM} R^3$ .

Circular orbit are intrinsically STABLE.

mark_alec · « **Reply #5 on:** June 06, 2009, 12:04:43 am »

Out of curiosity, what do you mean when you are describing an orbit as stable?

naved_s9994 · « **Reply #6 on:** June 06, 2009, 12:07:08 am »

my teacher said remember it as SPA....and its on my cheat sheet, which i dont have at this stage, but tommorow morning i can post it here.

Mao · « **Reply #7 on:** June 06, 2009, 01:51:30 am »

Quote from: /0 on June 05, 2009, 11:48:55 pm

The satellite must be in CIRCULAR orbit. This can be seen from its derivation (I think it's worth learning this):

$\frac{mv^2}{R} = G\frac{Mm}{R^2}$ (here mv^2/r implies it MUST be circular motion)

$\Rightarrow v^2 = G\frac{M}{R}$

But $v = \frac{2\pi R}{T}$ (again, 2piR/T implies circular motion)

So $\left(\frac{2\pi R}{T} \right)^2 = G\frac{M}{R}$

Simplify gives:

$\Rightarrow T^2 = \frac{4\pi^2}{GM} R^3$ .

Circular orbit are intrinsically STABLE.

That's a simplified derivation of Kepler's Third Law of Planetary Motion, assuming uniform circular motion.

Kepler's third law actually works for all orbital paths, including highly elliptical ones.

Circular motion is not 'intrinsically' stable. So long as the motion of the body follows Kepler's three laws (which you don't learn in VCE), it's not headed for a collision course..

naved_s9994 · « **Reply #8 on:** June 06, 2009, 10:43:30 am »

k thanks, mao

/0 · « **Reply #9 on:** June 06, 2009, 01:37:03 pm »

fail :S
at least elliptical orbits aren't in the course

Wait, but in elliptical orbits T stays constant while R varies! How can this be the case, if only one combination of R and T will work?

Mao · « **Reply #10 on:** June 07, 2009, 03:17:35 am »

Quote from: /0 on June 06, 2009, 01:37:03 pm

fail :S
at least elliptical orbits aren't in the course

Wait, but in elliptical orbits T stays constant while R varies! How can this be the case, if only one combination of R and T will work?

T is orbital period, R is sub-major axis of the ellipse.

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Author Topic: Keplers (Read 1808 times) Tweet Share

naved_s9994

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